Question

A horizontal $90\text{kg}$ merry-go-round is a solid disc of radius $1.50\text{m}$ is started from rest by a constant horizontal force of $50\text{N}$ applied tangentially to the edge of the disc. The kinetic energy of the disc after $3\text{s}$ is

• A. $125\text{J}$
• B. $500\text{J}$
• C. $250\text{J}$
• D. $150\text{J}$

Answer 1

The correct option is C $250\text{J}$

Given, mass of the disc $m=90\text{kg}$, radius of disc $r=1.50\text{m}$, initial angular velocity ${\omega }_i=0$ and force applied $F=50\text{N}$
As we know that, $I_{\text{disc}}=\frac{1}{2}M{R}^2$
Thus, for the rigid object under a net torque model.
$\sum \tau =I\alpha \Rightarrow \alpha =\frac{\sum \tau }{I}=\frac{FR}{\frac{1}{2}M{R}^2}=\frac{2F}{MR}$
From the definition of rotational kinetic energy and the rigid object under constant angular acceleration model,
$K.E_f=\frac{1}{2}I{\omega }_f^2=\frac{1}{2}I({\omega }_i+\alpha t{)}^2$
$\Rightarrow K.E_f=\frac{1}{2}I{\alpha }^2{t}^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{2F}{MR}\right)}^2{t}^2=\frac{F^2{t}^2}{M}$
On substituting values, $K.E_{|@t=3s}=\frac{\left(50{)}^2\right(3{)}^2}{90}=250\text{J}$