System of Particles and Rotational Motion: NEET and JEE questions with solutions

Answer: (B) $\tfrac27$

For a solid sphere $I = \tfrac25 mR^2$ and rolling gives $v = \omega R$.
$$K_{rot} = \tfrac12 \cdot \tfrac25 mR^2 \omega^2 = \tfrac15 mv^2, \qquad K_{total} = \tfrac12 mv^2 + \tfrac15 mv^2 = \tfrac{7}{10} mv^2$$
$$\frac{K_{rot}}{K_{total}} = \frac{1/5}{7/10} = \frac27$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) see figure

By symmetry, the two particles are always at angles $\theta$ and $\pi - \theta$, mirror images about the line AB. Each has speed $v$ along the tangent.

A particle at angle $\theta$ moving along the circle has velocity components $(\pm v\sin\theta,\ -v\cos\theta)$. For the mirror-image particle, the components along the horizontal diameter are opposite and cancel, while the components along AB add:
$$P = 2mv\,|\cos\theta|$$

As $\theta$ goes from $\dfrac{\pi}{2}$ to $-\dfrac{\pi}{2}$:

• $P = 0$ at the start (at A the velocities are equal and opposite),
• $P$ is maximum ($2mv$) at $\theta = 0$,
• $P = 0$ again at B.

So $P$ rises from zero to a maximum and falls back to zero along a smooth $\cos\theta$ curve, which is option (3).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $(m - M)(R + r)^2$

The spheres touch, so their centres are $d = R + r$ apart.

About the axis through the centre of A (parallel axis theorem for B):
$$I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R + r)^2$$

About the axis through the centre of B:
$$I_B = \frac{2}{5}MR^2 + M(R + r)^2 + \frac{2}{5}mr^2$$

$$I_A - I_B = (m - M)(R + r)^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1$

The angular momentum about any point P is
$$\vec{L}_P = \vec{L}_{cm} + \vec{r}_{P\to cm} \times M\vec{v}_{cm}$$

The disc rotates about its fixed centre, so its centre of mass is at rest: $\vec{v}_{cm} = 0$. The second term vanishes, and the angular momentum is the same ($I\omega$) about every point.

$$\frac{L_A}{L_B} = 1$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $108$ days

With no external torque, angular momentum is conserved: $I\omega = \text{constant}$.

For a uniform sphere, $I = \dfrac{2}{5}MR^2$, so $R^2\omega$ is constant and $T \propto R^2$.

$$T_2 = T_1\left(\frac{R_2}{R_1}\right)^2 = 27 \times 4 = 108\ \text{days}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $100\sqrt{3}$ N

The smooth wall gives only a horizontal normal force $N_w$. Horizontal equilibrium: friction $f = N_w$.

The rod makes $60°$ with the wall. Its top is at height $L\cos 60°$, and its centre is a horizontal distance $\dfrac{L}{2}\sin 60°$ from the foot.

Torques about the foot of the rod:
$$N_w \cdot L\cos 60° = mg\cdot\frac{L}{2}\sin 60°$$
$$N_w = \frac{mg}{2}\tan 60° = \frac{200}{2}\sqrt{3} = 100\sqrt{3}\ \text{N}$$

So the friction force is $f = 100\sqrt{3}$ N.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{7}{57}$

Let the small sphere have mass $m$. The large sphere has 8 times its volume, so mass $8m$.

The Y-axis passes through the centre of the large sphere. The small sphere's centre is a distance $R$ from it.

Small sphere: $I_s = \dfrac{2}{5}mR^2 + mR^2 = \dfrac{7}{5}mR^2$

Full large sphere: $I_L = \dfrac{2}{5}(8m)(2R)^2 = \dfrac{64}{5}mR^2$

Remaining part: $I_{\text{rest}} = \dfrac{64}{5}mR^2 - \dfrac{7}{5}mR^2 = \dfrac{57}{5}mR^2$

$$\frac{I_s}{I_{\text{rest}}} = \frac{7}{57}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Point $P$ moves faster than point $Q$.

In rolling without slipping, the velocity of any point is the velocity of the centre plus the velocity due to rotation, with $\omega R = v$.

Top point $P$: $v + \omega R = 2v$.

Bottom point $Q$ (in contact with the road): $v - \omega R = 0$.

So $P$ moves faster than $Q$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $8.5$ cm

$$I = \frac{ML^2}{12} \;\Rightarrow\; L^2 = \frac{12I}{M} = \frac{12 \times 2400}{400} = 72\ \text{cm}^2$$

$$L = \sqrt{72} \approx 8.5\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $3:5$

$I = Mk^2$.

Solid sphere: $\dfrac{2}{5}MR^2 = Mk_1^2$, so $k_1^2 = \dfrac{2}{5}R^2$.

Hollow sphere: $\dfrac{2}{3}MR^2 = Mk_2^2$, so $k_2^2 = \dfrac{2}{3}R^2$.

$$\frac{k_1^2}{k_2^2} = \frac{2/5}{2/3} = \frac{3}{5} \;\Rightarrow\; \frac{k_1}{k_2} = \sqrt{\frac{3}{5}}$$

Strictly, the ratio of the radii of gyration is $\sqrt{3} : \sqrt{5}$, which is not among the options. Option (1), $3 : 5$, is the ratio of their squares, $k_1^2 : k_2^2$, and is the intended answer.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) along the axis of rotation

Angular velocity and angular acceleration are axial vectors. For motion in a circle they point along the axis of rotation (perpendicular to the plane of the circle), with the direction given by the right-hand rule.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{3}{4}$

The remaining $270°$ arc has mass $\dfrac{3}{4}M$, and all of it is at distance $R$ from the axis:
$$I = \frac{3}{4}MR^2 \;\Rightarrow\; K = \frac{3}{4}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{1}{12}$ kg

Take torques about the wedge at the $40$ cm mark.

Anticlockwise: the $2$ kg mass at $20$ cm, lever arm $20$ cm: $2g \times 20$.

Clockwise: the rod's weight at its centre ($100$ cm), lever arm $60$ cm: $0.5g \times 60$. And $m$ at $160$ cm, lever arm $120$ cm: $mg \times 120$.

$$2 \times 20 = 0.5 \times 60 + 120m \;\Rightarrow\; 40 = 30 + 120m \;\Rightarrow\; m = \frac{1}{12}\ \text{kg}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $67$ cm

Taking the $5$ kg particle as origin:
$$x_{cm} = \frac{5 \times 0 + 10 \times 1}{5 + 10} = \frac{2}{3}\ \text{m} \approx 67\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $-6\hat{i}$ N m

$$\vec{\tau} = \vec{r} \times \vec{F} = 2\hat{k} \times 3\hat{j} = 6(\hat{k} \times \hat{j}) = -6\hat{i}\ \text{N m}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) see figure

Concept. For a rod with linear density $\lambda(x)$, $X_{cm}=\dfrac{\int x\lambda\,dx}{\int\lambda\,dx}$.

Step 1. Integrate with $\lambda\propto x^n$.
$$X_{cm}=\frac{\int_0^Lx^{n+1}dx}{\int_0^Lx^ndx}=\frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)}=\frac{n+1}{n+2}L$$

Step 2. Shape. At $n=0$: $L/2$. As $n$ grows the mass crowds towards $x=L$: $X_{cm}$ rises and approaches $L$, flattening out.

Answer. Option A (rising from $L/2$ towards $L$).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $I_{AC} = I_{EF}$

Concept. For a square lamina, by symmetry and the perpendicular-axis theorem, every axis in its plane through the centre has the same moment of inertia, $\tfrac1{12}Ma^2$.

Step 1. Why. $I_z=I_{EF}+I_{\perp EF}=2I_{EF}$ and also $I_z=I_{AC}+I_{BD}=2I_{AC}$ (each pair is perpendicular and equal by symmetry). So $I_{AC}=I_{EF}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{2}{3}ma^2$

Concept. Find $I$ about the normal axis through the centre with the perpendicular-axis theorem, then shift to the corner with the parallel-axis theorem.

