Question

A ramp has an angle of θ = 24° above the horizontal. A uniform, solid cylinder...

A ramp has an angle of θ = 24° above the horizontal. A uniform, solid cylinder with mass 1.3 kg and radius 3.7 cm is placed on the ramp.
What is the torque that will rotate the cylinder when it is released?
  τ =  N·m

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Answer #1

Free body diagram

R is radius of cylinder. mg is weight of cylinder. N is normal reaction.
For cylinder to start rolling, point of contact acts as axis of rotation.
Now torque = Force * perpendicular distantance
Here only weight force contributes as it is not passing through the point of contact unlike normal reaction.
small r is perpendicular distance of wieght from point of contact

r= R*sin(24)=3.7*0.4067=1.5049 cm = 0.015049 m

Therefore torque = mg * r = 1.3*9.8 * 0.015049 = 0.191728 Nm

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