Rotational Motion Question 12
Question 12 - 30 January - Shift 2
A uniform disc of mass $0.5 kg$ and radius $r$ is projected with velocity $18 m / s$ at $t=0 s$ on a rough horizontal surface. It starts off with a purely sliding motion at $t=0 s$. After $2 s$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2 s$ will be J
(given, coefficient of friction is 0.3 and $.g=10 m / s^{2})$.
Show Answer
Answer: (54)
Solution:
Formula: Centre of Mass
Rotational Motion
$a=-\mu_k g=-3$
$V=18-3 \times 2$
$V=12 m / s$
$KE=\frac{1}{2} mv^{2}+\frac{1}{2} \frac{mr^{2}}{2} \frac{v^{2}}{r^{2}}$
$KE=\frac{3}{4} mv^{2}$
$KE=3 \times 18=54 J$
