Question

M, a solid cylinder (M=2.23 kg, R=0.131 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. Calculate the angular acceleration of the cylinder. 5.84×101 rad/s^2

If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder.

How far does m travel downward between 0.510 s and 0.710 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 0.355 m in a time of 0.450 s. Find Icm of the new cylinder.

Answer 1

Torque = R × F

= 0.131 × 8.535

= 1.12 Nm

mg - T = ma ........................eq(1)

As, torque = R×T

Ia/R = R×T { Where I is moment of inertia of the cylinder and a is the acceleration of the block of mass m }

T = Ia/R²

Putting the value of T in eq (1)

mg - Ia/R² = ma

a = mg / {m + (I/R²)}

Putting ,

I = MR²/2 = 1.91×10^-2 kgm²

mg = 8.535 N

m = 0.87 kg

Acceleration of mass m is given by

a = 4.30 m/s²

Angular acceleration = a /R

Angular acceleration = 4.30 / 0.131 = 32.79 rad/s²

Distance travelled in time 0.51 s

S₁ = ut + (1/2)at²

= 0 + (1/2)×4.3×(0.51)²

= 0.559m

Distance travelled in time 0.71 s

S₂ = ut + (1/2)at²

= 0 + (1/2)×4.30×(0.71)²

= 1.084m

Distance travelled between 0.51s to 0.71s

S = S₂ - S₁

S = 1.084 - 0.559

S = 0.525m

To find new moment of inertia of the cylinder

S = ut + (1/2)at²

S = 0 + (1/2) × [ mg / { m + (I/R²)} ] × t²

On putting , S = 0.355 m

mg = 8.535 N

m = 0.87 kg

R = 0.131 m

t = 0.45 s

0.355 = (1/2)×[ 8.535 / { 0.87 + (I / R²)} ] × (0.45)²

I = 0.027 kgm²​​​​​​