Question

A point mass m moves in a circle of radius R with speed v on a smooth horizontal table, with the help of a string whose other end passes through a hole. By pulling the string, the radius of circle is reduced to r/2 The final angular momentum of mass is​

Answer 1

Answer:

The angular momentum of mass about center will remain constant as the torque of tension is zero.

mv°

r =mv

2/r

v′ =2v°

Final value of K.E.= 2/1

m(2v°) 2=2mv°

Answer 2

There is reduction of radius of circle, but it will not effect the force acting on the body.

• As according the statement R be the radius and  v be the speed.
• Here in the statement the speed is  changing.
• It also describes that the torque acting on the body is constant.
• Which means the torque is Zero.
• That shows that the angular momentum at the center of the mass remains constant.
• The formula is
• mvR=mVR/2
• v=V/2
• 2v=V
• Using kinetic energy formula
• K.E=(1/2)m(2v)²
• K.E=2mv²