A car with tires whose radius is 32 cm is traveling along a highway at 24.5 m/s. The mass of the tire is equal to 12.0 kg and it's mass is distributed around the outer diameter so that it's moment of inertia is I=MR². (a) What is the angular velocity of the tires? (b) Compute the rotational kinetic energy of the tire (c) Compute the total kinetic energy of the tire (d) How many revolutions will the tires make in six minutes? (e) Consider now that the car brakes with an angular acceleration of -25.5 rad/sec^2 and comes to a stop; how many revolutions will the tires have made between the time the brakes are applied and when the car stops?
A car with tires whose radius is 32 cm is traveling along a highway at 24.5 m/s. The mass of the tire is equal to 12.0 kg and it's mass is distributed around the outer diameter so that it's moment of inertia is I=MR². (a) What is the angular velocity of the tires? (b) Compute the rotational kinetic energy of the tire (c) Compute the total kinetic energy of the tire (d) How many revolutions will the tires make in six minutes? (e) Consider now that the car brakes with an angular acceleration of -25.5 rad/sec^2 and comes to a stop; how many revolutions will the tires have made between the time the brakes are applied and when the car stops?
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Parts D and E
![### Rotational Motion of Tires: Educational Analysis
A car with tires whose radius is 32 cm is traveling along a highway at 24.5 m/s. The mass of the tire is equal to 12.0 kg and its mass is distributed around the outer diameter so that its moment of inertia is I = MR².
Given the information above, let's solve the following questions:
(a) **What is the angular velocity of the tires?**
(b) **Compute the rotational kinetic energy of the tire.**
(c) **Compute the total kinetic energy of the tire.**
(d) **How many revolutions will the tires make in six minutes?**
(e) **Consider now that the car brakes with an angular acceleration of -25.5 rad/sec² and comes to a stop; how many revolutions will the tires have made between the time the brakes are applied and when the car stops?**
---
### Solutions:
**(a) Angular Velocity (ω):**
\[ v = r\omega \]
Where:
- \( v \) is the linear velocity (24.5 m/s)
- \( r \) is the radius (0.32 m)
Solving for \( \omega \):
\[ \omega = \frac{v}{r} = \frac{24.5 \text{ m/s}}{0.32 \text{ m}} \approx 76.56 \text{ rad/s} \]
**(b) Rotational Kinetic Energy (KE\(_\text{rot}\)):**
\[ KE_\text{rot} = \frac{1}{2} I \omega^2 \]
Where:
- \( I = MR^2 \) \( (I = 12.0 \text{ kg} \times (0.32 \text{ m})^2) = 1.2288 \text{ kg}\cdot\text{m}^2 \)
- \( \omega = 76.56 \text{ rad/s} \)
\[ KE_\text{rot} = \frac{1}{2} \times 1.2288 \times (76.56)^2 \approx 3602.27 \text{ J} \]
**(c) Total Kinetic Energy (KE):**
The total kinetic energy is the sum of the translational and rotational kinetic energy components. However, in this context, we consider only the rotational movement.
\[ KE_\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff63a57a-dcd1-4774-86da-c01ab69a7706%2Fee648ecf-58db-45c5-b079-cbc30b308d47%2Fa4jkk0d_processed.png&w=3840&q=75)
Transcribed Image Text:### Rotational Motion of Tires: Educational Analysis
A car with tires whose radius is 32 cm is traveling along a highway at 24.5 m/s. The mass of the tire is equal to 12.0 kg and its mass is distributed around the outer diameter so that its moment of inertia is I = MR².
Given the information above, let's solve the following questions:
(a) **What is the angular velocity of the tires?**
(b) **Compute the rotational kinetic energy of the tire.**
(c) **Compute the total kinetic energy of the tire.**
(d) **How many revolutions will the tires make in six minutes?**
(e) **Consider now that the car brakes with an angular acceleration of -25.5 rad/sec² and comes to a stop; how many revolutions will the tires have made between the time the brakes are applied and when the car stops?**
---
### Solutions:
**(a) Angular Velocity (ω):**
\[ v = r\omega \]
Where:
- \( v \) is the linear velocity (24.5 m/s)
- \( r \) is the radius (0.32 m)
Solving for \( \omega \):
\[ \omega = \frac{v}{r} = \frac{24.5 \text{ m/s}}{0.32 \text{ m}} \approx 76.56 \text{ rad/s} \]
**(b) Rotational Kinetic Energy (KE\(_\text{rot}\)):**
\[ KE_\text{rot} = \frac{1}{2} I \omega^2 \]
Where:
- \( I = MR^2 \) \( (I = 12.0 \text{ kg} \times (0.32 \text{ m})^2) = 1.2288 \text{ kg}\cdot\text{m}^2 \)
- \( \omega = 76.56 \text{ rad/s} \)
\[ KE_\text{rot} = \frac{1}{2} \times 1.2288 \times (76.56)^2 \approx 3602.27 \text{ J} \]
**(c) Total Kinetic Energy (KE):**
The total kinetic energy is the sum of the translational and rotational kinetic energy components. However, in this context, we consider only the rotational movement.
\[ KE_\
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