Tuesday, October 12, 2010

Ch8 3 #1

This problem is like the conservation of energy problems in Ch6, but now there is also kinetic energy of rotation, equation 8.15

4 comments:

Anonymous said...

Can't figure this out. I used KE(1)+PE(1) = KE(2)+PE(2) and let PE(1)=0 and KE(2)=0.
I let PE=Mgh and KE=(1/2)*Mv^2 + (1/2)fMr^2 + (v^2)/(r^2).
So I have Mgh = (1/2)*Mv^2 + (1/2)fMr^2 + (v^2)/(r^2)

Because gravity, radius, and velocity are the same for all objects I canceled those terms in the equations. I used the correct moment of inertia (f) for all objects, but I still can't get the answer. Can anyone help?

Anonymous said...

Remember, when putting in the answer, if two have the same value, then you place one on top of the other. The moment of inertia for each object varies, and for example, a solid cylinder or disk has a moment of inertia as expressed by:
I=(1/2)MR^2, while the hoop retains I=MR^2. I hope this helps!

Anonymous said...

To the first commenter:

Your error is that you added the omega instead of multiplying it. The equation should be:

KE=(1/2)*Mv^2 + (1/2fMr^2) * (V^2)/(r^2).

So Mgh = (1/2)*Mv^2 + (1/2)fMr^2 * (v^2)/(r^2).

The terms M and r^2 should cancel out. You will be eventually left with the equation:

v^2 * (1+f) = 2gh.

Notice that v and g are same for all of the objects. Hope this helps.

Anonymous said...

This one really got me messed up and frustrated. The idea is that you get h=v^2(1+f)/2g. However, sinve v and g are the same, just find which objects have the highest f. The higher 1+f is, the higher the object will go. In order it is; hoop; hollow sphere; solid disk and solid cylinder stacked; and solid sphere.