Answer :
To tackle the given physics problems, we will break down each section separately and provide detailed explanations and solutions.
### 1. Definition of Terms
(a) Definitions:
i. Angular Acceleration:
Angular acceleration is the rate of change of angular velocity with respect to time. It describes how quickly an object is rotating faster or slower. It is usually measured in radians per second squared (rad/s²).
ii. Centrifugal Force:
Centrifugal force is an apparent force that acts outward on a body moving around a center, arising from the body's inertia. It is not an actual force but rather a result of the inertia of the body moving in a curved path.
iii. Uniform Acceleration:
Uniform acceleration refers to a constant acceleration, meaning that the velocity of an object changes at a constant rate over time. It implies that the net force acting on the object is constant in magnitude and direction.
### 2. Equations and Problems
(b) Equations Relating Linear and Angular Motion:
The relationship between linear velocity (v) and angular velocity (ω) with the radius (r) of the path in circular motion is given by the equation:
[tex]\[ v = \omega \cdot r \][/tex]
i. Calculation of Angular Acceleration:
To find the angular acceleration (α), we use the formula:
[tex]\[ \alpha = \frac{\Delta \omega}{\Delta t} \][/tex]
where [tex]\( \Delta \omega \)[/tex] is the change in angular velocity and [tex]\( \Delta t \)[/tex] is the time over which this change occurs.
Given the angular velocities (initial and final):
- Initial angular velocity [tex]\( \omega_i = 5.8 \, \text{revs/s} \)[/tex]
- Final angular velocity [tex]\( \omega_f = 16.4 \, \text{revs/s} \)[/tex]
- The time is not given in this instance, so the formula can't be completed without it.
ii. An Object on a Circular Path:
Given:
- Mass [tex]\( m = 5.0 \, \text{kg} \)[/tex]
- Radius [tex]\( r = 6.0 \, \text{m} \)[/tex]
- Constant speed [tex]\( v = 10 \, \text{m/s} \)[/tex]
a. Angular Velocity Calculation:
Using the linear and angular velocity relation:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{10 \, \text{m/s}}{6.0 \, \text{m}} \][/tex]
[tex]\[ \omega \approx 1.67 \, \text{rad/s} \][/tex]
b. Centripetal Force Calculation:
Centripetal force [tex]\( F_c \)[/tex] is given by:
[tex]\[ F_c = m \cdot a_c \][/tex]
where [tex]\( a_c \)[/tex] is the centripetal acceleration:
[tex]\[ a_c = \frac{v^2}{r} \][/tex]
[tex]\[ a_c = \frac{(10 \, \text{m/s})^2}{6.0 \, \text{m}} \][/tex]
[tex]\[ a_c \approx 16.67 \, \text{m/s}^2 \][/tex]
Thus,
[tex]\[ F_c = 5.0 \, \text{kg} \cdot 16.67 \, \text{m/s}^2 \][/tex]
[tex]\[ F_c \approx 83.35 \, \text{N} \][/tex]
### 3. Circular Motion of a Body
(c) Motion Analysis:
Given:
- A body makes six complete revolutions in 4.0 s.
- Radius [tex]\( r = 25.0 \, \text{cm} = 0.25 \, \text{m} \)[/tex]
i. Linear Velocity:
First, find the total distance traveled, which is the circumference times the number of revolutions:
[tex]\[ \text{Total distance} = 2 \pi r \times \text{number of revolutions} \][/tex]
[tex]\[ \text{Total distance} = 2 \pi \times 0.25 \, \text{m} \times 6 \][/tex]
[tex]\[ \text{Total distance} = 3 \pi \, \text{m} \approx 9.42 \, \text{m} \][/tex]
Now, calculate the linear velocity (v) by dividing the total distance by the total time:
[tex]\[ v = \frac{\text{Total distance}}{\text{Total time}} \][/tex]
[tex]\[ v = \frac{9.42 \, \text{m}}{4.0 \, \text{s}} \][/tex]
[tex]\[ v = 2.36 \, \text{m/s} \][/tex]
ii. Angular Velocity:
Using the linear and angular velocity relation:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{2.36 \, \text{m/s}}{0.25 \, \text{m}} \][/tex]
[tex]\[ \omega \approx 9.44 \, \text{rad/s} \][/tex]
This concludes the detailed, step-by-step solutions for the given physics problems.
