Question

During the launch from a board, a diver’s angular speed about her center of mass changes from zero to 6.20rad/sin 220ms. Her rotational inertia about her center of mass is 12.0kg.m². During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?

Short answer

1. Average angular acceleration is, 28.2rad/s² .
2. The average external torque on the diver from the board is, $3.38\times {10}^{2}N.m$..

Step-by-step solution

Understanding the given information

1. The difference between initial and final velocity is, $\omega -{\omega }_0=6.20rad/s$.
2. The time is, $t=220ms=220\times {10}^{-3}s$.
3. The rotational inertia is, $l=12kg.m^2$.

Concept and formula used in the given question

The diver’s angular speed about her center of masschanges from zero to some value. Her initial angular velocity and final angular velocity is given in the problem. You can find angular acceleration by using kinematic equation. Also, as the diver was diving from the board, you may consider their moment of inertia in a rotational frame. Therefore, the average external torque on her, from the board, can be calculated. The formulas required are given below.

1 Kinematic equation,
$\mathit{\omega }\mathbf{=}{\mathit{\omega }}_{\mathbf{0}}\mathbf{+}\mathit{\alpha }\mathit{t}$

2 External Torque,

$\mathit{\tau }\mathbf{=}\mathit{l}\mathit{\alpha }$

(a) Calculation for the magnitudes of her average angular acceleration

Calculating average angular acceleration by using kinematic equation,

We have,

$$\begin{gathered} \omega ={\omega }_0+\alpha t \\ \alpha =\frac{\omega -{\omega }_0}{t} \end{gathered}$$

Substitute all the value in the above equation

$\begin{array}{rcl}\alpha & =& \frac{6.20rad/s}{220\times {10}^{-3}s}\\ & =& 28.2rad/s^2\end{array}$

Hence the acceleration is, 28.2rad/s² .

(b) Calculation for the magnitudes of the average external torque on her from the board

Now calculating magnitude of torque acting onthediver’s body,

We have,

$\tau =l\alpha$

Substitute all the value in the above equation

$\begin{array}{rcl}\tau & =& 12kg.m^2\times 28.2rad/s^2\\ & =& 3.38\times {10}^{2}N.m\end{array}$

Hence the torque is, $\begin{array}{rcl}& & 3.38\times {10}^{2}N.m\end{array}$.