Question

A small block of mass $100g$ moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius $25cm$. If the block takes $2.0s$ to complete one ground, find the normal contact force by the side wall of the groove.

• A. $10N$
• B. $20N$
• C. $30N$
• D. None

Answer 1

The correct option is D None

The normal reaction by the side wall is responsible for providing the necessary centripetal force for circular motion.

Thus, $N_{wall}=m{\omega }^{2}r$

(where $m=0.1kg$ is the mass of the block, $\omega$ is the angular velocity and $r=0.25m$ is the radius of circular motion.)

$\omega =\frac{2\pi }{T}$, where $T$ is the time required to complete one revolution. We're given that $T=2s$.

$\therefore \omega =\pi rad/s$

$\therefore N_{wall}=\left(0.1\right)\left(\pi {)}^2\right(0.25)\approx (0.1\left)\right(10\left)\right(0.25)$

$\therefore N_{wall}\approx 0.25N\to$ option $\text{(D)}$

$\to QED.$