### 1. Definition of Terms
(a) Definitions:
i. Angular Acceleration:
Angular acceleration is the rate of change of angular velocity with respect to time. It describes how quickly an object is rotating faster or slower. It is usually measured in radians per second squared (rad/s²).
ii. Centrifugal Force:
Centrifugal force is an apparent force that acts outward on a body moving around a center, arising from the body's inertia. It is not an actual force but rather a result of the inertia of the body moving in a curved path.
iii. Uniform Acceleration:
Uniform acceleration refers to a constant acceleration, meaning that the velocity of an object changes at a constant rate over time. It implies that the net force acting on the object is constant in magnitude and direction.
### 2. Equations and Problems
(b) Equations Relating Linear and Angular Motion:
The relationship between linear velocity (v) and angular velocity (ω) with the radius (r) of the path in circular motion is given by the equation:
[tex]\[ v = \omega \cdot r \][/tex]
i. Calculation of Angular Acceleration:
To find the angular acceleration (α), we use the formula:
[tex]\[ \alpha = \frac{\Delta \omega}{\Delta t} \][/tex]
where [tex]\( \Delta \omega \)[/tex] is the change in angular velocity and [tex]\( \Delta t \)[/tex] is the time over which this change occurs.
Given the angular velocities (initial and final):
- Initial angular velocity [tex]\( \omega_i = 5.8 \, \text{revs/s} \)[/tex]
- Final angular velocity [tex]\( \omega_f = 16.4 \, \text{revs/s} \)[/tex]
- The time is not given in this instance, so the formula can't be completed without it.
ii. An Object on a Circular Path:
Given:
- Mass [tex]\( m = 5.0 \, \text{kg} \)[/tex]
- Radius [tex]\( r = 6.0 \, \text{m} \)[/tex]
- Constant speed [tex]\( v = 10 \, \text{m/s} \)[/tex]
a. Angular Velocity Calculation:
Using the linear and angular velocity relation:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{10 \, \text{m/s}}{6.0 \, \text{m}} \][/tex]
[tex]\[ \omega \approx 1.67 \, \text{rad/s} \][/tex]
b. Centripetal Force Calculation:
Centripetal force [tex]\( F_c \)[/tex] is given by:
[tex]\[ F_c = m \cdot a_c \][/tex]
where [tex]\( a_c \)[/tex] is the centripetal acceleration:
[tex]\[ a_c = \frac{v^2}{r} \][/tex]
[tex]\[ a_c = \frac{(10 \, \text{m/s})^2}{6.0 \, \text{m}} \][/tex]
[tex]\[ a_c \approx 16.67 \, \text{m/s}^2 \][/tex]
Thus,
[tex]\[ F_c = 5.0 \, \text{kg} \cdot 16.67 \, \text{m/s}^2 \][/tex]
[tex]\[ F_c \approx 83.35 \, \text{N} \][/tex]
### 3. Circular Motion of a Body
(c) Motion Analysis:
Given:
- A body makes six complete revolutions in 4.0 s.
- Radius [tex]\( r = 25.0 \, \text{cm} = 0.25 \, \text{m} \)[/tex]
i. Linear Velocity:
First, find the total distance traveled, which is the circumference times the number of revolutions:
[tex]\[ \text{Total distance} = 2 \pi r \times \text{number of revolutions} \][/tex]
[tex]\[ \text{Total distance} = 2 \pi \times 0.25 \, \text{m} \times 6 \][/tex]
[tex]\[ \text{Total distance} = 3 \pi \, \text{m} \approx 9.42 \, \text{m} \][/tex]
Now, calculate the linear velocity (v) by dividing the total distance by the total time:
[tex]\[ v = \frac{\text{Total distance}}{\text{Total time}} \][/tex]
[tex]\[ v = \frac{9.42 \, \text{m}}{4.0 \, \text{s}} \][/tex]
[tex]\[ v = 2.36 \, \text{m/s} \][/tex]
ii. Angular Velocity:
Using the linear and angular velocity relation:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{2.36 \, \text{m/s}}{0.25 \, \text{m}} \][/tex]
[tex]\[ \omega \approx 9.44 \, \text{rad/s} \][/tex]
This concludes the detailed, step-by-step solutions for the given physics problems.
