Question

The cable lifting an elevator is wrapped around a 1.0-m-diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 1.7 m/s . It then slows to a stop, while the cylinder turns one complete revolution.

How long does it take for the elevator to stop?

Express your answer to two significant figures and include the appropriate units.

Thanks for the help!

Answer 1

According to the question, the initial angular velocity of the cylinder is;

$\small \omega _{i}=\frac{v}{r}=\frac{v}{\frac{d}{2}}=\frac{1.7}{\frac{1.0}{2}}=3.4\: rad/s$

Final angular velocity of the cylinder is;

$\small \omega _{f}=0\: rad/s$

Angular displacement of the cylinder is;

$\small \theta =1\: rev=2\pi \: rad$

Thus the angular acceleration can be calculated as;

$\small \omega _{f}^{2}-\omega _{i}^{2}=2\alpha \theta$

$\small \Rightarrow \alpha =\frac{0-3.4}{2(2\pi )}=-0.92\: rad/s^{2}$

Therefore the time taken to come to stop can be calculated as;

   $\small t=\frac{\omega _{f}-\omega _{i}}{\alpha }=\frac{0-3.4}{-0.92}=3.7\: s$