Step 1. Centre, normal axis. For a square plate, the axis through the centre parallel to a side gives $\dfrac{ma^2}{12}$ (like a rod of length $a$). Two such axes are perpendicular, so
$$I_{O}=\frac{ma^2}{12}+\frac{ma^2}{12}=\frac{ma^2}{6}$$

Step 2. Distance from the centre to a corner: $d=\dfrac{a}{\sqrt2}$ (half the diagonal).

Step 3. Parallel axis.
$$I_{\text{corner}}=\frac{ma^2}{6}+m\frac{a^2}{2}=\frac{2}{3}ma^2$$

Answer. $\dfrac23ma^2$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\left(\dfrac{\rho_B}{\rho_A}\right)^{2/3}$

Equal masses: $\dfrac{4}{3}\pi R_A^3\rho_A = \dfrac{4}{3}\pi R_B^3\rho_B$, so $\dfrac{R_A}{R_B} = \left(\dfrac{\rho_B}{\rho_A}\right)^{1/3}$.

$I = \dfrac{2}{5}mR^2$ with equal $m$:
$$\frac{I_A}{I_B} = \left(\frac{R_A}{R_B}\right)^2 = \left(\frac{\rho_B}{\rho_A}\right)^{2/3}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{5}{2}I$

$I = \dfrac{MR^2}{2}$, so $MR^2 = 2I$.

About a diameter: $\dfrac{MR^2}{4}$. About a parallel tangent (parallel axis theorem):
$$I' = \frac{MR^2}{4} + MR^2 = \frac{5}{4}MR^2 = \frac{5}{4}(2I) = \frac{5}{2}I$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{\rho_A}{\rho_B}$

Same radius, so $I = \dfrac{2}{5}mR^2 \propto m \propto \rho$:
$$\frac{I_A}{I_B} = \frac{\rho_A}{\rho_B}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{1}{6}\dfrac{l^2\omega^2}{g}$

Concept. At the lowest point the rod's KE is greatest; it all turns into PE at the top of the swing: $\tfrac12I\omega^2=mgh_{cm}$, with $I=\tfrac13ml^2$ about the end.

Step 1. Energy. $\tfrac12\cdot\tfrac13ml^2\omega^2=mgh$.

Step 2. Solve. $h=\dfrac{l^2\omega^2}{6g}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) more than $3$ but less than $6$

Concept. A time-varying torque: $\alpha(t)=\dfrac{F(t)R}{I}$. Integrate for $\omega(t)$; the direction reverses when $\omega=0$. Integrate again for the angle turned.

Step 1. Angular acceleration. $\alpha=\dfrac{2(20t-5t^2)}{10}=4t-t^2$.

Step 2. Angular velocity. $\omega=2t^2-\dfrac{t^3}{3}$, which is zero again at $t=6$ s.

Step 3. Angle turned.
$$\theta=\int_0^6\left(2t^2-\frac{t^3}{3}\right)dt=\frac{2(216)}{3}-\frac{1296}{12}=144-108=36\ \text{rad}$$

Step 4. Rotations. $\dfrac{36}{2\pi}\approx5.7$.

Answer. More than 3 but less than 6 (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) zero torque

A central force always acts along the line joining the particle to the centre, i.e. along $\vec{r}$. So its torque about the centre is $\vec{\tau} = \vec{r}\times\vec{F} = 0$.

With zero torque, $\dfrac{d\vec{L}}{dt} = 0$ and the angular momentum is conserved.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $V_Q > V_C > V_P$

In rolling without slipping, every point moves as if rotating about the contact point with angular speed $\omega$, so its speed is $\omega \times$ (its distance from the contact point).

Q is above C, so it is farther from the contact point than C. P is below C, so it is nearer. Hence
$$V_Q > V_C > V_P$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\left(\dfrac{2}{3},\ 1\right)$ m

$$x_{cm} = \frac{1(0) + 2(2) + 3(0)}{6} = \frac{2}{3}\ \text{m}, \qquad y_{cm} = \frac{1(0) + 2(0) + 3(2)}{6} = 1\ \text{m}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{R}{6}$

Treat the hole as negative mass. Areas are in the ratio $1 : \dfrac{1}{4}$, so take masses $4m$ (full disc) and $m$ (hole). The hole's centre is at $\dfrac{R}{2}$ from O.
$$x_{cm} = \frac{4m(0) - m\left(\frac{R}{2}\right)}{4m - m} = -\frac{R}{6}$$

The centre of mass shifts $\dfrac{R}{6}$ away from the hole.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) continues along the original parabolic path

The explosion forces are internal, so they cannot change the motion of the centre of mass. The only external force is still gravity, so the centre of mass keeps following the same parabola (until a fragment hits the ground).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{3}{2}MR^2$

About a diameter: $\dfrac{MR^2}{2}$ (perpendicular axis theorem: two diameters add to $MR^2$).

Parallel axis theorem for the tangent in the plane:
$$I = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{L}{\sqrt{3}}$

$I = \dfrac{ML^2}{3} = Mk^2$, so $k = \dfrac{L}{\sqrt{3}}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5\ \text{m/s}^2$

Block: $mg - T = ma$. Pulley: $TR = I\alpha = \dfrac{1}{2}MR^2\dfrac{a}{R} \Rightarrow T = \dfrac{Ma}{2}$.
$$a = \frac{mg}{m + M/2} = \frac{10}{1 + 1} = 5\ \text{m/s}^2$$

(The radius cancels.)

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{1}{3}$

$K_{rot} = \dfrac{1}{2}\left(\dfrac{1}{2}MR^2\right)\omega^2 = \dfrac{1}{4}Mv^2$ and $K_{trans} = \dfrac{1}{2}Mv^2$.
$$\frac{K_{rot}}{K_{total}} = \frac{1/4}{3/4} = \frac{1}{3}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{10}{3}\ \text{m/s}^2$

For rolling, $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$. For a solid cylinder $\dfrac{k^2}{R^2} = \dfrac{1}{2}$:
$$a = \frac{10 \times 0.5}{1.5} = \frac{10}{3}\ \text{m/s}^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) solid sphere

$a = \dfrac{g\sin\theta}{1 + k^2/R^2}$: the smaller $\dfrac{k^2}{R^2}$, the larger the acceleration.

Solid sphere $\dfrac{2}{5}$ < disc $\dfrac{1}{2}$ < hollow sphere $\dfrac{2}{3}$ < ring $1$.

The solid sphere wins; the ring comes last. (Mass and radius do not matter.)

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $10$ m/s

$$mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2}mv^2 \cdot \frac{7}{5} \;\Rightarrow\; v^2 = \frac{10gh}{7} = 100$$
$$v = 10\ \text{m/s}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $90$ rpm, $3$

No external torque, so $I_1\omega_1 = I_2\omega_2$: $\omega_2 = \dfrac{6}{2} \times 30 = 90$ rpm.

$K = \dfrac{L^2}{2I}$ with $L$ constant, so $\dfrac{K_2}{K_1} = \dfrac{I_1}{I_2} = 3$. The extra energy comes from the work the man does pulling his arms in.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $18$ J

$$\omega_f = \frac{2 \times 6}{2 + 2} = 3\ \text{rad/s}$$
$$K_i = \frac{1}{2}(2)(36) = 36\ \text{J}, \qquad K_f = \frac{1}{2}(4)(9) = 18\ \text{J}$$

Lost $= 18$ J, turned into heat by friction between the discs.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $60\ \text{kg m}^2\text{/s}$ and $450$ J

Angular impulse: $L = \tau t = 20 \times 3 = 60\ \text{kg m}^2/\text{s}$.
$$K = \frac{L^2}{2I} = \frac{3600}{8} = 450\ \text{J}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{3g}{2L}$

Torque of gravity about the pivot: $\tau = mg\dfrac{L}{2}$. $I = \dfrac{mL^2}{3}$.
$$\alpha = \frac{\tau}{I} = \frac{mgL/2}{mL^2/3} = \frac{3g}{2L}$$

(The free end then starts with acceleration $\alpha L = 1.5g$, more than $g$.)

