Question

A 2.7-kg 12-cm-radius cylinder, initially at rest, is free to rotate about the axis of the cylinder. A rope of negligible mass is wrapped around it and pulled with a force of 18 N.

(a) Find the magnitude of the torque exerted by the rope.
  N · m

(b) Find the angular acceleration of the cylinder.
  rad/s²

(c) Find the angular velocity of the cylinder at t = 0.70 s.
rad/s

Answer 1

Torque about the axis of the cylinder is given by the distance from the center of the cylinder to the force (radius) multiplied by the force. Since answer needs to be in N.m distance needs to be in meter so

$\tau=r.F\rightarrow \tau=(0.12m)(18N)\rightarrow \mathbf{\tau=2.16Nm}$

Thats answer to part A

For part B we have that the angular acceleration is given by

$\alpha =\frac{\tau}{I}$

But we dont know the moment of inertia (I) which is given by:

$I =\frac{1}{2}m.r^2$

Substituting this on the first equation and solving for the answer:

$\alpha =\frac{2\tau}{m.r^2}\rightarrow \alpha =\frac{2(2.16Nm)}{(2.7Kg).(0.12m)^2}\rightarrow \mathbf{\alpha =111.11\frac{rad}{s^2}}$

Thats answer B

Finally for part C we have that the angular velocity at a certain time is given by:

$\omega =\omega _0+\alpha .t$

Since it was initially at rest we have that $\omega _0=0$ so we solve for the angular velocity

$\omega = (111.11\frac{rad}{s^2}) .(0.7s)\rightarrow\mathbf{ \omega =77.7\frac{rad}{s}}$

And thats answer C

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