A figure skater is spinning with an angular velocity of +18.8 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +4.94 rad. Determine (a) her average angular acceleration and (b) the time during which she comes to rest. (a) Number i Units (b) Number i
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![### Physics Problem: Angular Motion of a Figure Skater
**Problem Statement:**
A figure skater is spinning with an angular velocity of +18.8 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is +4.94 rad. Determine:
- (a) Her average angular acceleration.
- (b) The time during which she comes to rest.
**Solution Approach:**
**(a) Average Angular Acceleration**
To find the average angular acceleration (\(\alpha\)), we can use the formula:
\[
\theta = \omega_0 t + \frac{1}{2} \alpha t^2
\]
Where:
- \(\theta\) is the angular displacement (+4.94 rad)
- \(\omega_0\) is the initial angular velocity (+18.8 rad/s)
- \(t\) is the time
- \(\alpha\) is the angular acceleration
Since the skater comes to rest, the final angular velocity (\(\omega\)) is 0. The formula connecting initial angular velocity, final angular velocity, angular acceleration, and time is:
\[
\omega = \omega_0 + \alpha t
\]
Given:
- \(\omega = 0\) (comes to rest)
- \(\omega_0 = +18.8 \, \text{rad/s}\)
Rearranging the second equation to solve for \(\alpha\):
\[
\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 18.8}{t} = \frac{-18.8}{t}
\]
Now, substituting \(\alpha = \frac{-18.8}{t}\) into the first equation:
\[
4.94 = 18.8 t + \frac{1}{2} \left(\frac{-18.8}{t}\right) t^2
\]
This is a quadratic equation in \(t\):
\[
4.94 = 18.8 t - 9.4t
\]
Solving for \(t\):
\[
4.94 = 9.4t
\]
\[
t = \frac{4.94}{9.4}
\]
**(b) Time**
Using the derived time (t), substitute it to find \(\alpha\):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2402ca11-e4b3-4540-8aaf-9a16624dfe94%2Ffefa6f13-924f-46da-bf3c-957a715effd8%2F6xp30tk_processed.jpeg&w=3840&q=75)
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