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $11\hat{k}$ N m

$$\vec{\tau} = \vec{r}\times\vec{F} = (r_xF_y - r_yF_x)\,\hat{k} = \big(2 \times 4 - (-1) \times 3\big)\hat{k} = 11\hat{k}\ \text{N m}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 12

$$K = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2}(4)(4)\left(1 + \frac{1}{2}\right) = 12\ \text{J}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 8

$$\theta = \frac{\omega_0 + \omega}{2}t = \frac{0 + 20}{2}(5) = 50\ \text{rad}$$
$$\text{Revolutions} = \frac{50}{2\pi} \approx 7.96 \approx 8$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $5$

Concept. While slipping, kinetic friction $\mu mg$ acts backwards at the contact point. It decelerates the centre and increases the spin. Rolling starts when $v=\omega R$.

Step 1. Translation. $a=-\mu g=-0.25\times9.8=-2.45$ m s$^{-2}$, so $v=49-2.45t$.

Step 2. Rotation. Torque $\mu mgR=\tfrac12mR^2\alpha$ gives $\alpha R=2\mu g=4.9$ m s$^{-2}$. Starting from $\omega_0R=\dfrac{v_0}{4}=12.25$ m s$^{-1}$:
$$\omega R=12.25+4.9t$$

Step 3. Rolling condition.
$$49-2.45t=12.25+4.9t\ \Rightarrow\ 36.75=7.35t\ \Rightarrow\ t=5\ \text{s}$$

Answer. $5$ s (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0.0128\pi$ Nm, $100$

Concept. Constant torque gives constant $\alpha=\Delta\omega/t$, and $\tau=I\alpha$. The number of turns is the average angular speed (in rev/s) times the time.

Step 1. Moment of inertia. $I=\tfrac25(5)(0.04)^2=3.2\times10^{-3}$ kg m$^2$.

Step 2. Angular speed. $1200$ rpm $=20$ rev/s $=40\pi$ rad/s; $\alpha=\dfrac{40\pi}{10}=4\pi$ rad s$^{-2}$.

Step 3. Torque. $\tau=3.2\times10^{-3}\times4\pi=0.0128\pi$ N m.

Step 4. Rotations. Average $10$ rev/s for $10$ s: $100$.

Answer. $0.0128\pi$ N m, $100$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\left(\dfrac{\sqrt{3}}{4}, 1.25\right)$

Concept. Write each mass's coordinates with the origin at $P$, the midpoint of the altitude, then use $\vec r_{cm}=\sum m\vec r/\sum m$.

Step 1. Geometry. Sides $10$ m with an apex angle of $120^\circ$: altitude $10\cos60^\circ=5$ m, half-base $10\sin60^\circ=5\sqrt3$ m. With $P$ at the middle of the altitude: apex $(0,2.5)$, base corners $(-5\sqrt3,-2.5)$ (2 kg) and $(5\sqrt3,-2.5)$ (3 kg), as in the figure.

Step 2. $x$-coordinate. $\dfrac{-2(5\sqrt3)+3(5\sqrt3)+0}{20}=\dfrac{5\sqrt3}{20}=\dfrac{\sqrt3}{4}$ m.

Step 3. $y$-coordinate. $\dfrac{15(2.5)+5(-2.5)}{20}=\dfrac{25}{20}=1.25$ m.

Answer. $\left(\dfrac{\sqrt3}{4},1.25\right)$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5 : 21$

Concept. A force at the top of a rolling body: write $F+f=ma$ (friction $f$ unknown) and $(F-f)R=I\alpha$ with $a=\alpha R$. Adding gives $a=\dfrac{2F}{m(1+k^2/R^2)}$.

Step 1. Sphere A ($5m$, $k^2/R^2=\tfrac25$). $a_A=\dfrac{2F}{5m\cdot\tfrac75}=\dfrac{2F}{7m}$.

Step 2. Shell B ($m$, $k^2/R^2=\tfrac23$). $a_B=\dfrac{2F}{m\cdot\tfrac53}=\dfrac{6F}{5m}$.

Step 3. Ratio. $\dfrac{a_A}{a_B}=\dfrac{2/7}{6/5}=\dfrac{10}{42}=\dfrac5{21}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) straight line

Concept. The centre of mass moves like a single particle with velocity $\vec v_{cm}=\dfrac{\sum m\vec v}{\sum m}$ and acceleration $\vec a_{cm}=\dfrac{\sum m\vec a}{\sum m}$. If $\vec a_{cm}$ is constant and parallel to $\vec v_{cm}$, the path is a straight line; if not parallel, a parabola.

Step 1. Initial velocity of the centre of mass (equal masses, so a plain average):
$$\vec v_{cm}=\frac{4\hat i+4\hat j}{2}=2\hat i+2\hat j$$

Step 2. Acceleration of the centre of mass:
$$\vec a_{cm}=\frac{(6\hat i+6\hat j)+\vec 0}{2}=3\hat i+3\hat j$$

Step 3. Compare directions. Both vectors point along $\hat i+\hat j$. A constant acceleration along the velocity only changes the speed, not the direction.

Answer. Straight line (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $3$

Concept. From rest with constant $\alpha$, $\theta=\tfrac12\alpha t^2$: the angles in successive equal intervals go as $1:3:5:\dots$

Step 1. Angles. $\theta_1=\tfrac12\alpha(2)^2=2\alpha$; $\theta_1+\theta_2=\tfrac12\alpha(4)^2=8\alpha\Rightarrow\theta_2=6\alpha$.

Step 2. Ratio. $\dfrac{\theta_2}{\theta_1}=3$.

Answer. $3$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) solid sphere

Concept. Rolling up without slipping (static friction does no work): all the KE, translational and rotational, becomes PE: $h=\dfrac{v_0^2}{2g}\left(1+\dfrac{k^2}{R^2}\right)$.

Step 1. Compare. $\dfrac{v_0^2}{2g}\left(1+\dfrac{k^2}{R^2}\right)=\dfrac{7v_0^2}{10g}\Rightarrow1+\dfrac{k^2}{R^2}=\dfrac75$.

Step 2. Identify. $\dfrac{k^2}{R^2}=\dfrac25$: a solid sphere.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $70$

Same density, $\dfrac{1}{8}$ of the volume: $r = \dfrac{R}{2}$.

$I_1 = \dfrac{2}{5}\cdot\dfrac{M}{8}\cdot\dfrac{R^2}{4} = \dfrac{MR^2}{80}$.

Disc of mass $\dfrac{7M}{8}$, radius $2R$, about a diameter: $I_2 = \dfrac{1}{4}\cdot\dfrac{7M}{8}\cdot 4R^2 = \dfrac{7MR^2}{8}$.

$\dfrac{I_2}{I_1} = \dfrac{7}{8} \times 80 = 70$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) A, B and D only

Concept. Differentiate $\vec r$ for $\vec v$ and $\vec a$; then $\vec p=m\vec v$, $\vec F=m\vec a$, $\vec L=\vec r\times\vec p$, $\vec\tau=\vec r\times\vec F$.

Step 1. At $t=1$ s. $\vec r=(10,5)$, $\vec v=(20t,15t^2)=(20,15)$, $\vec a=(20,30t)=(20,30)$.

Step 2. Check each.

• A: $\vec p=0.1(20,15)=(2,1.5)$. Correct.
• B: $\vec F=0.1(20,30)=(2,3)$. Correct.
• C: $L_z=10(1.5)-5(2)=5$, not 15. Wrong.
• D: $\tau_z=10(3)-5(2)=20$. Correct.

