Physical Pendulum Question for Moment of Inertia

In summary: So using the parallel axis theorem, I about P = Icenter + Md2 = (1/12)ML2 + M(L/2)2 = (1/12)ML2 + 1/4ML2 = (1/3)ML2No, what you are doing is adding (the moment of inertia of rod about its own center) + (moment of inertia of the sphere about P).You need to get the the moment of inertia of rod about P.For the rod, Icenter=(1/12)ML2and P is at L/2 from the center. So using the parallel axis theorem, I about P = Icenter + Md2 = (1/12)ML2 +
  • #1
AdanSpirus
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Homework Statement


Ok So here's my question.
The physical Pendulum consists of a thin rod of mass m = 100 g and length 80 cm, and a spherical bob of mass M = 500 g and radius R = 25 cm. There a pivot P at the top of the rod.
(Sorry, I don't have a picture >.<)
a) It asks for the center of mass(I already got it)

b) It asks for the moment of Inertia

c)Calculate Net Torque

d) Find Angular Freq. of small oscillations

e) At t= 0, the pendulum position is theta = (theta)max / 2, where (theta)max is the amplitude , and theta is increasing. When is the next instance where the particle will have a speed that is one third of its maximum?

Homework Equations


I = Icm + Md^2
Sphere = 2/5MR^2
Rod = 1/2MR^2

Rcm = (m*l + M(R+l)/(m+M)

s = Acos(wt + phi)
v = -wAsin(wt + phi)
a = -w^2Asin(wt+ phi)

The Attempt at a Solution



I have got the center of mass which results to be 0.605 m.
The only problem I have is that I am not sure about the moment of Inertia considering there are 2 objects and I am not exactly sure of how to set up the moment of inertia equation with the Sphere and the Rod. This is what is only bugging me atm, I probably can do the rest if I figure out the moment of inertia >.< But I might post back after if I don't get the rest of the question.
 
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  • #2
AdanSpirus said:

The Attempt at a Solution



I have got the center of mass which results to be 0.605 m.
The only problem I have is that I am not sure about the moment of Inertia considering there are 2 objects and I am not exactly sure of how to set up the moment of inertia equation with the Sphere and the Rod. This is what is only bugging me atm, I probably can do the rest if I figure out the moment of inertia >.< But I might post back after if I don't get the rest of the question.

The moment of inertia of the rod+bob about the pivot would just the sum of the moments of inertia of the rod and bob about the point P

AdanSpirus said:

Homework Equations


I = Icm + Md^2
Sphere = 2/5MR^2
Rod = 1/2MR^2

Rod should be (1/12)ML2

Since the sphere will not be rotating you can consider it as a point mass such that the distance of its center to P is (L+R).

moment of ineria of a point mass = mass*distance2
 
  • #3
rock.freak667 said:
Since the sphere will not be rotating you can consider it as a point mass such that the distance of its center to P is (L+R).

moment of ineria of a point mass = mass*distance2

So is the moment of Inertia then I = 1/12ml^2 + M(R+l)^2 ?
 
  • #4
AdanSpirus said:
So is the moment of Inertia then I = 1/12ml^2 + M(R+l)^2 ?

For the rod rotating about its own center the inertia is (1/12)ML2, so to get it about P, you need to use the parallel axis theorem.

The inertia for the bob about P is correct as M(R+L)2
 
  • #5
rock.freak667 said:
For the rod rotating about its own center the inertia is (1/12)ML2, so to get it about P, you need to use the parallel axis theorem.

The inertia for the bob about P is correct as M(R+L)2

I still don't understand >.<
Since the Parallel Axis Theorem states that its I = I(center of mass) + Md^2, so wouldn't my statement be correct?
I = (1/12)ml^2 + M(L+R)^2?
 
  • #6
AdanSpirus said:
I still don't understand >.<
Since the Parallel Axis Theorem states that its I = I(center of mass) + Md^2, so wouldn't my statement be correct?
I = (1/12)ml^2 + M(L+R)^2?

Ibob/SUB] (about P) = M(L+R)2

Irod(about its center)=(1/12)ML2

You need to apply the parallel axis theorem for the rod to get it about P.
 
  • #7
rock.freak667 said:
Ibob/SUB] (about P) = M(L+R)2

Irod(about its center)=(1/12)ML2

You need to apply the parallel axis theorem for the rod to get it about P.


Uhhmmmm so is that what I said? I = (1/12)ML^2 + M(L+R)^2?

Because the parallel axis theorem is the following equation:I = Icm + Md^2
 
  • #8
AdanSpirus said:
Uhhmmmm so is that what I said? I = (1/12)ML^2 + M(L+R)^2?

Because the parallel axis theorem is the following equation:I = Icm + Md^2

No, what you are doing is adding (the moment of inertia of rod about its own center) + (moment of inertia of the sphere about P).

You need to get the the moment of inertia of rod about P.

For the rod, Icenter=(1/12)ML2

and P is at L/2 from the center.
 

FAQ: Physical Pendulum Question for Moment of Inertia

1. What is a physical pendulum?

A physical pendulum is a rigid body that is free to rotate about a fixed axis. It consists of a mass suspended from a pivot point and is subject to the force of gravity. The motion of a physical pendulum can be described by its moment of inertia.

2. What is the moment of inertia of a physical pendulum?

The moment of inertia of a physical pendulum is a measure of its resistance to rotational motion. It is calculated by multiplying the mass of the pendulum by the square of its distance from the pivot point.

3. How is the moment of inertia of a physical pendulum determined?

The moment of inertia of a physical pendulum can be determined experimentally by measuring the period of oscillation and using it to calculate the moment of inertia. It can also be calculated using the pendulum's dimensions and the mass distribution of the pendulum.

4. Why is the moment of inertia important in a physical pendulum?

The moment of inertia is important in a physical pendulum because it determines the rate at which the pendulum will oscillate. A larger moment of inertia will result in a slower oscillation, while a smaller moment of inertia will result in a faster oscillation.

5. How does the moment of inertia affect the motion of a physical pendulum?

The moment of inertia affects the motion of a physical pendulum by determining its period of oscillation. A larger moment of inertia will result in a longer period of oscillation, while a smaller moment of inertia will result in a shorter period of oscillation. Additionally, a larger moment of inertia will require more energy to change the motion of the pendulum, while a smaller moment of inertia will require less energy.

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