Question

A circle disc is rotating is horizontal plane about its own axis at the rate of 210 rpm. A coin of mass 10 gm is placed on the disc at distance 20 cm from the axis. If the coin doses not skid, the frictional force between the coin and the disc is ,

• A. 0.967 N
• B. 0.484N
• C. 0.844N
• D. 1.26N

Answer 1

The correct option is A 0.967 N

The disc rotates by a frequency of $v=210rpm$

which means,

$v=\frac{210}{60}=3.5Hz\Rightarrow w=2\pi v$ (Angular frequency )

$=2\times 3.5\times \pi$

$=7\pi rad/s$

When a coin of man $m=0.0LKg$ in placed at a radial distance $r=0.2m$ experiences a centripetal force :

$F_e=mr{w}^2$

$=0.01\times 0.2\times (\pi {)}^2$

$\Rightarrow F_e=0.967N$.

As the coin does't said. It means the centripetal force $\left(F_e\right)$ in counteracted by friction $\left(f\right)$. so friction :

$f=F_e=0.967N$