Answer. A, B and D only (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $130$

$\theta=\dfrac{t^4}{8}-\dfrac{t^3}{3}$

$\omega=\dfrac{d\theta}{dt}=\dfrac{t^3}{2}-t^2,\qquad\alpha=\dfrac{d\omega}{dt}=\dfrac{3t^2}{2}-2t$

At $t=10$ s: $\alpha=150-20=130\ \text{rad/s}^2$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2.22$

Concept. The centre of mass of the two blocks moves with $a_{cm}=\dfrac{m_1a_1+m_2a_2}{m_1+m_2}$, where the accelerations are taken with signs. Here the blocks move in opposite directions, so their contributions partly cancel.

Step 1. Acceleration of each block (Atwood machine).
$$a=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{(2-1)(10)}{3}=\frac{10}{3}\ \text{m s}^{-2}$$
The $2$ kg block goes down, the $1$ kg block goes up.

Step 2. Acceleration of the centre of mass (downward positive):
$$a_{cm}=\frac{2\left(\frac{10}{3}\right)+1\left(-\frac{10}{3}\right)}{3}=\frac{10}{9}\ \text{m s}^{-2}\ \text{(downward)}$$

Step 3. Distance in $2$ s from rest. Both blocks start at rest, so the centre of mass does too:
$$s=\tfrac12a_{cm}t^2=\tfrac12\cdot\frac{10}{9}\cdot4=\frac{20}{9}=2.22\ \text{m}$$

Answer. $2.22$ m (Option C). Shortcut: for an Atwood machine $a_{cm}=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)^2g$.

Solution by Sreeraj P, M.Sc Physics

Answer: 15

The axis $AB$ passes through one end of the rod, perpendicular to it:

$$I_{rod}=\frac{ML^2}{3}=\frac{40\times9}{3}=120\ \text{kg m}^2$$

The sphere rotates about an axis parallel to $AB$ and $3$ m from its centre, so by the parallel axis theorem:

$$I_{sphere}=\frac25MR^2+Md^2=\frac25(10)R^2+10(3)^2=4R^2+90$$

Setting them equal: $4R^2+90=120\Rightarrow R^2=7.5=\dfrac{15}{2}$, so $R=\sqrt{\dfrac{15}{2}}$ and $\alpha=15$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Angular momentum

Under $\vec r\to -\vec r$: $\vec v = \dfrac{d\vec r}{dt}\to -\vec v$, $\vec p = m\vec v \to -\vec p$ and $\vec a \to -\vec a$ all flip.

Angular momentum $\vec L = \vec r\times\vec p \to (-\vec r)\times(-\vec p) = \vec r\times\vec p$, so it does not flip (it is an axial vector).

Solution by Sreeraj P, M.Sc Physics

Answer: 30

Treat the flywheel as a uniform disc: $I = \frac12MR^2$. With no slipping, $v = \omega R$.

Energy conservation:
$$mgh = \frac12mv^2 + \frac12I\omega^2 = \frac12mv^2 + \frac14Mv^2$$
$$3\times10\times3 = \frac12(3)v^2 + \frac14(3)v^2 = \frac94v^2 \Rightarrow v^2 = 40\ \text{m}^2/\text{s}^2$$
Kinetic energy of the wheel:
$$K = \frac14Mv^2 = \frac14\times3\times40 = 30\ \text{J}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 11

The axis $AB$ is perpendicular to the rod, $10$ cm from the left disc's centre and $20$ cm from the right disc's centre. Each disc's plane is perpendicular to the rod, so $AB$ is parallel to a diameter of each disc. Use CGS units ($M = 600$ g, $R = 10$ cm).

Disc about its diameter: $\frac14MR^2 = \frac14(600)(100) = 15000$ g·cm².

Left disc: $15000 + 600(10)^2 = 75000$ g·cm²

Right disc: $15000 + 600(20)^2 = 255000$ g·cm²

Rod ($L = 30$ cm, centre at $5$ cm from $AB$): $\frac{1}{12}(600)(30)^2 + 600(5)^2 = 45000 + 15000 = 60000$ g·cm²

$$I = 75000 + 255000 + 60000 = 3.9\times10^5\ \text{g·cm}^2$$
$$\alpha = \frac{\tau}{I} = \frac{43\times10^5}{3.9\times10^5} \approx 11\ \text{rad/s}^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: 265

Concept. Parallel-axis theorem: $I=I_{cm}+Md^2$, so $k^2=k_{cm}^2+d^2$. For a solid sphere $k_{cm}^2=\tfrac25R^2$.

Step 1. Substitute (cm). $k^2=\tfrac25(100)+15^2=40+225=265$.

Answer. $n=265$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\sqrt{\dfrac{3g}{L}}$

Concept. The rod turns about the pivot from vertical to horizontal; the loss of PE of its centre of mass becomes rotational KE about the pivot.

Step 1. CM height above the pivot. From the figure, $\tfrac{2L}{3}$ is above the pivot and $\tfrac L3$ below. The CM (at $\tfrac L2$ from the bottom) is $\tfrac L2-\tfrac L3=\tfrac L6$ above the pivot; it ends level with the pivot, a drop of $\tfrac L6$.

Step 2. Moment of inertia about the pivot. $\tfrac1{12}ML^2+M\left(\tfrac L6\right)^2=\tfrac{ML^2}{12}+\tfrac{ML^2}{36}=\tfrac{ML^2}{9}$.

Step 3. Energy. $\tfrac12\cdot\tfrac{ML^2}{9}\omega^2=Mg\tfrac L6\Rightarrow\omega^2=\dfrac{3g}{L}$.

Answer. $\sqrt{\dfrac{3g}{L}}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 162

Mass $\propto$ volume $\propto LR^2$. The small cylinder has mass
$$m = M\times\frac12\times\frac19 = \frac M{18}$$
$$I_1 = \frac12MR^2,\qquad I_2 = \frac12\cdot\frac M{18}\cdot\frac{R^2}{9} = \frac{MR^2}{324}$$
$$\frac{I_1}{I_2} = \frac{1/2}{1/324} = 162$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $9.5\times10^{-3}$

From $s = \frac12at^2$: $a = \dfrac{2\times0.81}{81} = 0.02\ \text{m/s}^2$.

For an Atwood machine with a pulley of moment of inertia $I$:
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/r^2}\Rightarrow 0.02 = \frac{0.05\times9.8}{0.75 + I/r^2}$$
$$0.75 + \frac{I}{r^2} = 24.5\Rightarrow\frac I{r^2} = 23.75\ \text{kg}$$
$$I = 23.75\times(0.02)^2 = 9.5\times10^{-3}\ \text{kg·m}^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{13}{32}MR^2$

Concept. $I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed}}$, all about the same axis. The removed disc is off-centre, so its $I$ needs the parallel-axis theorem. Its mass follows from area.

Step 1. Mass removed. Radius halves, so area (and mass) is $\dfrac14$: $m=\dfrac M4$.

Step 2. Where is it? The hole touches the rim and has radius $\dfrac R2$, so its centre is $\dfrac R2$ from $O$.

Step 3. Removed piece about the axis through $O$:
$$I_{\text{rem}}=\frac12\cdot\frac M4\left(\frac R2\right)^2+\frac M4\left(\frac R2\right)^2=\frac{MR^2}{32}+\frac{MR^2}{16}=\frac{3MR^2}{32}$$

Step 4. Subtract from the full disc ($\tfrac12MR^2=\tfrac{16}{32}MR^2$):
$$I=\frac{16}{32}MR^2-\frac{3}{32}MR^2=\frac{13}{32}MR^2$$

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 90

Concept. Use the two conditions at $t=1$ s ($|\vec V_A|=|\vec V_B|$ and $\vec V_A\perp\vec V_B$) to find $n$ and $p$. Then $\vec L=\vec r_{AB}\times m\vec V_A$, with $\vec r_{AB}=\vec r_A-\vec r_B$.

Step 1. Velocities at $t=1$. $\vec V_A=(2,\,3n,\,2)$, $\vec V_B=(2,\,-2,\,4p)$.

Step 2. Conditions. Equal magnitudes: $9n^2=16p^2$, so $3n=\pm4p$. Perpendicular: $4-6n+8p=0$. With $3n=4p$ this gives $4=0$ (impossible), so $3n=-4p$: $4+16p=0\Rightarrow p=-\tfrac14$, $n=\tfrac13$.

