A disc, which has a mass of 12 kg and a radius of .65 m sits at the top of an in

Question

A disc, which has a mass of 12 kg and a radius of .65 m sits at the top of an inclined plane, which is 8.4 meters long and 1.5 meters high. At t=0 the disc is released and is allowed to roll to the bottom of the incline without slipping.

What will be the linear velocity of the disc when it reaches the bottom of the inclined plane? (Ans: 4.42)

What would the linear velocity be if the disc is replaced by a sphere? (Ans: 4.58)

What would the velocity be if the object was a ring? (Ans: 3.83)
I know the answers but how do you get there?

Explanation / Answer

We first need to use conservation of energy. The potential energy from being 1.5m high will turn into both translation and rotational kinetic energy

PE=KE+KE
mgh=(1/2)mv2+(1/2)I2

mgh=(1/2)mv2+(1/2)[(1/2)mr2]2 ((1))

gh=(1/2)v2+(1/4)v2

v2=(4/3)gh=(4/3)(9.81)(1.5)

v=4.429 m/s

We now do the same thing excepet the moment of inertia of a disk will be replaced by the moment of inertia of a sphere. I will start where I marked ((1)) above by changing that equation

mgh=(1/2)mv2+(1/2)[(2/5)mr2]2

gh=(1/2)v2+(1/5)v2

v2=(7/10)gh=(10/7)(9.81)(1.5)

v=4.585 m/s

We will once again do the same thing, but use the moment of inertia of a ring and start by chaning ((1))

mgh=(1/2)mv2+(1/2)[mr2]2

gh=(1/2)v2+(1/2)v2

v2=gh=(9.81)(1.5)

v=3.836 m/s

Hope that helps

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