# Rotational energy

1. Nov 17, 2013

### MathewsMD

60. A ship’s anchor weighs 5000 N. Its cable passes over a roller of negligible mass and is wound around a hollow cylindrical drum of mass 380 kg and radius 1.1 m, mounted on a frictionless axle. The anchor is released and drops 16 m to the water. Use energy considerations to determine the drum’s rotation rate when the an- chor hits the water. Neglect the cable’s mass.

Solution:

Under "Evaluate" they say the kinetic energy of the anchor is 1/2m(ωR)2 but why do we assume R to be the radius of the cylinder and why do we consider the angular velocity of both the cylinder and anchor to be the same? Why can't the angular velocities be different? I may be visualizing it incorrectly, but I am slightly confused on why we assume R and ω to be the same for both the anchor and cylinder when they are not even in contact anymore.

2. Nov 17, 2013

### tiny-tim

Hi MathewsMD!
no, they aren't interested in the angular velocity of the anchor, only its actual velocity, v …

if you think about it, v must equal ωr whatever the angle that the cable makes!

3. Nov 18, 2013

### MathewsMD

Hmmm...Well r would be different, wouldn't it? I can kind of understand how ω would be the same, but wouldn't it change as it is falling? If anyone could provide a more in-depth explanation that would be great!

4. Nov 18, 2013

### Staff: Mentor

The tangential speed of the rotating cylinder is ωr, where ω is the angular velocity of the cylinder and r is the cylinder radius.

The cable meets the cylinder tangentially. Thus the cable speed is equal to the tangential speed of the cylinder.

This is an important relationship that will come up quite often where you need to relate a linear speed or acceleration to a rotational speed or rotation.

5. Nov 18, 2013

### tiny-tim

yup …

forget about ω changing, it's only v (of the anchor) that we're interested in …

and v of the anchor = v of the rim