College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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### Problem Statement

A solid sphere (moment of inertia \( I = 0.4 mR^2 \)) with radius \( R = 0.14 \) meters starts from rest and rolls without slipping down a ramp through a vertical height of 0.74 meters. At the bottom, where it is rolling horizontally, what are its linear velocity \( v \) and its rotational velocity \( \omega \)?

### Diagram Description

The diagram accompanying the problem shows a solid sphere at the top of a curved ramp. The ramp descends to the right from a height \( h \) which is marked to be 0.74 meters. The radius of the sphere is indicated to be R which is specified as 0.14 meters.

The diagram visually represents:
- A solid sphere at rest at the top of the ramp.
- A curved ramp that leads to a horizontal section.
- An indication of the height \( h \) of the ramp’s starting point (0.74 meters).

### Calculation Hints

To solve this problem, we apply the principles of conservation of energy and the relationship between linear and rotational motion for a rolling object.

1. **Conservation of Energy:**
   The gravitational potential energy at the top will convert into translational and rotational kinetic energy at the bottom:
   
   \[
   mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
   \]
   
2. **Moment of Inertia (\( I \)) for a solid sphere:**
   
   \[
   I = \frac{2}{5}mR^2
   \]
   
3. **Relationship Between Linear and Rotational Motion (For Rolling Without Slipping):**
   
   \[
   v = \omega R
   \]

By substituting the given values and relationships into the conservation of energy equation, one can solve for the linear and rotational velocities.
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Transcribed Image Text:### Problem Statement A solid sphere (moment of inertia \( I = 0.4 mR^2 \)) with radius \( R = 0.14 \) meters starts from rest and rolls without slipping down a ramp through a vertical height of 0.74 meters. At the bottom, where it is rolling horizontally, what are its linear velocity \( v \) and its rotational velocity \( \omega \)? ### Diagram Description The diagram accompanying the problem shows a solid sphere at the top of a curved ramp. The ramp descends to the right from a height \( h \) which is marked to be 0.74 meters. The radius of the sphere is indicated to be R which is specified as 0.14 meters. The diagram visually represents: - A solid sphere at rest at the top of the ramp. - A curved ramp that leads to a horizontal section. - An indication of the height \( h \) of the ramp’s starting point (0.74 meters). ### Calculation Hints To solve this problem, we apply the principles of conservation of energy and the relationship between linear and rotational motion for a rolling object. 1. **Conservation of Energy:** The gravitational potential energy at the top will convert into translational and rotational kinetic energy at the bottom: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] 2. **Moment of Inertia (\( I \)) for a solid sphere:** \[ I = \frac{2}{5}mR^2 \] 3. **Relationship Between Linear and Rotational Motion (For Rolling Without Slipping):** \[ v = \omega R \] By substituting the given values and relationships into the conservation of energy equation, one can solve for the linear and rotational velocities.
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