Question

A block of mass $m=20\text{kg}$ is kept at a distance $R=1\text{m}$ from the central axis of rotation of a round turn table. Table start from rest and rotates with constant angular acceleration $\alpha =3{\text{rad/s}}^2$. The frictional coefficient between block and table is ${\mu }_s=0.5$ and ${\mu }_k=0.4$. At time $t=x/30\text{s}$ from starting of motion, the block is just about to slip. The value of $x$ is _______.

Answer 1

The angular velocity of the block at any time is given by,

$\omega ={\omega }_0+\alpha t$

Where, ${\omega }_0$ is the initial angular velocity.

$\Rightarrow \omega =0+3t=3t...\left(i\right)$

Also,

$a_t=3{\text{rad/s}}^2$

$a_c={\omega }^{2}R=(3t{)}^{2}1=9t^2[\text{From}\left(i\right)]$

Now,
$a_{net}=\sqrt{(a_t{)}^2+(a_c{)}^2}=\sqrt{9+81t^4}$

From equilibrium of forces,

$m{a}_t=f$

$\Rightarrow m{a}_t={\mu }_{s}N$

$\Rightarrow 20\times \left(\sqrt{9+81t^4}\right)=0.5\times 20\times 10$

On solving this, we get,

$t=\frac{2}{3}\text{s}=\frac{20}{30}\text{s}=\frac{x}{30}\text{s}$

$\therefore x=20$