M, a solid cylinder (M=1.83 kg, R=0.123 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.83 kg, R=0.123 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.810 kg mass, i.e., F = 7.946 N. Calculate the angular acceleration of the cylinder.

Answer: 7.06×101 rad/s2

If instead of the force F an actual mass m = 0.810 kg is hung from the string, find the angular acceleration of the cylinder.

Answer: 3.75×101 rad/s2

How far does m travel downward between 0.590 s and 0.790 s after the motion begins?
Answer: 6.36×10-1 m

The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.568 m in a time of 0.570 s. Find Icm of the new cylinder.

I've done the last part 10 times but I still can't figure out what's wrong. Please Help.

Explanation / Answer

(part A)

T*r = I*

so, Tr = .5*m*r^2 *

so, = 2T/mr

also T = F=7.946N

so = 70.6 rad/s2



(part B)

if an actual mass is hung then by drawing the free body diagram of tthe mass, we get,

mg - T = ma

also T*r = I* = .5*M*r^2 *a/r

so, T =.5Ma

so, substituting the force equation,we get

mg - Ma/2 = ma

so , (m+M/2)a = mg

so, a= mg/(m+M/2) =4.606 m/s2

so, angular acceleration = a/r = 37.45 rad/s2



(part C)

a =4.606 m/s2 (obtained from above)

using the equation, s =ut +.5*at^2

for start from rest, u =0

so, s= .5*a*t^2

so, s1= .5*a*t1^2

and , s2 = .5*a*t2^2

so, s2-s1 = .5*a*(t2^2-t1^2) = 0.636m

part(D)
using the same equation

.568 = .5*a*.57^2
so, a = 3.496 m/s2



solving backwards, we get
Tr = Ia/r
so, T = Ia/r^2

substituting in the main force equation, we get,
mg - Ia/r^2 = ma
so, Ia/r^2 = m(g-a)

so, I = m(g-a)*r^2/a = 0.022 kg.m2

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