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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Bowling on an incline

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