Question

a small block of mass 100 gram with move with uniform speed in a horizontal circular groove with a vertical side wall of radius 25 cm is blocked take 2 seconds to complete one round find the contact force by the side wall of Gurgaon

Answer 1

v=2π(25)/2
v=0.785m/s

a=v^2/r
a=0.785/0.25
a=2.5m/s^2

n=ma
n=0.100×2.5
n=0.25N

Answer 2

Speed of the block of mass. v = \frac{2\Pi(25 cm) }{2 s} = 0.785 metre/sec.

Acceleration of the block a = \frac{v^{2}}{r} = \frac{0.785}{0.25} = 2.5 m/s^{2}.

When moving towards the centre only a single force is present in this direction and it is the normal contact force.

Hence according to Newton's second law, N = ma = (0.100)(2.5 m/s^{2}) = 0.25 N