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Finding the period

  • Thread starter zumi78878
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  • #1
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Homework Statement



The pendulum shown below right consists of a uniform disk with radius r = 20.0 cm and mass m = 1.2 kg attached to a uniform rod with length L = 40 cm and mass M = 0.8 kg. What is the period of the motion?

3508mqs.jpg


Homework Equations


T = 2∏√(I/mgL)


The Attempt at a Solution



I dont know what to plug into what.
 

Answers and Replies

  • #2
Doc Al
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To start with, you need to calculate the rotational inertia (I) of the system about the axis of rotation. How would you do that?

Also, in your formula for T what does L represent? You'll have to find that as well.
 
  • #3
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To start with, you need to calculate the rotational inertia (I) of the system about the axis of rotation. How would you do that?

Also, in your formula for T what does L represent? You'll have to find that as well.

T = 2∏√(I/mgL)


L represents the distance to the center of mass of the system?
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?


using parallel axis theorem:

I = 1/2MR2 + ML2
= (0.5)(1.2)(0.2)2 + 0.8(0.2+0.4)2
=0.312?
 
  • #4
Doc Al
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T = 2∏√(I/mgL)


L represents the distance to the center of mass of the system?
Right.
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?
What's the distance from the center of the disk to the axis?


using parallel axis theorem:

I = 1/2MR2 + ML2
= (0.5)(1.2)(0.2)2 + 0.8(0.2+0.4)2
=0.312?
Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
 
  • #5
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Right.

What's the distance from the center of the disk to the axis?

T = 2∏√(I/mgL)

L represents the distance to the center of mass of the system?
L = [(1.2*(0.4+0.6)/2)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.38m?

the distance from the center of the disk to the pivot is 0.6m.


Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
L = [(1.2*0.6)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.44m?


using parallel axis theorem:

I of disk = 1/2MR2 + ML2= 1/2(1.2)(0.2)2 + (1.2)(0.6)2= 0.456?
I of thin rod = 1/3ML2 = 1/3(0.8)(0.4)2= 0.0427?

I total = 0.4987?

I have an exam in 45 minutes so i really need to get this cleared up before that.
 
Last edited:
  • #6
Doc Al
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L = [(1.2*0.6)+(0.8*(0.4/2)]/(1.2+0.8)
= 0.44m?


using parallel axis theorem:

I of disk = 1/2MR2 + ML2= 1/2(1.2)(0.2)2 + (1.2)(0.6)2= 0.456?
I of thin rod = 1/3ML2 = 1/3(0.8)(0.4)2= 0.0427?

I total = 0.4987?
Looks good.
 
  • #7
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Right.

What's the distance from the center of the disk to the axis?



Things are a bit mixed up here. Find the rotational inertia for each piece separately, then add them up. To find the I of the disk about the axis, you'll need the parallel axis theorem. What's the I of a thin rod about one end?
Looks good.
wen i plug m into the formula, T = 2∏√(I/mgL)

do i use the mass of the entire system? so 1.2 + 0.8 = 2kg?
 
  • #8
Doc Al
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wen i plug m into the formula, T = 2∏√(I/mgL)

do i use the mass of the entire system? so 1.2 + 0.8 = 2kg?
Yes.

For more insight into the physical pendulum and how that equation comes about, review this: Physical Pendulum
 
  • #9
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Yes.

For more insight into the physical pendulum and how that equation comes about, review this: Physical Pendulum
thanks so much for ur help! really appreicate it. glad there are people like you out there.

but last question, can you explain why it has a negative sign in

Torque = -mgLsinθ?
 
  • #10
Doc Al
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but last question, can you explain why it has a negative sign in

Torque = -mgLsinθ?
Because the torque is in the direction of decreasing θ. (It's a restoring force, tending to bring the system back to the equilibrium point.)

Good luck on your exam!
 

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