Please forgive me if this is sloppy. And sorry for not posting this thread here in the first place (I didn't know about this section of the forums). Thanks in advance for any help you can provide.(adsbygoogle = window.adsbygoogle || []).push({});

A 2.0kg cylinder (radius=.1m and length=.5m) is released from rest at the top of a ramp, and is allowed to roll without slipping. The ramp is .75 m high and 5.0m long. When the cylinder reaches the bottom of the ramp, what is

(a) it's total KE (b) its rotational KE and (c) its translational KE?

Here's what I've got so far. Let me know if I'm off course.

I decided to use the Total Mechanical Energy Formula, and the conservation of energy:

E(total) = 1/2*m*v^2 + 1/2*I*ω^2 + m*g*h

E(total) = TranslationalKE + RotationalKE + GravitationalPE

Since the cylinder is at rest initially:

E(initial)=0+0+ 2.0kg*9.8*.75m ; E=14.7 J

From here I assume that at the end of the ramp, when the PE becomes 0, the total KE will equal 14.7 J as well. So there's (a) solved.

Next I plug in the knowns into the E(final) equation, and come up with:

14.7 J = (1/2)*2kg*v^2 + (1/2)*(1/2*2kg*.1^2)*ω^2 +0

which simplifies to:

14.7 J = V^2 + .005*ω^2

Here's my problem. First the equation for the E(final) seems off to me. And second, I'm not sure how to go about solving for ω^2 AND v^2 (I can't think of a third formula to throw in). Any advice? Thank You!

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# Homework Help: Rotational Kinematics and Energy Problem.

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