M , a solid cylinder (M= 1.75kg ,R=0.135 m ) pivots on a thin fixed frictionless

Question

M , a solid cylinder (M= 1.75kg ,R=0.135 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 .870 kg mass , I.e , F=8.535N . Calculate the angular acceleration of the cylinder.



M , a solid cylinder (M= 1.75kg ,R=0.135 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 .870 kg mass , I.e , F=8.535N . Calculate the angular acceleration of the cylinder.

Explanation / Answer

Here ,

for the cyclinder ,

moment of inertia , I = 0.5 * m *r^2

Now, let the angular acceleration is a ,

Using second law of motion ,

Torque = I * a

F*r = 0.5 * m *r^2 * a

8.535 = 0.5 * 1.75 * 0.135 * a

a = 72.25 rad/s^2

the angular acceleration is 72.25 rad/s^2

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