Question

A cube of mass M starts at rest from point 1 at a height 4R, where R is the radius of the circular track. The cube slides down the frictionless track and around the loop. The force which the track exerts on the cube at point 2 is:

Answer 1

A cube of mass M starts at rest from point -1 at a height 4R.

speed of cube by which it reaches the bottom point , v

initial speed of cube, u = 0

so, using formula, v² = u² + 2as

or, v² = 0 + 2(-g)(-4R) = 8gR

hence, v = √(8gR)...........(1)

now, cube moves in circular loop of radius R and it has to reach the topmost point-2 as shown in figure.

Let speed of cube at topmost point 2 is v'

displacement covered by cube from bottom to topmost point 2 = 2R

so, using formula , v² = u² + 2aS

here, v = v' , u = √(8gR) [ from equation (1), ] , S = 2R and a = -g

so, v'² = 8gR +2(-g)2R = 4gR

hence, v' = √(4gR).........(2)

now, net force acts on point-2 , F = mv'²/R - mg

= m(4gR)/R - mg = 3mg

hence, force which the track exerts on the cube at point 2 is 3mg.

Answer 2

Answer:

Hope this helps you....