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# A particle (P) of mass is released from rest at a point A on a curved smooth wedge of mass M,free to move on smooth horizontal surface.Velocity of wedge w.r.t ground is V1 and velocity of particle w.r.t wedge is V2 at,when particle is point B.Momentum conservation equation in horizontal direction can be writren as MV1+m(?)=0.Find (?)

Asked by napaliwal 18th March 2018, 12:36 PM
Answered by Expert
Answer:
There are two questions in this post.
First question is not clear because figure is not provided. it is not clear how points A and B are located.

Answer to Second Question:-
When the cylinder rolls down with acceleration a due to the net force of its own weigth mg and the tension in the string .

m×a = m×g - T ...................(1)

Torque due to Tension can be written as τ = T×R = I α =  ( mR2/2 ) (a/R)  .........(2)
in the abobe equation we have used the relation between linear acceleration and angular acceleration :  a = R×α

from (2) we get T = (m×a) /2 ............(3)

from (1) and (3) , we get   a = (2/3) g

Answered by Expert 22nd March 2018, 4:10 PM
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