Step 3. Positions at $t=1$. $\vec r_A=(1,1,2)$, $\vec r_B=(2,-1,-1)$, so $\vec r_{AB}=(-1,2,3)$; $\vec V_A=(2,1,2)$.

Step 4. Angular momentum.
$$\vec L=\begin{vmatrix}\hat i&\hat j&\hat k\\-1&2&3\\2&1&2\end{vmatrix}=(4-3)\hat i-(-2-6)\hat j+(-1-4)\hat k=(1,8,-5)$$
$|\vec L|^2=1+64+25=90$.

Answer. $90$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\hat i - \hat k$

Concept. $\vec\tau=\vec r\times\vec F$ by the determinant.

Step 1. Determinant.
$$\vec\tau=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\2&1&2\end{vmatrix}=\hat i(2-1)-\hat j(2-2)+\hat k(1-2)=\hat i-\hat k$$

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.0\ \text{cm}$

Concept. A body with a hole = full body minus the removed piece. Treat the hole as a "negative mass" at its own centre: $x_{cm}=\dfrac{M x_1-m x_2}{M-m}$. For a uniform plate, mass is proportional to area.

Step 1. Locate the hole. Disc radius $R=20$ cm at the origin. The hole (radius $5$ cm) touches the rim, so its centre is at $x_2=20-5=15$ cm.

Step 2. Masses by area. Take the mass per unit area as $\sigma$.
$$M=\sigma\pi(20)^2=400\pi\sigma,\qquad m=\sigma\pi(5)^2=25\pi\sigma$$

Step 3. Apply the negative-mass formula. The full disc's centre is at $x_1=0$:
$$x_{cm}=\frac{400\pi\sigma(0)-25\pi\sigma(15)}{400\pi\sigma-25\pi\sigma}=\frac{-375}{375}=-1\ \text{cm}$$

Answer. $1.0$ cm from the origin, on the side opposite the hole (Option C). The centre of mass always shifts away from a hole.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1 : \sqrt2$

Concept. Rolling down from rest: $v^2=\dfrac{2gh}{1+k^2/R^2}$ with $h=L\sin\theta$. Same body and length, so $v^2\propto\sin\theta$.

Step 1. Ratio. $\dfrac{v_1^2}{v_2^2}=\dfrac{\sin30^\circ}{\sin45^\circ}=\dfrac{1/2}{1/\sqrt2}=\dfrac1{\sqrt2}$.

Answer. $1:\sqrt2$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $60MR^2$

$\omega = \dfrac{d\theta}{dt} = 10t - 8$, so $\omega(2) = 12\ \text{rad/s}$; $\alpha = \dfrac{d\omega}{dt} = 10\ \text{rad/s}^2$.

Taking the axis through the centre, $I = \tfrac{1}{2}MR^2$:
$$\tau = I\alpha = \frac{1}{2}MR^2\times10 = 5MR^2$$
$$P = \tau\omega = 5MR^2\times12 = 60MR^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{\sqrt2}{3}g$

Concept. Rolling without slipping down an incline: $a=\dfrac{g\sin\theta}{1+k^2/R^2}$; solid cylinder $k^2/R^2=\tfrac12$.

Step 1. Substitute. $a=\dfrac{g/\sqrt2}{3/2}=\dfrac{\sqrt2}{3}g$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{mv_0^3}{4\sqrt2g}$ along negative $z$-axis

Concept. At the top the velocity is horizontal, $v_0\cos45^\circ$, at height $H$. About the origin, $L=mvH$ (perpendicular distance $H$); the direction follows from $\vec r\times\vec v$.

Step 1. Height. $H=\dfrac{v_0^2\sin^245^\circ}{2g}=\dfrac{v_0^2}{4g}$.

Step 2. Magnitude. $L=m\dfrac{v_0}{\sqrt2}\cdot\dfrac{v_0^2}{4g}=\dfrac{mv_0^3}{4\sqrt2g}$.

Step 3. Direction. $L_z=xp_y-yp_x=0-H(mv)\lt0$: along $-z$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $t_1 < t_2$

Rolling down an incline, $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$.

• Solid sphere: $k^2/R^2 = \tfrac{2}{5}$, $a_1 = \tfrac{5}{7}g\sin\theta$
• Hollow sphere: $k^2/R^2 = \tfrac{2}{3}$, $a_2 = \tfrac{3}{5}g\sin\theta$

$a_1 > a_2$, and for the same length $t = \sqrt{2L/a}$, so $t_1 < t_2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{5}{2}$

Concept. Rolling: $K_{\text{trans}}=\tfrac12mv^2$, $K_{\text{rot}}=\tfrac12I\omega^2=\tfrac12\cdot\tfrac25mR^2\cdot\dfrac{v^2}{R^2}=\tfrac15mv^2$.

Step 1. Ratio. $\dfrac{1/2}{1/5}=\dfrac52$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\left(\dfrac{2}{3}a, \dfrac{b}{2}\right)$

The density depends only on $x$, so by symmetry $y_{cm} = b/2$.

Take strips of width $dx$: $dm = \sigma b\,dx \propto x\,dx$.
$$x_{cm} = \frac{\int_0^a x\cdot x\,dx}{\int_0^a x\,dx} = \frac{a^3/3}{a^2/2} = \frac{2a}{3}$$

Centre of mass: $\left(\dfrac{2a}{3}, \dfrac{b}{2}\right)$

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. Same material and thickness: $M=\rho t\pi R^2$, so $I=\tfrac12MR^2\propto R^4$.

Step 1. Ratio. $\dfrac{I_1}{I_2}=\left(\dfrac{R_1}{R_2}\right)^4=\dfrac1{16}$.

Answer. $x=16$.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

About a diameter: disc $\tfrac{1}{4}MR_d^2$, ring $\tfrac{1}{2}MR_r^2$, solid sphere $\tfrac{2}{5}MR^2$.

$$\frac{1}{4}MR_d^2 = 2.5\times\frac{1}{2}MR_r^2 \Rightarrow R_d^2 = 5R_r^2$$

Sphere with radius $R_d$:
$$I_s = \frac{2}{5}M(5R_r^2) = 2MR_r^2 = 4\times\frac{1}{2}MR_r^2$$

$n = 4$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $190\ \text{g}$

Concept. A body balanced on a pivot has zero net torque about that pivot: clockwise moments = anticlockwise moments. The rod's own weight acts at its centre (the $50$ cm mark).

Step 1. Lever arms about the pivot at $40$ cm.

• $400$ g at $10$ cm: $30$ cm to the left
• rod ($250$ g) at $50$ cm: $10$ cm to the right
• unknown $m$ at $90$ cm: $50$ cm to the right

Step 2. Balance moments ($g$ cancels):
$$400\times30=250\times10+m\times50\ \Rightarrow\ 12000=2500+50m\ \Rightarrow\ m=190\ \text{g}$$

Answer. $190$ g (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. The $z$ component of $\vec r\times\vec F$ is $\tau_z=xF_y-yF_x$.

Step 1. Substitute. $\tau_z=(1)(-1)-(1)(1)=-2$ N m.

Answer. Magnitude $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. $\vec r_{cm}=\dfrac{\sum m_i\vec r_i}{\sum m_i}$. Put the origin at the right-angle corner, where one mass sits.

Step 1. Positions. $(0,0)$, $(4,0)$, $(0,4)$ m, all of mass $2M$.

Step 2. Centre of mass. Equal masses: the average position $\left(\tfrac43,\tfrac43\right)$ m.

Step 3. Magnitude. $\sqrt{\tfrac{16}9+\tfrac{16}9}=\dfrac{4\sqrt2}{3}$ m, so $x=3$.

Answer. $3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{1}{2}\dfrac{v^2}{g}$

Concept. On a smooth incline there is no friction, so nothing can change the disc's spin (gravity and the normal force pass through the centre). The rotational energy stays locked in rotation; only the translational energy converts into height.

