Q46 / 50 · Manipal (MET) 2023 · Physics
A disc of mass 5 kg and radius 50 cm rolls on the ground at the rate of $10 \mathrm{~ms}^{-1}$. Find the kinetic energy of the disc in J .
Correct answer: 375
Solution
Here, mass of the disc, $M=5 \mathrm{~kg}$
Radius of the disc, $R=50 \mathrm{~cm}=1 / 2 \mathrm{~m}$ Linear velocity of the disc, $v=10 \mathrm{~ms}^{-1}$
As, $\quad v=R \omega$
$\therefore \quad 10=\frac{1}{2} \omega$ or $\omega=10 \times 2=20 \mathrm{rad} \mathrm{s}^{-1}$
Also, moment of inertia of disc about an axis through its centre,
$I=\frac{1}{2} M R^2=\frac{1}{2} \times 5 \times\left(\frac{1}{2}\right)^2=\frac{5}{8} \mathrm{~kg}^2-\mathrm{m}$
$\begin{aligned} & \therefore \mathrm{KE} \text { of translation }=\frac{1}{2} m v^2=\frac{1}{2} \times 5 \times(10)^2=250 \mathrm{~J} \\ & \text { Rotational kinetic energy }=\frac{1}{2} / \omega^2=\frac{1}{2}\left(\frac{5}{8}\right)(20)^2=125 \mathrm{~J}\end{aligned}$
$\begin{aligned} \text { Total kinetic energy } & =\text { Translational }+ \text { Rotational } \\ & =(250+125)=375 \mathrm{~J}\end{aligned}$
Hence, the correct answer is 375.