Hobbit__song
New member
Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed omega_0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.
1) Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U = Mgy_cm. . .cm=center of mass)
I found an expression for omega (final) but it was not right because the angular speed is supposed to be in terms of R, m, and M.
This was my answer:
omega(final) = sqrt[(omega_0)^2 +[(4mg(pi)R)/((M+2m)R^2)]]
Thank you SO SO SO much for your help!!! (please try to explain clearly or tell me what I'm missing
) )
1) Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U = Mgy_cm. . .cm=center of mass)
I found an expression for omega (final) but it was not right because the angular speed is supposed to be in terms of R, m, and M.
This was my answer:
omega(final) = sqrt[(omega_0)^2 +[(4mg(pi)R)/((M+2m)R^2)]]
Thank you SO SO SO much for your help!!! (please try to explain clearly or tell me what I'm missing
) )