A solid sphere of mass $m$ rolls without slipping down a rough inclined plane of inclination $\theta$. Consider the following statements:
A. The friction on the sphere acts up the incline and has magnitude $\dfrac27 mg\sin\theta$.
B. Friction does no work on the sphere.
C. The acceleration of the centre of the sphere is $\dfrac57 g\sin\theta$.
D. Rolling without slipping is possible only if the coefficient of friction $\mu \ge \dfrac27\tan\theta$.
E. At every instant the translational and rotational kinetic energies of the sphere are equal.
Choose the correct answer from the options given below:
Answer: (C) A, B, C and D only
Concept. For rolling, $a = \dfrac{g\sin\theta}{1 + I/mR^2}$ and friction supplies the torque. With no slipping the contact point is at rest, so static friction does no work.
Step 1. Acceleration (C). $I = \dfrac25mR^2$:
$$a = \frac{g\sin\theta}{1 + \frac25} = \frac57g\sin\theta \quad\text{(C true)}$$
Step 2. Friction (A). Torque $fR = I\dfrac aR$:
$$f = \frac25 ma = \frac27 mg\sin\theta,\ \text{up the incline} \quad\text{(A true)}$$
Step 3. Work (B). The contact point has zero velocity, so friction does no work (B true).
Step 4. Condition (D). $f \le \mu mg\cos\theta \Rightarrow \mu \ge \dfrac27\tan\theta$ (D true).
Step 5. Energies (E).
$$\frac{K_{rot}}{K_{tr}} = \frac{\frac12\cdot\frac25mR^2\omega^2}{\frac12mv^2} = \frac25 \ne 1 \quad\text{(E false)}$$
Answer. A, B, C and D only (Option C).
Solution by Sreeraj P, M.Sc Physics