Problem 13
Question
A turntable rotates with a constant 2.25 rad/s\(^2\) angular acceleration. After 4.00 s it has rotated through an angle of 30.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00-s interval?
Step-by-Step Solution
Verified Answer
The initial angular velocity is 3.0 rad/s.
1Step 1: Identify the Known Variables
We are given:- Angular acceleration, \( \alpha = 2.25 \, \text{rad/s}^2 \)- Time interval, \( t = 4.00 \, \text{s} \)- Angular displacement, \( \theta = 30.0 \, \text{rad} \)We need to find the initial angular velocity \( \omega_0 \) of the wheel.
2Step 2: Use the Angular Kinematics Equation
The angular kinematics equation that relates initial angular velocity (\( \omega_0 \)), angular acceleration (\( \alpha \)), time (\( t \)), and angular displacement (\( \theta \)) is:\[\theta = \omega_0 t + \frac{1}{2} \alpha t^2\]We'll use this formula to solve for \( \omega_0 \).
3Step 3: Plug in Known Values and Solve for \(\omega_0\)
Substitute the known values into the equation: \[30.0 = \omega_0 (4.00) + \frac{1}{2} (2.25) (4.00)^2\]Calculate \( \frac{1}{2} \times 2.25 \times 16.0 \):\[= \frac{1}{2} \times 36.0\]\[= 18.0 \text{ rad}\]Thus, the equation becomes:\[30.0 = 4.00\omega_0 + 18.0\]
4Step 4: Isolate \(\omega_0\) and Solve
Subtract 18.0 from both sides:\[30.0 - 18.0 = 4.00\omega_0\]\[12.0 = 4.00\omega_0\]Divide both sides by 4.00:\[\omega_0 = \frac{12.0}{4.00} = 3.0 \, \text{rad/s}\]
5Step 5: Conclusion
The initial angular velocity of the wheel at the beginning of the 4-second interval is \( 3.0 \, \text{rad/s} \). Ensure the result makes sense in the context of the given data.
Key Concepts
Angular AccelerationAngular DisplacementInitial Angular Velocity
Angular Acceleration
Angular acceleration is a concept that describes the rate of change of angular velocity over time. Imagine it as how quickly something is spinning up or slowing down as it rotates. In this exercise, the turntable has a constant angular acceleration of \(2.25 \ \text{rad/s}^2\). That means with each passing second, the velocity increases by \(2.25 \ \text{rad/s}\). Angular acceleration can be positive, causing the object to spin faster, or negative, causing it to slow down.
Key points about angular acceleration include:
Key points about angular acceleration include:
- It is typically denoted by the symbol \(\alpha\).
- Its units are radians per second squared \((\text{rad/s}^2)\).
- It is a vector, meaning it has both magnitude and direction.
Angular Displacement
Angular displacement refers to the change in angle as an object moves along a circular path. In the context of this problem, the turntable rotates through an angle of \(30.0 \ \text{rad}\) over a period of 4 seconds, which is its angular displacement during the time interval. This concept is similar to linear displacement in straight-line motion but for rotational movement.
Important aspects of angular displacement are:
Important aspects of angular displacement are:
- It is represented by the symbol \(\theta\).
- It is measured in radians, a unit derived from the radius and circumference of a circle.
- Unlike angular velocity or acceleration, which are rates of change, angular displacement is a measure of distance in angular terms.
Initial Angular Velocity
Initial angular velocity is the angular speed at which a rotating object starts its motion. In the exercise, it represents the speed of the turntable at the very beginning of the 4-second period. It's what we're trying to find out from the given values using the angular kinematics equation.
Some essentials about initial angular velocity are:
Some essentials about initial angular velocity are:
- It is denoted by \(\omega_0\).
- Measured in radians per second \((\text{rad/s})\).
- It can be determined using the equation \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\), given angular displacement, time, and angular acceleration.
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