Question

M, a solid cylinder (M=1.59 kg, R=0.111 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 0.870 kg
mass, i.e., F = 8.535 N. How far does m travel downward between
0.530 s and 0.730 s after the motion begins? 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 0.495 m in a time of 0.450 s. Find I_{cm}
of the new cylinder?

the angular acceleration of the cylinder is 96.7 rad/s^2

Answer #1

here,

the mass of cylinder , m1 = 1.59 kg

radius , R = 0.111 m

F = 8.535 N

m2 = 0.87 kg

the angular acceleration , alpha' = net torque /moment of inertia

alpha' = F * R /(0.5 * m1 * R^2 + m2 * R^2)

alpha' = 8.535 * 0.111 /(0.5 * 0.87 * 0.111^2 + 0.87 * 0.111^2) = 46.2 rad/s^2

acceleration , a = alpha' * R = 5.13 m/s^2

t1 = 0.53 s, t2 = 0.73 s

the distance traveled between t1 and t2 , s = (0 + 0.5 * a * t2^2) - (0 + 0.5 * a * t1^2)

s = 0.5 * 5.13 * ( 0.73^2 - 0.53^2) m

**s = 0.65 m**

when

s3 = 0.495 m

t3 = 0.45 s

let the acceleration be a3

using second equation of motion

s3 = 0 + 0.5 * a3 * t3^2

0.495 = 0 + 0.5 * a3 * 0.45^2

solving for a3

a3 = 4.89 m/s^2

let the moment of inertia of cylinder be I

the acceleration of system , a3 = Fnet /(m2 + I/R^2)

4.89 = 8.535 /(0.87 + I/0.111^2)

solving for I

**I = 1.08 * 10^-2 kg.m^2**

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