A solid cylinder is given a speed of $10$ m/s up a rough incline of inclination $30^\circ$ and rolls without slipping all the way up and back down. The time after which it returns to its starting point is ______ s. (Take $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 6
Concept. While rolling without slipping, the acceleration along the incline is $\dfrac{g\sin\theta}{1 + I/mR^2}$, directed down the incline both on the way up and on the way down (static friction adjusts its direction so the cylinder keeps rolling). So the motion is uniformly accelerated, like a ball thrown up.
Step 1. Acceleration. For a solid cylinder $\dfrac{I}{mR^2} = \dfrac12$:
$$a = \frac{g\sin30^\circ}{1 + \frac12} = \frac{5}{1.5} = \frac{10}{3}\ \text{m/s}^2$$
Step 2. Time up. $t_1 = \dfrac{v_0}{a} = \dfrac{10}{10/3} = 3$ s.
Step 3. Time down. Same acceleration and distance, so $t_2 = 3$ s.
Answer. $6$.
Solution by Sreeraj P, M.Sc Physics