How Is Rotational Kinetic Energy Calculated for a Cylinder Rolling Down a Ramp?

[map7s]

A 2.4 kg cylinder (radius = 0.09 m, length = 0.50 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.74 m high and 5.0 m long. What is its rotational kinetic energy?

I just wanted to make sure that I was on the right track...

I=2/5 mr^2
KE=1/2 mv^2 + 1/5 mv^2

Using conservation of energy: mgh=1/2 mv^2 + 1/5 mv^2 and then solve for v and plug into equation for KE.

Does this sound like I'm doing this correctly?

 

[Doc Al]

I just wanted to make sure that I was on the right track...

I=2/5 mr^2
KE=1/2 mv^2 + 1/5 mv^2

You are on the right track, but you are using the wrong formula for rotational inertia. (2/5 mr^2 is for a rolling ball, not a cylinder.)

 

[kyle8921]

Yes, your approach is correct. To find the rotational kinetic energy, you need to use the formula KE = 1/2 Iω^2, where I is the moment of inertia (in this case, 2/5 mr^2 for a solid cylinder) and ω is the angular velocity. You can find the angular velocity by using the conservation of energy equation, as you mentioned. Once you have the angular velocity, you can plug it into the rotational kinetic energy formula to get the final answer. Keep up the good work!