Question

A particle of mass 1 kg slides in a horizontal circle of radius 20 cm with a constant speed of 1 m/s. The only forces in the vertical direction acting on the particle are its weight and the normal reaction, however no information is available about the forces in the horizontal plane. The coefficient of friction is μ = 0.5. Then

• A. the magnitude of frictional force due to ground acting on particle must be 5 N.
• B. the frictional force due to ground must be in tangential direction.
• C. the frictional force due to ground must be towards the centre.
• D. no comment can be made about the direction or magnitude of friction force due to ground based on the given data.

Answer 1

Answer: (A), (C)

The magnitude of frictional force due to ground acting on particle must be 5 N. The frictional force due to ground must be towards the centre.

Answer 2

The particle is undergoing uniform circular motion, implying a centripetal acceleration towards the center. The centripetal acceleration ($a_c$) is given by $a_c = \frac{v^2}{r}$.

Given:

• Mass ($m$) = 1 kg
• Radius ($r$) = 20 cm = 0.2 m
• Speed ($v$) = 1 m/s

$a_c = \frac{(1 \text{ m/s})^2}{0.2 \text{ m}} = \frac{1}{0.2} \text{ m/s}^2 = 5 \text{ m/s}^2$

The centripetal force ($F_c$) required to maintain this motion is:

$F_c = m \cdot a_c = 1 \text{ kg} \cdot 5 \text{ m/s}^2 = 5 \text{ N}$

Since friction is the sole horizontal force, it provides the centripetal force. Therefore, the frictional force is 5 N and directed towards the center.