Rotational vs Translational Kinetic Energy of a Baseball

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The discussion focuses on calculating the ratio of rotational kinetic energy to translational kinetic energy for a baseball, treated as a uniform sphere. Initial calculations using the moment of inertia as I = mr² yielded a ratio of 1:64. However, upon correcting the moment of inertia to I = (2/5)mr², the revised ratio became 1:160. Participants noted that the translational kinetic energy is significantly larger than the rotational energy, prompting questions about the underlying reasons for this disparity. The calculations and corrections were confirmed as accurate.
UrbanXrisis
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The center of mass of a pitched base ball (radius=3.8cm) moves at 38m/s. The ball spins about an axis through its center of mass with an angular speed of 125 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

0.5Iw^2:0.5mv^2
I=mr^2=m(.038m)^2=m0.001444
0.5m0.001444(125rad/s)^2:0.5m(39m/s)^2
22.5625:1444
1:64

is this correct?
 
Last edited:
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The moment of inertia for a uniform sphere is as follows:

I = \frac{2}{5}m\,r^{2}
 
thanks:

0.5Iw^2:0.5mv^2
I=(2/5)mr^2=m(2/5)(.038m)^2=m5.776E-4
0.5m(5.776E-4)(125rad/s)^2:0.5m(39m/s)^2
9.025:1444
1:160

is this correct?
 
Looks good to me..
 
why is the kinetic energy so much larger than the rotational energy?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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