### Problem StatementA 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 DescriptionThe 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 HintsTo 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.

Write the given values with the suitable variables. R=0.14 mh=0.74 mI=0.4mR2 Here, R, h, and I are…

2. A solid sphere (I = 0.4 mR2) with radius R = 0.14m starts from rest and rolls without slipping down a ramp through a vertical height of 0.74m. At the bottom where it is rolling horizontally, what are its velocity v and its rotational velocity wo? h

2. A solid sphere (I = 0.4 mR2) with radius R = 0.14m starts from rest and rolls without slipping down a ramp through a vertical height of 0.74m. At the bottom where it is rolling horizontally, what are its velocity v and its rotational velocity wo? h

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

### 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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