Question

A car moving at a speed of ${\text{36 kmph}}^{-1}$ is taking a turn on a circular road of radius $\text{50 m}$. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road. A small block of mass $\text{100 g}$ is kept on the seat which rests against the plate. Coefficient between the block and the plate is $\text{0.58}$. The plate is slowly turned so that the angle between the normal to the plate and the radius of the road is slowly increased. Find the angle at which the block will just start sliding on the plate.

• A. ${\text{30}}^{\circ }$
• B. ${\text{45}}^{\circ }$
• C. ${\text{60}}^{\circ }$
• D. ${\text{75}}^{\circ }$

Answer 1

The correct option is A ${\text{30}}^{\circ }$

Given: $v=$ Velocity of car $=\text{36 km/hr= 10 m/s}$;
$r=$ Radius of circular path $=\text{50 m}$;
$\text{m}=$ mass of small body $=\text{100 g= 0.1 kg}$;
$\mu =$ friction coefficient between plate and body $=\text{0.58}$

Suppose the normal to the plate makes an angle $\theta$ with the radius of circle. Then the components of the centripetal force will be provided by both normal reaction and friction.



The normal contact force exerted by the plate on the block
$N=\frac{m{v}^2}{r}\cos \theta$ .... (1)
and friction
$f=\mu N=\frac{m{v}^2}{r}\sin \theta$ ......(2)
(when the block just starts to slide)

Putting value of $N$ from (1) in (2)
$\mu \frac{m{v}^2}{r}\cos \theta =\frac{m{v}^2}{r}\sin \theta$
$\Rightarrow \mu =\tan \theta$
$\Rightarrow \theta ={\tan }^{-1}\mu$
$={\tan }^{-1}\text{0.58}={\text{30}}^{\circ }$
(as $\frac{1}{\sqrt{3}}=0.58$)