Question

A figure skater is spinning with an angular velocity of + 12.8 rad/s. She then comes to a stop over a brief period of time. During this time, her angular displacement is 4.38 rad. Determine (a) her average angular acceleration and (b) the time during which she comes to rest.

Answer 1

(a) The average angular acceleration can be determined using the formula:

$\text{average angular acceleration} = \dfrac{\text{change in angular velocity}}{\text{change in time}}$

Since the skater has come to a stop, her change in angular velocity is equal to her initial angular velocity of +12.8 rad/s. Thus, the average angular acceleration is:

$\text{average angular acceleration} = \dfrac{0 - 12.8}{t} = -12.8/t$

(b) The angular displacement can be related to the angular velocity and the time using the formula:

$\text{angular displacement} = \text{angular velocity} \times \text{time}$

In this case, the angular displacement is given as 4.38 rad and the angular velocity is given as 12.8 rad/s. Solving for time, we get:

$\text{time} = \dfrac{\text{angular displacement}}{\text{angular velocity}} = \dfrac{4.38}{12.8} \approx 0.342\ \text{s}$

Therefore, the time during which the skater comes to rest is approximately 0.342 seconds.