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Question
Sun February 17, 2013

# Numerical question in Physics

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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