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A solid disk of 10 kg ..?
Yes! A solid disk of 10 kg and a radius R=1m, 12.5 cm wide starts rolling down a 10 degree incline when it is 1 m above the base of the incline. Assuming no slippage compute:
(a) Translational velocity when it gets to the bottom of the incline?
(b) It’s potential energy?
(c) It’s total energy?
(d) How old is the observer?
1 Answer
- Anonymous1 decade agoFavorite Answer
CONSERVATION OF ENERGY
initial PE = mgh
=
final KE (translational and rotational)
= 1/2mv^2 + 1/2 I (omega)^2
omega is just v*r
Your book should have a table that tells you the moment of inertia of the disk in terms of m and r.
So they give you m. You know g. They give you h. You can express I and omega in terms of m,r, and v. I don't see that the width of the disk is relevant.
The only thing you don't know in that equation is v. Solve for it.
PE = mgh at the top
That is also the total energy--it's all you had up at the top.
The observer is ancient.