Question

Two blocks tied with a massless string of length $3\text{m}$ are placed on a table rotating at an angular speed $\omega =4\text{rad/s}$. The axis of rotation is $1\text{m}$ from the block having $1\text{kg}$ mass. Assume that the surface below $2\text{kg}$ block is smooth and that below the $1\text{kg}$ block is rough. Find the tension $\left(T\right)$ in the string and the frictional force$\left(f\right)$ acting on the $1\text{kg}$ block. Take $g=10{\text{m/s}}^2$

• A. $T=64\text{N},f=48\text{N}$
• B. $T=48\text{N},f=64\text{N}$
• C. $T=8\text{N},f=4\text{N}$
• D. $T=4\text{N},f=8\text{N}$

Answer 1

The correct option is A $T=64\text{N},f=48\text{N}$

F.B.D. of $2\text{kg}$ block,


Centripetal force act towards the centre. So,
$T=\frac{m{v}^2}{R}=m{w}^{2}R$
$T=2\times 4^2\times 2=64\text{N}$
$T=64\text{N}$

F.B.D for $1\text{kg}$ block


Here, tension in the string is $64\text{N}$ and
Centripetal force is $=m{w}^{2}r$
$=1\times 4^2\times 1$
$=16\text{N}$, which is opposing the tension in the string.

So, the remaining $64-16=48\text{N}$ force is balanced by the frictional force.
So, $f=48\text{N}$