Question

a) M, a solid cylinder (M=2.43 kg, R=0.137 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 0.710 kg mass, i.e., F = 6.965 N. Calculate the angular acceleration of the cylinder.

b)If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder.

c)How far does m travel downward between 0.690 s and 0.890 s after the motion begins?

d)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 0.386 m in a time of 0.490 s. Find I_cm of the new cylinder.

Answer 1

Moment of inertia for a solid disk = I = mr^2/2

A) Sum of the moments about the center of the disk = I*alpha = F*r (torque)

I = m*r^2/2

m*r^2/2 * alpha = F*r

m*r / 2 * alpha = F

alpha = angular acceleration = 2*6.965/(2.43*0.137)

alpha = 41.843 rad/s^2

B) mg-T = ma

Tr = I (alpha)

a = acceleration = alpha *r

alpha = mgr/mr^2 +I

I = 2.43*.137*.137*.5 = 0.0228 kg-m2

alpha = 0.71*9.8*0.137/(0.71*0.137*0.137+0.0228)

alpha = 26.387 rad/s2

C) a = 26.387*0.137 = 3.615 m/s2

velocity = 3.615*0.69 = 2.494 m/s

y = vt +0.5 a t^2 = 2.494* (0.89-0.69) +0.5 * 3.615 *0.2*0.2

y = 0.5711 m

D) y = 0.386 = 0.5 a *0.49*0.49

a = 3.2153 m/s2

alpha = mgr/mr^2 +I

3.2153/0.137 = 0.71*9.8*0.137/(0.71*0.137*0.137)+I

I = 0.02729 kg-m2