Question

A $3-4-5$ (or $3cm\times 4cm\times 5cm$) inclined plane is fixed to a rotating turntable. A block B rests on the inclined plane and the coefficient of static friction between the inclined plane and the block is ${\mu }_s=\frac{1}{4}$. The block is to remain at a position $40cm$ from the centre C of rotation of the turntable.

The minimum angular velocity $\omega$ to keep the block from sliding down the plane (towards the centre) is ($g=10m/s^2$) (in rad/s):

• A. $3.2$
• B. $1.6$
• C. $6.4$
• D. $0.8$

Answer 1

The correct option is A $3.2$

The FBD is drawn as below.
$N$ is normal reaction
$f$ is static friction
$P$ is centrifugal force
For equilibrium
The equations of motion are:

mg sin $\theta$ = P cos $\theta$ + ${\mu }_s$ N
and, $N=mgcos$ $\theta$ + $Psin$ $\theta$

$\Rightarrow$ mg sin $\theta$ = P cos $\theta$ + ${\mu }_s$ mg cos $\theta$ +${\mu }_s$ P sin $\theta$

$\Rightarrow P=m{\omega }^{2}r=\left(\frac{sin\theta -{\mu }_{s}cos\theta }{cos\theta +{\mu }_{s}sin\theta }\right)mg$

$\Rightarrow {\omega }^2=\left(\frac{sin\theta -{\mu }_{s}cos\theta }{cos\theta +{\mu }_{s}sin\theta }\right)\frac{g}{r}$

=$\left(\frac{\frac{3}{5}-\frac{1}{4}.\frac{4}{5}}{\frac{4}{5}+\frac{1}{4}.\frac{3}{5}}\right)\times \frac{10}{0.4}=10.3$

$\Rightarrow \omega \simeq$ 3.2 rad/s