Rotational kinetic energy with no friction

[rfkstata]

Homework Statement

Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s^2. In the Earth tests, when m is set to 17.0 kg and allowed to fall through 6.00 m, it gives 200.0 J of kinetic energy to the drum.

a. If the system is operated on Mars, through what distance would the 17.0-kg mass have to fall to give the same amount of kinetic energy to the drum?
- I figured out this answer to be 15.8 m. I just can't seem to figure out this next part:

b. How fast would the 17.0-kg mass be moving on Mars just as the drum gained 200.0 J of kinetic energy?

Homework Equations

v=omega*r
Potential energy = Kinetic rotational + kinetic linear
mgh = rotational + translation kinetic energy

The Attempt at a Solution

Kinetic energy = potential energy
mgh= .200
For Mars, replace 9.8 with 3.71.

Can't seem how to put everything together, haven't been able to come up with the right answer according to mastering physics.

 

[gneill]

Homework Statement

Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s^2. In the Earth tests, when m is set to 17.0 kg and allowed to fall through 6.00 m, it gives 200.0 J of kinetic energy to the drum.

a. If the system is operated on Mars, through what distance would the 17.0-kg mass have to fall to give the same amount of kinetic energy to the drum?
- I figured out this answer to be 15.8 m. I just can't seem to figure out this next part:

b. How fast would the 17.0-kg mass be moving on Mars just as the drum gained 200.0 J of kinetic energy?

Homework Equations

v=omega*r
Potential energy = Kinetic rotational + kinetic linear
mgh = rotational + translation kinetic energy

The Attempt at a Solution

Kinetic energy = potential energy
mgh= .200
For Mars, replace 9.8 with 3.71.

Can't seem how to put everything together, haven't been able to come up with the right answer according to mastering physics.

Think about conservation of energy. Where is the energy coming from? If the drum ends up with 200 J of it, where did the rest go?