Step 1. What converts. $\tfrac12Mv^2=Mgh$.

Step 2. Solve. $h=\dfrac{v^2}{2g}$.

Answer. Option C. On a rough incline (rolling up) the disc would rise higher, to $\dfrac{3v^2}{4g}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

Concept. On a smooth surface the rod is free. An impulse $J$ at the end gives $v_{cm}=J/M$ and an angular impulse about the CM $J\tfrac l2=I_{cm}\omega$. The rod then turns at constant $\omega$.

Step 1. Angular speed.
$$\omega=\frac{J\,l/2}{Ml^2/12}=\frac{6J}{Ml}=\frac{6(0.2)}{2(0.3)}=2\ \text{rad/s}$$

Step 2. Time for a right angle. $t=\dfrac{\pi/2}{2}=\dfrac\pi4$ s.

Answer. $x=4$.

Solution by Sreeraj P, M.Sc Physics

Answer: 7

Concept. As above, $h=\dfrac{v^2}{2g}\left(1+\dfrac{k^2}{R^2}\right)$: solid sphere $\tfrac75$, hollow cylinder $2$.

Step 1. Ratio. $\dfrac{h_1}{h_2}=\dfrac{7/5}{2}=\dfrac7{10}$.

Answer. $n=7$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. The centre of mass stays where it is only if the mass-weighted shifts cancel: $m_1\,\Delta x_1+m_2\,\Delta x_2=0$. So the two particles must move in opposite directions along the line joining them.

Step 1. Write the condition. Take the line through the particles as the $x$-axis. Moving $m_1$ towards the centre of mass by $2$ cm gives $\Delta x_1=+2$ cm (say). For the centre of mass not to move,
$$m_1\Delta x_1+m_2\Delta x_2=0\ \Rightarrow\ 3(2)+2\,\Delta x_2=0\ \Rightarrow\ \Delta x_2=-3\ \text{cm}$$

Step 2. Read the sign. $\Delta x_2$ is opposite to $\Delta x_1$. Since $m_2$ lies on the other side of the centre of mass, moving opposite to $m_1$ means moving towards the centre of mass.

Answer. $m_2$ moves $3$ cm towards the centre of mass. The lighter particle has to move farther: shifts are in the inverse ratio of the masses.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $67$

Hollow sphere about a diameter: $k_1^2 = \dfrac23R^2$.

Solid cylinder (radius $R$, length $L = 4R$) about a diameter of its end face (perpendicular to its length):
$$I = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right) = \frac{67}{12}MR^2 \Rightarrow k_2^2 = \frac{67}{12}R^2$$
$$\frac{k_1}{k_2} = \sqrt{\frac{2/3}{67/12}} = \sqrt{\frac{8}{67}} \Rightarrow x = 67$$

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. $\dfrac{K_{\text{rot}}}{K_{\text{total}}}=\dfrac{k^2/R^2}{1+k^2/R^2}$; hollow sphere $k^2/R^2=\tfrac23$.

Step 1. Ratio. $\dfrac{2/3}{5/3}=\dfrac25$.

Answer. $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 13

Concept. No external torque: $I\omega$ is conserved. With $I=\tfrac25MR^2$ and fixed $M$, $T\propto R^2$.

Step 1. New day. $T'=24\times\left(\tfrac34\right)^2=24\times\tfrac9{16}=13.5$ h $=13$ h $30$ min.

Answer. $13$.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

Concept. For point masses $I=\sum mr^2$. The midpoint of each side of an equilateral triangle is the inradius $r=\dfrac{a}{2\sqrt3}$ from the centroid.

Step 1. Distance. $r=\dfrac{2}{2\sqrt3}=\dfrac1{\sqrt3}$ m, so $r^2=\tfrac13$.

Step 2. Sum. $I=(2+4+6)\times\tfrac13=4$ kg m$^2$.

Answer. $4$.

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. For point masses, $I=\sum m r^2$, where $r$ is each mass's perpendicular distance from the axis.

Step 1. Distances from the axis through one corner (side $2$ m):

• that corner: $0$
• two adjacent corners: $2$ m each
• the opposite corner: diagonal $2\sqrt2$ m

Step 2. Add.
$$I=1(0)^2+1(2)^2+1(2)^2+1(2\sqrt2)^2=0+4+4+8=16\ \text{kg m}^2$$

Answer. $16$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $3\ \text{kg-wt}$

Concept. The man's force is normal to the bar. Take moments about the end on the ground: the weight acts at the middle, with horizontal arm $\tfrac L2\cos60^\circ$; the man's force acts at the far end with arm $L$ (it is perpendicular to the bar).

Step 1. Moments. $N\cdot L=W\cdot\dfrac L2\cos60^\circ$.

Step 2. Solve. $N=\dfrac{12}{2}\times\dfrac12=3$ kg-wt.

Answer. $3$ kg-wt (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 7

In rolling without slipping, static friction does no work, so mechanical energy is conserved. Taking bodies of equal mass that descend the same height $h$, each gains kinetic energy (translational + rotational) equal to the loss in potential energy:
$$K_{ring} = K_{sphere} = mgh$$
Only the split between translational and rotational energy differs. The ratio is $1 = \dfrac77$, so $x = 7$.

Solution by Sreeraj P, M.Sc Physics

Answer: 10

For a solid cylinder $\dfrac{k^2}{R^2} = \dfrac12$:
$$a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}} = \frac23\times10\times\frac{\sqrt3}{2} = \frac{10}{\sqrt3}\ \text{m s}^{-2}$$
So $x = 10$.

Solution by Sreeraj P, M.Sc Physics

Answer: 60

Concept. For a particle moving in a straight line, $L=mv\,d$, where $d$ is the perpendicular distance from the reference point to the line of motion. It stays constant along the line.

Step 1. Perpendicular distance from the origin to $x-y+4=0$:
$$d=\frac{|0-0+4|}{\sqrt{1^2+1^2}}=\frac{4}{\sqrt2}=2\sqrt2\ \text{m}$$

Step 2. Angular momentum.
$$L=mvd=5\times3\sqrt2\times2\sqrt2=60\ \text{kg m}^2\text{s}^{-1}$$

Answer. $60$.

Solution by Sreeraj P, M.Sc Physics

Answer: 15

Concept. A cut-out acts as negative mass (mass $\propto$ area for a uniform plate). Use the whole rectangle minus the square.

Step 1. Pieces. Rectangle $3\times2$: area $6$, centre $(1.5,1)$. Square from $(1,1)$ to $(2,2)$: area $1$, centre $(1.5,1.5)$.

Step 2. Coordinates. $x_{cm}=1.5$ by symmetry.
$$y_{cm}=\frac{6(1)-1(1.5)}{6-1}=\frac{4.5}{5}=0.9$$

Step 3. Ratio. $\dfrac{1.5}{0.9}=\dfrac{15}{9}$, so $n=15$.

Answer. $15$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac23\omega$

Concept. Placing a disc gently on a rotating disc involves only internal torques (friction between them): $I\omega$ is conserved.

Step 1. Moments of inertia. $I_1=\tfrac12MR^2$, $I_2=\tfrac14MR^2$ (half the mass, same radius).

Step 2. Conserve. $\tfrac12MR^2\omega=\tfrac34MR^2\omega'\Rightarrow\omega'=\tfrac23\omega$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac W2$

Take the person's supporting force $F$ as vertical. Taking torques about the end on the ground (bar length $L$, weight acting at the middle):
$$F\,(L\cos\theta) = W\left(\frac L2\cos\theta\right) \;\Rightarrow\; F = \frac W2$$

Solution by Sreeraj P, M.Sc Physics

Answer: 100

$$\tau = FR = 40\times0.1 = 4\ \text{N m},\qquad \alpha = \frac\tau I = \frac{4}{0.4} = 10\ \text{rad s}^{-2}$$
$$\omega = \alpha t = 10\times10 = 100\ \text{rad/s}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. Down a plane of length $l$ from rest, $l=\tfrac12at^2$, so $t\propto\dfrac1{\sqrt a}$. Sliding (no friction): $a=g\sin\theta$. Rolling: $a=\dfrac{g\sin\theta}{1+k^2/R^2}$.

