What is the translational speed of a bowling ball encountering a vertical rise?

[MKM]

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 5.64 m/s at the bottom of the rise. Find the translational speed at the top.
I have been working this one for the longest. I thought I was supposed to use the formula v= the square root of 2*gh but I think I am missing something I get 3.13 m/s however this is incorrect I belive.

 

[Fermat]

The gain in PE (mgh) is the difference between the KE (0.5mv²) at the bottom of the rise and the KE at the top of the rise.

Is that what you did ?

 

[Doc Al]

Use conservation of energy. (Don't forget rotational energy. Assume it rolls without slipping.)

 

[MKM]

I thought about using the kinematics but I am confused about where to find the mass. SHould I just cancel it out?

 

[Doc Al]

Try it and find out. (Maybe you don't need the mass. )

 

[MKM]

I think I might have to use the conservation of total mechanical energy and try and find w and I? does that sound right?

 

[Doc Al]

That's the idea.

 

[MKM]

Okay I keep pluging in my numbers for mgh=1/2mvsquared -1/2mvsqaured I get 4.11 and it's wrong am I usign the write formula I am confused about what to get rid of and what I don;t need . I know I don;t need the mass.

 

[Fermat]

As Doc Al pointed out, and I forgot, sorry, you should also include the difference in rotational energy as well.

 

[minhtrietle]

i'll give you somemore hints. Helps this can help:

K1 + U1 = K2 + U2

where K1 = (1/2)*I*w^2 + (1/2)*m*v^2
the same to K2

for the ball: I = (2/5)*m*r^2

You should choose OGPE at the bottom where the ball first starts (It's just easier ). So U2 = mgh.

Use conservation energy to solve this prob.

Minh T. Le

PS: I got the result for v final = 4.601 m/s. If this is not correct, don't follow my advice HAHAHA ... and sorry.

 

[radou]

And don't forget to use the relation between translational and rotational speed.

P.S. I got the same answer.