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

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To calculate the rotational kinetic energy of a cylinder rolling down a ramp, the moment of inertia formula I = 1/2 mr^2 should be used, as it applies to cylinders, not I = 2/5 mr^2, which is for spheres. The total mechanical energy conservation principle states that the potential energy (mgh) at the top of the ramp converts into both translational and rotational kinetic energy as the cylinder rolls down. The correct kinetic energy equation combines translational KE (1/2 mv^2) and rotational KE (1/2 Iω^2), where ω relates to v through the radius. After finding the velocity using energy conservation, substituting it into the kinetic energy equations will yield the correct rotational kinetic energy. Properly applying these formulas ensures accurate calculations for the cylinder's motion.
map7s
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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?
 
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map7s said:
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.)
 
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