Step 1. Accelerations. For a disc $\dfrac{k^2}{R^2}=\dfrac12$:
$$a_{\text{slip}}=g\sin\theta,\qquad a_{\text{roll}}=\frac{g\sin\theta}{3/2}=\frac23g\sin\theta$$

Step 2. Times.
$$\frac{t_{\text{roll}}}{t_{\text{slip}}}=\sqrt{\frac{a_{\text{slip}}}{a_{\text{roll}}}}=\sqrt{\frac32}$$

Step 3. Match the form. $\left(\dfrac\alpha2\right)^{1/2}=\left(\dfrac32\right)^{1/2}$, so $\alpha=3$.

Answer. $3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{\sqrt3}{16}\dfrac{mu^3}{g}$

Concept. $L=mv_\perp\,d$. At the top of the flight the velocity is horizontal ($u\cos\theta$), and its line is at height $H$ above the point of projection. So $L=m(u\cos\theta)H$.

Step 1. Maximum height.
$$H=\frac{u^2\sin^2 30^\circ}{2g}=\frac{u^2}{8g}$$

Step 2. Speed at the top. $u\cos30^\circ=\dfrac{\sqrt3}{2}u$.

Step 3. Angular momentum.
$$L=m\cdot\frac{\sqrt3}{2}u\cdot\frac{u^2}{8g}=\frac{\sqrt3}{16}\frac{mu^3}{g}$$

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 250

Concept. Angular momentum is conserved; kinetic energy is not (friction between the discs). Identical disc: $\omega$ halves and $I$ doubles, so half the KE is lost.

Step 1. Initial KE. $I=\tfrac12(5)(4)=10$ kg m$^2$; $K=\tfrac12(10)(100)=500$ J.

Step 2. Final KE. $\omega'=5$ rad/s, $I'=20$: $K'=\tfrac12(20)(25)=250$ J.

Step 3. Dissipated. $500-250=250$ J.

Answer. $250$.

Solution by Sreeraj P, M.Sc Physics

Answer: 24

Assuming both rotate in the same sense, angular momentum is conserved:
$$\omega = \frac{4\times10 + 2\times4}{6} = 8\ \text{rad s}^{-1}$$
$$K_i = \tfrac12(4)(100) + \tfrac12(2)(16) = 216\ \text{J},\qquad K_f = \tfrac12(6)(64) = 192\ \text{J}$$
Loss $= 24\ \text{J}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

For a rolling disc $K = \tfrac12mv^2\left(1 + \dfrac{k^2}{R^2}\right) = \tfrac34mv^2$.
$$K = \tfrac34\times50\times0.16 = 6\ \text{J}$$
By the work–energy theorem $|W| = 6\ \text{J}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 53

Concept. Each sphere: $I_{cm}=\tfrac25mR^2$ plus $md^2$ (parallel axes), with $d$ half the centre separation.

Step 1. One sphere. $\tfrac25(2)(0.5)^2+2(0.75)^2=0.2+1.125=1.325$ kg m$^2$.

Step 2. Two spheres. $2.65=\dfrac{53}{20}$ kg m$^2$.

Answer. $53$.

Solution by Sreeraj P, M.Sc Physics

Answer: 8

At the top, $v = u\cos45^\circ = \dfrac{u}{\sqrt2}$ (horizontal) and $H = \dfrac{u^2\sin^245^\circ}{2g} = \dfrac{u^2}{4g}$.
$$L = mvH = m\frac{u}{\sqrt2}\cdot\frac{u^2}{4g} = \frac{mu^3}{4\sqrt2\,g} = \frac{\sqrt2\,mu^3}{8g}$$
$X = 8$.

Solution by Sreeraj P, M.Sc Physics

Answer: 40

For rolling, $K=\tfrac12mv^2\left(1+\dfrac{k^2}{R^2}\right)=\tfrac12mv^2\left(1+\tfrac25\right)=\tfrac{7}{10}mv^2$.
$$2240=\frac{7}{10}\times2\times v^2\ \Rightarrow\ v^2=1600\ \Rightarrow\ v=40\ \text{m s}^{-1}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 800

The angular momentum about the launch point is $L=m\,|\vec r\times\vec v|$. With $\vec r=(u_xt,\ u_yt-\tfrac12gt^2)$ and $\vec v=(u_x,\ u_y-gt)$:
$$L=m\left[u_xt(u_y-gt)-u_x\left(u_yt-\tfrac12gt^2\right)\right]=-\tfrac12mgu_xt^2$$
Magnitude: $L=\tfrac12mg\,u\cos45^\circ\,t^2=\tfrac12\times0.1\times10\times\dfrac{20}{\sqrt2}\times4=20\sqrt2=\sqrt{800}$.

So $K=800$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. Rotational KE $E=\tfrac12I\omega^2$ with $I=\tfrac1{12}ML^2$ about the centre; the mass is $M=AdL$.

Step 1. Mass and $I$. $L=2$ m: $M=2Ad$, $I=\tfrac1{12}(2Ad)(4)=\dfrac{2Ad}{3}$.

Step 2. Energy. $E=\tfrac12\cdot\dfrac{2Ad}{3}\omega^2=\dfrac{Ad\,\omega^2}{3}\Rightarrow\omega=\sqrt{\dfrac{3E}{Ad}}$.

Answer. $\alpha=3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 54

Concept. At first the disc slides, so kinetic friction $\mu mg$ acts backwards. It slows the centre and spins the disc up. Pure rolling starts when $v=\omega R$. Angular momentum about the contact point is conserved (friction passes through it), which gives the final speed directly.

Step 1. Final speed from angular momentum about the contact point. Initially $L=mv_0R$ (no spin). Finally $L=mvR+\tfrac12mR^2\cdot\dfrac vR=\tfrac32mvR$.
$$mv_0R=\frac32mvR\ \Rightarrow\ v=\frac23v_0=12\ \text{m s}^{-1}$$

Step 2. Check the time. Friction decelerates the centre: $v=18-\mu gt=18-3t$. Its torque $\mu mgR=\tfrac12mR^2\alpha$ gives $\alpha R=2\mu g$, so $\omega R=2\mu gt=6t$. They are equal at $t=2$ s, as given.

Step 3. Kinetic energy when rolling.
$$K=\frac12mv^2\left(1+\frac12\right)=\frac34mv^2=\frac34\times0.5\times144=54\ \text{J}$$

Answer. $54$ J.

Solution by Sreeraj P, M.Sc Physics

Answer: 10

For a rolling solid sphere $K=\dfrac7{10}mv^2$:
$$7\times10^{-3}=0.7\times1\times v^2\Rightarrow v=0.1\ \text{m s}^{-1}=10\ \text{cm s}^{-1}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Equal masses: $\rho_1\pi r_1^2t_1=\rho_2\pi r_2^2t_2$, so
$$\frac{r_1^2}{r_2^2}=\frac{\rho_2t_2}{\rho_1t_1}=\frac{5\times0.5}{3\times1}=\frac56$$
About a diameter $I=\dfrac{mr^2}{4}$, so $\dfrac{I_1}{I_2}=\dfrac{r_1^2}{r_2^2}=\dfrac56$ and $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 110

Concept. Parallel axes: $k^2=\tfrac25R^2+d^2$ for a solid sphere whose centre is $d$ from the axis (from the figure, $d=10$ cm).

Step 1. Substitute. $k^2=\tfrac25(25)+100=110$ cm$^2$.

Answer. $x=110$.

Solution by Sreeraj P, M.Sc Physics

Answer: 32

The carved part has $\dfrac14$ of the cross-section and $\dfrac12$ the length, so its mass is $\dfrac M8$.
$$I_1=\frac12MR^2,\qquad I_2=\frac12\cdot\frac M8\cdot\frac{R^2}{4}=\frac{MR^2}{64}$$
$\dfrac{I_1}{I_2}=32$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $300\ \text{N}$

Concept. For the rod in equilibrium, take torques about the hinge $C$. The hinge force then has zero lever arm and drops out, leaving one equation for the tension.

