Question

A 50 kg cylinder with diameter of ,12 m has a cable wrapped around it with a force F of 9 newtons applied to the cable so that a point on the horizontal part of the cable accelerates to the left at 0.60 m/s2. What is the magnitude of the angular acceleration of the cylinder? What is the magnitude of the torque that the cable exerts on the cylinder? What is the magnitude of the force that you exert on the cable?

Answer #1

What is the magnitude of the torque that the cable exerts on the cylinder

angular acceleration, = a / r where r is the radius of cylinder = 0.12 / 2 = 0.06 m

so.

= 0.60 / 0.06

= 10 rad/sec^{2}

Now,

= I

= 1/2 * m * r^{2} *

= 1/2 * 50 * 0.06^{2} * 10

= 0.9 N.m

__________________________________

What is the magnitude of the force that you exert on the cable?

Force = / r

Force = 0.9 / 0.06

Force = 15 N

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