Bowling on an incline

  • #1
The Flintstones and Rubbles decide to try out the new inclined bowling alley, ``Bedslant Bowling''. Fred's ball and Barney's ball have the same size, but Barney's ball is hollow. Betty's ball and Wilma's ball are scaled down versions of Fred's ball and Barney's ball respectively. They all place their bowling balls on the same pitch incline and release them from rest at the same time. For the questions below the choices are greater than, less than, or equal to. I put my answer beside them.

The time it takes for Betty's ball to hit the pins is .... that for Barney's ball to hit. *less than*
The time it takes for Wilma's ball to hit the pins is .... that for Fred's ball to hit. *greater than*
The time it takes for Barney's ball to hit the pins is .... that for Wilma's ball to hit. *greater than*
The time it takes for Betty's ball to hit the pins is .... that for Wilma's ball to hit. *less than*
The time it takes for Betty's ball to hit the pins is .... that for Fred's ball to hit. *equal*
The time it takes for Fred's ball to hit the pins is .... that for Barney's ball to hit *less than*

Here's what I did to solve the problem. First I found the moment of inertia for all of the balls. I used masses of 2 kg for the bigger ones, and 1 kg for the smaller ones. The radiuses I used were also 2 and 1 m for bigger & smaller balls.
Fred- (2/5)MR^2 = (2/5)(2)(2^2)= 3.2 kgm^2
Betty- (2/5)MR^2= (2/5)(1)(1^2)= .4 kgm^2
Barney- MR^2= (2)(2^2)= 8 kgm^2
Wilma- MR^2= (1)(1^2)= 1 kgm^2

I then used conservation of energy to find the velocity
[tex] mgH= (1/2)mv^2 + (1/2)I \omega^2 [/tex]
Solving for v gave me: [tex] \sqrt 2mgh/ m+ (I/R^2) [/tex]
I then plugged in my moments of inertia to get the velocities for the balls.
Fred's= 11.8 m/s
Betty= 11.8 m/s
Barney= 8.08 m/s
Wilma= 9.90 m/s
Then I compared the velocities to see which ball would get to the bottom faster, but not all of them are right. Can someone help me? Thanks.
 

Answers and Replies

  • #2
Integral
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You are going to numbers to fast. Do not compute a value for I, you do not have either radius or mass, perhaps the numbers you have picked are a special case. You need to keep your expression for the velocity general.

Hint: Velocity can be expressed in terms of M alone.
 
  • #3
the only way I can think of to express velocity with M is with kinetic energy. Is that the right approach?
 
  • #4
Integral
Staff Emeritus
Science Advisor
Gold Member
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56
Your approach is fine, just retain the variables and leave out the specific numbers, they do not help and may mislead.
 
  • #5
I got it. You're right it's easier with just variables. Thank you!
 

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