Rotational inertia - globes connected by a thin rod

In summary, the problem involves finding the rotational inertia of a rigid body consisting of two spheres and a thin rod connected in the middle. The spheres have radii of 0.18 m and 0.28 m, and masses of 193 kg and 726 kg, respectively. The rod has a mass of 10 kg and a length of 0.88 m. The body can rotate frictionlessly around the rotation axis in the middle of the rod. To calculate the overall inertia moment, the parallel axis theorem is used for the spheres while the standard formula for the MoI of a uniform stick is used for the rod. This results in an inertia moment of 476 kgm^2.
  • #1
naji0044
7
0
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 can't tell how the inertiamoment of the rod would look. Ignore the F(force)
Udklip.PNG
 
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  • #2
Hint: Don't treat the spheres as if they were point masses. (Think parallel axis theorem.)
 
  • #3
naji0044 said:
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
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
naji0044 said:
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
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
 
  • #7
naji0044 said:
m3*1
Where does this term come from?
The rest is fine.
 
  • #8
haruspex said:
Where does this term come from?
The rest is fine.
haruspex said:
honestly i am
Where does this term come from?
The rest is fine.
 
  • #9
I am not sure how to calculate the rod when the rotation axis is right on the middle of it
 
  • #10
naji0044 said:
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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FAQ: Rotational inertia - globes connected by a thin rod

1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to rotational motion. It depends on the mass of the object and how it is distributed around the axis of rotation.

2. How is rotational inertia different from linear inertia?

Linear inertia is an object's resistance to linear motion, while rotational inertia is an object's resistance to rotational motion. They are both affected by an object's mass, but rotational inertia also takes into account the distribution of that mass around the axis of rotation.

3. How does the connection between the globes and the thin rod affect their rotational inertia?

The connection between the globes and the thin rod does not affect their individual rotational inertia. However, when connected, the combined system will have a higher rotational inertia due to the added mass and distribution of that mass around the axis of rotation.

4. How is rotational inertia related to angular momentum?

Rotational inertia and angular momentum are directly proportional, meaning that an increase in rotational inertia will result in an increase in angular momentum. This relationship is described by the equation L = Iω, where L is angular momentum, I is rotational inertia, and ω is angular velocity.

5. How can the rotational inertia of the globes and rod system be calculated?

The rotational inertia of the globes and rod system can be calculated using the parallel axis theorem. This theorem states that the rotational inertia of an object rotating around an axis is equal to the sum of its individual rotational inertia and the product of its mass and the square of the distance between the axis of rotation and the object's center of mass. In this case, the globes and rod system can be treated as two separate objects with their own rotational inertia, and the parallel axis theorem can be used to calculate the combined rotational inertia of the system.

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