Questions: Homework 7 Begin Date: 10/22/2024 12:01:00 AM Due Date: 11/1/2024 11:59:00 PM End Date: 12/6/2024 11:59:00 PM Problem 9: ( 10% of Assignment Value) Consider two cylindrical objects of the same mass and radius. Object A is a solid cylinder, whereas object B is a hollow cylinder. - Part (a) If these objects roll without slipping down a ramp, which one will reach the bottom of the ramp first? Object A Correct! Part (b) How fast, in meters per second, is object A moving at the end of the ramp if its mass is 330 g, its radius 33 cm, and the height of the beginning of the ramp is 13.5 cm? vA= Part (c) How fast, in meters per second, is object B moving at the end of the ramp if it rolls down the same ramp?

Solution Steps

We have two cylindrical objects, A (solid) and B (hollow), rolling down a ramp without slipping. We need to determine the speed of object A at the bottom of the ramp given its mass, radius, and the height of the ramp.

For a rolling object, the total mechanical energy is conserved. The potential energy at the top of the ramp is converted into translational and rotational kinetic energy at the bottom.

The potential energy at the top is: $$PE = mgh$$ where $m = 0.33 \, \text{kg}$, $g = 9.81 \, \text{m/s}^2$, and $h = 0.135 \, \text{m}$.

The kinetic energy at the bottom is the sum of translational and rotational kinetic energy: $$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a solid cylinder, the moment of inertia $I$ is: $$I = \frac{1}{2}mr^2$$

The angular velocity $\omega$ is related to the linear velocity $v$ by: $$\omega = \frac{v}{r}$$

Equating the potential energy at the top to the kinetic energy at the bottom: $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$

Simplifying: $$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$$

$$mgh = \frac{3}{4}mv^2$$

Canceling $m$ from both sides and solving for $v$: $$gh = \frac{3}{4}v^2$$

$$v^2 = \frac{4gh}{3}$$

$$v = \sqrt{\frac{4gh}{3}}$$

Substituting the known values: $$v = \sqrt{\frac{4 \times 9.81 \times 0.135}{3}}$$

$$v \approx \sqrt{1.7646}$$

$$v \approx 1.3285 \, \text{m/s}$$

Final Answer