Question

During the launch from aboard, a diver’s angular speed about her centre of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her centre of mass is 12.0 kg·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?

Answer 1

Hint:We are given with initial angular speed and final angular speed about the centre of mass. The time taken by the driver to change the speed is also given. Moment of inertia is also given. Angular acceleration can be determined by using the first equation of rotational motion and torque can find out using the relationship between torque, a moment of inertia and angular velocity

Complete step by step answer:
Initial angular velocity, \[{{\omega }_{0}}=0\]
Final angular velocity, \[\omega =6.2\]rad/s
Time is taken, t= 220 ms
t=0.22 s
Using the first equation, we get
\[\begin{align}
&\Rightarrow \omega ={{\omega }_{0}}+\alpha t \\
&\Rightarrow 6.2=0+0.22\alpha \\
&\Rightarrow 0.22\alpha =6.2 \\
&\therefore \alpha =28.2rad/{{s}^{2}} \\
\end{align}\]
Thus, the angular acceleration of the diver after 220 ms is 28.2 \[rad/{{s}^{2}}\]
Now to find out external torque, we use the formula \[\tau =I\alpha \], putting the values we get,
\[\begin{align}
&\Rightarrow \tau =12\times 28.2 \\
&\therefore =338.4Nm \\
\end{align}\]
So, the value of the angular acceleration of the diver is \[28.2rad/{{s}^{2}}\]and the value of external torque acting on the body is 338.4 Nm.

Note:Moment of Inertia plays the same role in rotation which is played by mass in translation. If a force acts on a body and the body starts rotating about its axis, then that force is termed as Torque. In angular motion, we talk about torque in place of force.