Question

A block of mass $m=20kg$ is kept at a distance $R=1m$ from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, $\alpha =3rad/s^2.$ The friction coefficient between block and table is $\mu =0.5$. At time $t=\frac{x}{30}s$ from starting of motion ($t=0$), the block is just about to slip. Find the value of $x(g=10m/s^2)$.

• A. 20
• B. 30
• C. 10
• D. 40

Answer 1

The correct option is A 20



Table starts rotation from rest
$\therefore {\omega }_0=0$
$\omega =$ Angular velocity at time t
$\therefore \omega ={\omega }_0+\alpha t$
$\omega =0+3\times t=3t$
Block is just about to slip when friction force is limiting.

$\therefore f_2=\sqrt{(m\alpha R{)}^2+(m{\omega }^{2}R{)}^2}$
Or $(\mu mg{)}^2=m^2{\alpha }^2{R}^2+m^2{\omega }^4{R}^2$
$\therefore {\mu }^2{g}^2=({\alpha }^2{R}^2+{\omega }^4{R}^2$)
Putting values :
$\frac{1}{4}\times 100=9\times 1+(3t{)}^4$
$\therefore (3t{)}^4=16$
$3t=2$
$t=\frac{20}{30}s$
$\therefore x=2$