Rotational inertia - globes connected by a thin rod

  • #1
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Homework Statement:
Finding the rotational inertia
Relevant Equations:
I=∑m*r^2
Problem Statement: Finding the rotational inertia
Relevant Equations: I=∑m*r^2

A rigid body of 2 massive globes with homogenous mass distribution and a thin rod is connecting the 2 globes. The globes has radius R1 = 0.18 m and R2 =0.28 m and masses m1=193 kg and m2=726 kg. The thin rod has the mass ms=10kg and the length L=0.88 m. THe body can rotate frictionless around the rotation axis in the middle of the thin rod.

a) Calculate the the body's overall inertiamoment and show that the inertiamoment is I=476 kgm^2


my calculations are m1*(1/2*L+r1)^2+m2*(1/2*L+r2)^2 and i doubt if this is even correct and cant tell how the inertiamoment of the rod would look. Ignore the F(force)
Udklip.PNG
 

Answers and Replies

  • #2
Doc Al
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Hint: Don't treat the spheres as if they were point masses. (Think parallel axis theorem.)
 
  • #3
haruspex
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cant tell how the inertiamoment of the rod would look
You should have a standard formula for the MoI of a uniform stick about its centre. Similarly for a uniform solid sphere.
 
  • #4
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okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?
 
  • #5
Doc Al
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okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?
Yes, about their respective centers.
 
  • #6
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Okay so you use the formula I=Icm+Md^2 and therefore:
2/5*m
1*r1^2+m1*(r1+1/2*L)^2 + 2/5*m2*r2^2+m2*(-r2-1/2*L)^2 + 1/12*m3*L+m3*1
 
  • #9
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I am not sure how to calculate the rod when the rotation axis is right on the middle of it
 
  • #10
haruspex
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I am not sure how to calculate the rod when the rotation axis is right on the middle of it
That's what the ##\frac 1{12}mL^2## formula gives you. There is nothing to add in this case.
 

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