Rotational motion and an incline

[Strontium90]

Hello, I have a problem that I have not been able to solve

Homework Statement

Here is the problem question:

A bowling ball rolling at 8 m/s begins to move up an inclined plane. What height does it reach?

Homework Equations

The equation that I used was the formula relating potential energy to kinetic energy. In the book it showed how the translational equivalent of kinetic energy related to angular kinetic energy. The equivalent of kinetic angular energy is (3/4)mv^2 while potential energy is mgh, with m is mass, g is gravity and h is height.

The Attempt at a Solution

take:

mgh = (3/4)mv^2

h = (3/4)(v^2/g)


The value for h I got is 4.90m

 

[Strontium90]

Oh and the books answer is 4.57m.

 

[barryj]

I think that (3/4)mv^2 is incorrect. Figure it out..
KE = KE(tranwslational) + KE(rotational)

KE(translational is (1/2)mv^2 , now go figure the KE(rotational) using the moment of inertia of a solid sphere.

 

[Strontium90]

So the Kinetic energy for any rotating object depends on it's Moment of Inertia? Taking the moment of inertia of a solid sphere and combining it with the value for translational KE, I get (9/10)MV^2 for the KE of a rolling bowling ball and soling for h in mgh = (9/10)mv^2, the value is 5.88m.

 

[consciousness]

The moment of inertia of a solid sphere about its COM is 0.4MR²

In general the kinetic energy of a body is nothing but the summation of the kinetic energy of its constituent particles.
This sum can be broken into two parts for extended bodies-
(i)The kinetic energy of the COM
(ii)The rotational kinetic energy about the COM
This energy depends on the work done by all the forces on the body and its initial kinetic energy.
(Work Energy Theorem)

 

[barryj]

I get .7MR^2 for the combined KE of the rolling sphere.
I for a sphere is .4mR^2

 

[haruspex]

So the Kinetic energy for any rotating object depends on it's Moment of Inertia? Taking the moment of inertia of a solid sphere and combining it with the value for translational KE, I get (9/10)MV^2 for the KE of a rolling bowling ball and soling for h in mgh = (9/10)mv^2, the value is 5.88m.

I'm guessing you forgot the 1/2 in (1/2)Iω².