Question

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

Answer 1

Since the table starts rotation from rest
${\omega }_{\text{initial}}=0$
$\omega =$ Angular velocity at time $t$
$\omega ={\omega }_{\text{initial}}+\alpha t$
$=0+3\times t=3t$

Force diagram for block as seen from above is



where $m{\omega }^{2}r$ and $m\alpha r$ are the centripetal and tangential accelerations of the block.

When the block is just about to slip, the friction $f$ should balance both these forces. Thus,

$f=\sqrt{(m{\omega }^{2}r{)}^2+(m\alpha r{)}^2}$
$(\mu mg{)}^2=m^2{\omega }^4{r}^2+m^2{\alpha }^2{r}^2$
${\mu }^2{g}^2={\omega }^4{r}^2+{\alpha }^2{r}^2$

On putting values.

$\frac{1}{4}\times 100={\omega }^4\times 1+3^2\times 1$

$16={\omega }^4$
$\Rightarrow \omega =2rad/s=3t$
$\Rightarrow t=\frac{2}{3}=\frac{20}{30}$