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M, a solid cylinder (M=2.15 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 0.730 kg mass, i.e., F = 7.161 N. If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.

Respuesta :

okpalawalter8

The angular acceleration of the cylinder is mathematically given as

[tex]\alpha = 60.2175 rad/s^2[/tex]

What is the angular acceleration of the cylinder.?

Question parameters

  • M=2.15kg
  • R = R=0m
  • W= 7.161 N

Generally, the equation for Torque generated in the cylinder is  mathematically given as

[tex]\sigma[/tex] = w×r

for circular motion torque is

[tex]\sigma = × \alpha[/tex]

t[tex]=0.5 × m ×r^2 × \alpha\\\\ =0.5 × 2.15 ×0.111 ^2 × \alpha[/tex]

t = 0.0132 ×[tex]\alpha[/tex]

Therefore

0.0132 × [tex]\alpha[/tex] = w × r

[tex]\\\\\alpha =\frac{ 7.161 0.111}{(0.0132)}[/tex]

[tex]\alpha = 60.2175 rad/s^2[/tex]

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