Question
Sun February 17, 2013 By: Ayan Dastidar

Numerical question in Physics

Expert Reply
Wed May 29, 2013
The initial moment of inertia of the cylinder when it rotates along the axis passing through its center,
I1 = MR2/2           (where symbols represent usual meaning)
intial angular velocity, w1 = 125 rad/s
thus inital angular momentum of cylinder, L1 = I1.w1
 
Now after the cylinder starts rolling, the moment of inertia, I2 = 3MR2/2
 
and let the translational velocity of its center mass be V
ths the angular velocity of the cylinder as it rotates along its point of contact will be, w2 = V/R
thus the final angular momentum will be L2  = I2.w2
 
the frictional force on cylinder acting at the point of contact, Fr = Mg?
since at the time of pure rolling, there is no sliding motion at the point of contact, thus the frictional acceleration provided a that point will stop any motion and that point will be at rest relatively. Thus its initial velocity will be,
v = w1.R = 15 m/s , and
the retarding acceleration prvided by the friction will be, a = Fr/M
let time to start pure rolling to be T
thus using the first equation of motion,
0 = 15 - 1.5T
T = 10 sec
 
and the rate of change of initial and final angular momentum will be equal to the torque applied by the frictional force.
Thus (L2 - L1)/T = Fr . R
 
thus solving we would get, V = 5 m/s
 
and the loss energy can be calculated.
(rotational energy can be calculated as Iw2/2  )
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