How Do You Calculate the Angular Momentum of a Rotating Bat?

AI Thread Summary
To calculate the angular momentum of a rotating bat, the moment of inertia must be correctly determined. The bat should be approximated as a long thin cylinder rather than a point mass at the end. The correct formula for moment of inertia is I = (1/3)mr^2 for a rod rotating about one end. Using this, the angular momentum can be calculated with L = Iω, where ω is the angular velocity. Properly applying these formulas leads to the correct solution for the problem.
Exustior
Messages
2
Reaction score
0

Homework Statement


A batter swings a 0.9 m long bat. Calculate the angular momentum of the bat relative to its axis of rotation. Assume the bat has a mass of 0.6 kg and rotates at 3 rad / s
(Hint: Approximate the bat by a rod, or a long thin cylinder, and find the moment of inertia.)
Answer in (N m)

Homework Equations


I=mr^2
ω=v/r
v=ωr
L=Iω

The Attempt at a Solution


L=Iω
I=mr^2
so L=(0.6)(.9^2)(ω)
ω=angular velocity is 3 rad/s
so L=(.6)(.9^2)(3rad/s)= Wrong answer
 
Last edited:
Physics news on Phys.org
Hey, welcome to physicsforums!
uh, the moment of inertia formula you have used is not correct for this problem. The inertia I=(0.6)(0.9)^2 would suggest the mass of the bat is concentrated at the end of the bat. But it looks like they want you to assume the bat is a long thin cylinder, of uniform density. So either you can look up the moment of inertia for this kind of object, or derive it yourself, if you have time.
 
BruceW said:
Hey, welcome to physicsforums!
uh, the moment of inertia formula you have used is not correct for this problem. The inertia I=(0.6)(0.9)^2 would suggest the mass of the bat is concentrated at the end of the bat. But it looks like they want you to assume the bat is a long thin cylinder, of uniform density. So either you can look up the moment of inertia for this kind of object, or derive it yourself, if you have time.
Thanks!
I got it I was using the wrong formula.
 
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 '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