Question

A sphere of mass 2 kg and radius 5 cm is rotating at the rate of 300 rev. per minute. Then the torque required to stop it in $2\pi $ revolution is:
A. $1.6\times {{10}^{5}}N-m$
B. $2.5\times {{10}^{-3}}N-m$
C. $1.6\times {{10}^{-3}}N-m$
D. $2.5\times {{10}^{-2}}N-m$

Answer 1

Hint: We can find Torque of sphere by using formula $\tau =I\alpha $ where I is moment of inertia, $\alpha $ is angular acceleration.
For the sphere moment of inertia it is $I=\dfrac{2}{5}M{{r}^{2}}$ where M is mass and r is radius of the sphere.

Complete step-by-step answer:
In the given question we have initial angular speed ${\omega}_0$ is 300 rev. per minute and revolution is $2\pi $.
We can change angular speed in radian per second by multiplying $\dfrac{2\pi }{60}$ .
Hence we can write
$\Rightarrow {{\omega }_{0}}=300\times \dfrac{2\pi }{60}rad/\sec $
$\Rightarrow {{\omega }_{0}}=10\pi rad/\sec $
Now we can find angular displacement $\left( \theta \right)$ as
$\Rightarrow \theta =2\pi \times revolution$
$\Rightarrow \theta =2\pi \times 2\pi \,radian$
$\Rightarrow \theta =4{{\pi }^{2}}\,radian$
Final angular speed $(\omega )$ of sphere is 0.
Now we can find angular acceleration by using third equation of angular motion which is $\Rightarrow {{\omega }^{2}}={{\omega }_{0}}^{2}+2\theta \alpha $
Where \[\omega \] is final angular speed, ${{\omega }_{0}}$ is initial angular speed , $\theta $ is angular displacement, $\alpha $ is angular acceleration.
Now we can substitute value of \[\omega ,{{\omega }_{0}},\theta \]
$\Rightarrow 0={{(10\pi )}^{2}}+2\times 4{{\pi }^{2}}\times \alpha $
$\Rightarrow 2\times 4{{\pi }^{2}}\times \alpha =100{{\pi }^{2}}$
$\Rightarrow \alpha =\dfrac{-100{{\pi }^{2}}}{2\times 4{{\pi }^{2}}}rad/{{\sec }^{2}}$
$\Rightarrow \alpha =\dfrac{-25}{2}\,rad/{{\sec }^{2}}$…………………………………………….(i)
We can ignore negative signs because the speed of the sphere is slowing down.
Moment of inertia of sphere is
$\Rightarrow I=\dfrac{2}{5}M{{r}^{2}}$
In question
$Mass(M)=2kg$
$radius(r)=5cm=5\times {{10}^{-2}}m$ $\left\{ \because 1cm={{10}^{-2}}m \right\}$
$\Rightarrow I=\dfrac{2}{5}\times 2\times {{(5\times {{10}^{-2}})}^{2}}$
$\Rightarrow I=\dfrac{4\times 25\times {{10}^{-4}}}{5}$
$\Rightarrow I=\dfrac{100\times {{10}^{-4}}}{5}$
$\Rightarrow I=\dfrac{{{10}^{-2}}}{5}$…………………………………………(ii)
Now we can find torque by using $\tau =I\alpha $
On substituting value of I and $\alpha $from equation (ii) and (i) respectively.
$\Rightarrow \tau =\dfrac{{{10}^{-2}}}{5}\times \dfrac{25}{2}\,N-m$
$\Rightarrow \tau =2.5\times {{10}^{-2}}\,N-m$
Hence option D is correct.

Note: In the given question we need to remember the formula of moment of inertia of sphere which is $I=\dfrac{2}{5}M{{r}^{2}}$ where M is mass of sphere and r is radius.
Also in this question we ignored negative signs of acceleration because it shows speed is slowing down. We can also say acceleration acts as deceleration in this question.