Rotational vs Translational Kinetic Energy of a Baseball

AI Thread Summary
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
Messages
1,192
Reaction score
1
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:
Physics news on Phys.org
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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top