Step 1. Clockwise torques (weights) about $C$.

• $8$ kg at $D$, $1$ m away: $80\times1=80$ N m
• rod's weight $20$ N at its centre, $0.5$ m: $10$ N m

Total: $90$ N m.

Step 2. Anticlockwise torque (cable). The cable meets the rod at $B$, $0.6$ m from $C$, at $30^\circ$. Only the component perpendicular to the rod turns it:
$$\tau_T=T\sin30^\circ\times0.6=0.3\,T$$

Step 3. Balance: $0.3\,T=90\Rightarrow T=300$ N.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 17

Parallel axis theorem:
$$I_{AB}=\frac12MR^2+M\left(\frac23R\right)^2=MR^2\left(\frac12+\frac49\right)=\frac{17}{18}MR^2$$
$\dfrac{I_{AB}}{I_{CM}}=\dfrac{17/18}{1/2}=\dfrac{17}{9}$, so $x=17$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Disc, tangent in its plane: $\dfrac14MR^2+MR^2=\dfrac54MR^2=\dfrac54(4)R^2=5R^2$.

Sphere, tangent: $\dfrac25MR^2+MR^2=\dfrac75(5)R^2=7R^2$.

Ratio $=\dfrac57$, so $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 15

$I=MR^2=5\times0.49=2.45\ \text{kg m}^2$, $\tau=FR=52.5\times0.7=36.75\ \text{N m}$.

$\alpha=\dfrac\tau I=\dfrac{36.75}{2.45}=15\ \text{rad s}^{-2}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

$k_{sph}=\sqrt{\tfrac25}R$, $k_{cyl}=\sqrt{\tfrac12}R$ (about their symmetry axes).
$$\frac{k_{sph}}{k_{cyl}}=\sqrt{\frac{2/5}{1/2}}=\sqrt{\frac45}=\frac2{\sqrt5}$$
So $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both A and R are correct and R is the correct explanation of A

Concept. A rotating body keeps rotating unless a torque stops it (rotational inertia, the analogue of Newton's first law). Friction at the bearings and air drag slow the fan only gradually.

Step 1. Assertion. True: the fan runs on after switch-off.

Step 2. Reason. True, and it is exactly why: inertia of (rotational) motion.

Answer. Both correct and R is the correct explanation of A (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Position of the point of application relative to $(2,-3)$: $\vec r=(0-2)\hat i+(0+3)\hat j=-2\hat i+3\hat j$.
$$\vec\tau=\vec r\times\vec F=(-2\hat i+3\hat j)\times(-P\hat k)=2P(\hat i\times\hat k)-3P(\hat j\times\hat k)=-3P\hat i-2P\hat j$$
$a=-3$, $b=-2$, $\dfrac ab=\dfrac32$, so $x=3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 35

$I_{tangent}=\dfrac75MR^2=1.4\times0.5\times0.0025=1.75\times10^{-3}\ \text{kg m}^2$.

$L=\dfrac25MR^2\omega=0.4\times0.5\times0.0025\times10=5\times10^{-3}\ \text{kg m}^2\,\text{s}^{-1}$.

Ratio $=\dfrac{1.75\times10^{-3}}{5\times10^{-3}}=0.35=35\times10^{-2}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Angular momentum is conserved and $K=\dfrac{L^2}{2I}$, so tripling $I$ makes $K=\dfrac E3$. $x=3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

For a shell $I=\tfrac23mR^2$: $K_{rot}=\tfrac13mv^2$, $K_{total}=\tfrac12mv^2+\tfrac13mv^2=\tfrac56mv^2$.

Ratio $=\dfrac{1/3}{5/6}=\dfrac25$, so $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $R\sqrt{\pi^2+4}$

In half a rotation the centre moves $\pi R$ forward and $A$ goes from the top to the bottom (down by $2R$):
$$d=\sqrt{(\pi R)^2+(2R)^2}=R\sqrt{\pi^2+4}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 4

$L=\tfrac25mR^2\omega$, $E=\tfrac7{10}mR^2\omega^2$, so $\dfrac LE=\dfrac{4}{7\omega}=\dfrac{\pi}{22}$.
$$\omega=\frac{88}{7\pi}=\frac{88}{22}=4\ \text{rad s}^{-1}\quad\left(\pi=\tfrac{22}{7}\right)$$

Solution by Sreeraj P, M.Sc Physics

Answer: 176

Concept. As above: each sphere $\tfrac25mR^2+md^2$ with $d=20$ cm.

Step 1. One sphere. $\tfrac25(2)(0.1)^2+2(0.2)^2=0.008+0.08=0.088$ kg m$^2$.

Step 2. Two spheres. $0.176=176\times10^{-3}$ kg m$^2$.

Answer. $176$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

$k_{ring}=R_r$, $k_{sphere}=\sqrt{\tfrac25}R_s$. Equal: $\dfrac{R_r}{R_s}=\sqrt{\dfrac25}$, so $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Every element of the semicircular ring is at distance $R$ from the axis, so $I=MR^2$ and $x=1$.

Solution by Sreeraj P, M.Sc Physics

Answer: 75

For a rolling hollow sphere $K=\tfrac12mv^2\left(1+\tfrac23\right)=\tfrac56mv^2=7.5m$ (J). It rolls up until all of this becomes potential energy:
$$mgh=7.5m\ \Rightarrow\ h=0.75\ \text{m}=75\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. Volume $\tfrac1{64}$ means radius $\tfrac14$ (volume $\propto R^3$). Angular momentum conservation gives $T\propto R^2$.

Step 1. New period. $T'=24\times\tfrac1{16}=\tfrac{24}{16}$ h.

Answer. $x=16$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Energy for rolling from rest: $v^2=\dfrac{2gh}{1+k^2/R^2}=\dfrac{4gh}{3}$ for a solid cylinder.

Step 1. Height. $h=0.6\sin30^\circ=0.3$ m.

Step 2. Speed. $v^2=\dfrac{4(10)(0.3)}{3}=4\Rightarrow v=2$ m s$^{-1}$.

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. To move an axis from the centre to a parallel axis at distance $d$, use the parallel-axis theorem $I=I_{cm}+Md^2$. The axis here is normal to the disc, so start from the normal axis through the centre.

Step 1. Normal axis through the centre. By the perpendicular-axis theorem, $I_z=I_x+I_y=\dfrac{MR^2}{4}+\dfrac{MR^2}{4}=\dfrac{MR^2}{2}$.

Step 2. Shift to the rim ($d=R$):
$$I=\frac{MR^2}{2}+MR^2=\frac32MR^2$$

Answer. $\dfrac x2=\dfrac32$, so $x=3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

Concept. Moments about the new balance point: the coins' moment equals the moment of the scale's weight, which acts at its centre ($50$ cm).

Step 1. Arms. Coins ($20$ g) at $10$ cm: arm $30$ cm. Scale's weight at $50$ cm: arm $10$ cm.

Step 2. Balance. $20\times30=M\times10\Rightarrow M=60$ g $=6\times10^{-2}$ kg.

Answer. $6$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $15\ \text{rad}$

Concept. Uniform angular acceleration obeys the same equations as uniform linear acceleration: $\theta=\omega_0t+\tfrac12\alpha t^2$. From rest, $\theta\propto t^2$.

Step 1. First second. $5=\tfrac12\alpha(1)^2\Rightarrow\alpha=10$ rad s$^{-2}$.

Step 2. First two seconds. $\theta(2)=\tfrac12(10)(2)^2=20$ rad.

Step 3. Next second only. $20-5=15$ rad.

Answer. $15$ rad (Option B). The angles in successive equal intervals from rest go as $1:3:5:\dots$

Solution by Sreeraj P, M.Sc Physics