Rotational dynamics about a moving axis: A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. How long does it take it to travel the first 3.1 m?
A) 1.4 s
B) 1.1 s
C) 2.1 s
D) 1.6 s
E) 1.8 s

Question

Quizplus · Accepted Answer

To solve this problem, we use the equations of motion for rolling without slipping, where the acceleration $a$ of the center of mass is given by $a = g \sin(\theta)$ for a slope of angle $\theta$. However, for a thin cylindrical shell, the moment of inertia $I$ about its axis is $I = mR^2$, where $m$ is the mass and $R$ is the radius. The acceleration down the incline is modified by the factor $\frac{1}{1 + \frac{I}{mR^2}}$ due to the rotational inertia. For a thin cylindrical shell, this factor becomes $\frac{1}{1 + 1} = \frac{1}{2}$, so the acceleration $a = \frac{1}{2}g \sin(\theta)$. With $\theta = 30^\circ$ and $g = 9.8 \, \text{m/s}^2$, $a = \frac{1}{2} \times 9.8 \times \sin(30^\circ) = \frac{1}{2} \times 9.8 \times 0.5 = 2.45 \, \text{m/s}^2$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ where $s = 3.1 \, \text{m}$, $u = 0$ (since it starts from rest), and $a = 2.45 \, \text{m/s}^2$, we solve for $t$:$$3.1 = 0 \times t + \frac{1}{2} \times 2.45 \times t^2$$$$3.1 = 1.225t^2$$$$t^2 = \frac{3.1}{1.225}$$$$t^2 = 2.53$$$$t = \sqrt{2.53} \approx 1.59 \, \text{s}$$Thus, the closest answer is 1.6 seconds.

Rotational Dynamics About a Moving Axis: a Thin Cylindrical Shell