# Disc rolling down hill

1. The problem statement, all variables and given known data
A solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 4.9 rad/s at the bottom, what is the height of the inclined plane?

mgh=1/2*I*W^2

## The Attempt at a Solution

i tried using conservation of energy

2.3*9.81*x=.5*(2.30*1.60^2)*4.99^2
x=3.19

but that is incorrect.

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1) Your formula for the inertia of a solid disk is incorrect.

2) What other energy is in your system that you have not yet accounted for?

1) Your formula for the inertia of a solid disk is incorrect.

2) What other energy is in your system that you have not yet accounted for?
ah your right about the inertia part, but im not sure what other energy? if the disc is at the bottom of the hill H=0 so mgh is not a factor, and i would use EITHER kinetic or rotational energy right? not both?

ah your right about the inertia part, but im not sure what other energy? if the disc is at the bottom of the hill H=0 so mgh is not a factor, and i would use EITHER kinetic or rotational energy right? not both?
Remember, your ball is rolling without slipping. So you do have to include both the rotational and kinetic.

The energy balance equation you wrote doesn't easily correspond to a physical situation, so imagine, instead, you wrote:

mgh = (1/2)mv^2

This would correspond to a ball sliding down the plane not slipping at all.

I suppose you could make a case for your energy balance equation being something like a disk sitting in an elevator going down with constant velocity, where the farther down it goes, the more your disk spins, in direct proportion to the lost gravitational potential.

But you definitely have to include both kinetic energies in your case.