Centre of Mass 2 Question 9

10. A block of mass M with a semicircular track of radius R, rests on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A (see fig.). The cylinder slips on the semicircular frictionless track.

(a) How far has the block moved when the cylinder reaches the bottom (point B ) of the track ?

(b) How fast is the block moving when the cylinder reaches the bottom of the track?

(1983,7 M)

Show Answer

Answer:

Correct Answer: 10. m(Rr)M+m,m2g(Rr)M(M+m)

Solution:

  1. (a) The centre of mass of M+m in this case will not move in horizontal direction. Let M moves towards left by a distance x then m will move towards right by a distance Rrx (with respect to ground). For centre of mass not to move along horizontal we should have

Mx=m(Rrx),x=m(Rr)M+m

(b) Let v1 be the speed of m towards right and v2 the speed of M towards left. From conservation of linear momentum.

mv1=Mv2(i)

From conservation of mechanical energy

mg(Rr)=12mv12+12Mv22

Solving these two equations, we get

v2=m2g(Rr)M(M+m)



NCERT Chapter Video Solution

Dual Pane