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). The table starts from rest and rotates with constant angular acceleration $\alpha =3rad/s^2$. The friction coefficient between the block and the table is $\mu =0.5$. At time $t=\frac{x}{3}$ second from starting of motion $(i.e.,t=0)$, the block is just about to slip. Find the value of $x$.

Answer 1

Draw a FBD of given situation.



Find the value of $x$.

Formula used:
$F_t=mr\alpha ,F_c=\frac{m{\omega }^2}{r}$

Let angular speed is $\omega$ when slipping starts.

Then, for just to slide resultant of $mr\alpha$ and $mr{\omega }^2$ will be equal to $f$.

$f=\sqrt{(mr\alpha {)}^2+(mr{\omega }^2{)}^2}$

$\Rightarrow \mu mg=\sqrt{(mr\alpha {)}^2+(mr{\omega }^2{)}^2}$

$\Rightarrow (0.5\times 10{)}^2=(1\times 3{)}^2+(1\times {\omega }^2{)}^2$

$\Rightarrow 25-9={\omega }^4$

$\Rightarrow \omega =2$

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

$\Rightarrow 2=0+3\times t$

$\Rightarrow t=\frac{2}{3}=\frac{x}{3}$

$\Rightarrow x=2$

Final Answer : $x=2$