Physical Pendulum Question for Moment of Inertia

[AdanSpirus]

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.

 

[rock.freak667]

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

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

 

[AdanSpirus]

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 ?

 

[rock.freak667]

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

 

[AdanSpirus]

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?

 

[rock.freak667]

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?

I<sub>bob/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.</sub>

 

[AdanSpirus]

I<sub>bob/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.</sub>

<sub>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</sub>

 

[rock.freak667]

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 = I_cm + 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, I_center=(1/12)ML²

and P is at L/2 from the center.