Center of mass of physical pendulum

In summary: Then, use the distance formula to find the distance between the pivot point and the center of mass. Finally, you can use the period formula to calculate the period of oscillation. In summary, to solve this problem, you need to use the parallel axis theorem, the center of mass formula, and the distance formula to find the rotational inertia, center of mass, and distance between the pivot point and center of mass, respectively. Then, you can use the period formula to calculate the period of oscillation.
  • #1
burianek
3
0

Homework Statement


A pendulum consists of a uniform disk with radius r=0.100m and mass 0.500 kg attached to the end of a uniform rod with length L=0.500 m and m 0.250 kg. It pivots at the other end of the rod. a) Calculate the rotational inertia of the pendulum about the pivot point. b) What is the distance between the pivot point and the center of mass of the pendulum? c)Calculate the period of oscillation


Homework Equations


I=Icom+mh^2
T=2pi(sqrt(I/(mgh)))


The Attempt at a Solution



I got the first part using the parallel axis theorum for both the disk and the rod and adding them together. I=0.205 kg*m^2 (checked, correct).
I can't figure out how to get the distance between the pivot point without having the period. I tried setting I=mL^2 and solving for L, but this was incorrect. The correct answer is 0.477 m. I can get part c) once I have some clues to b. I'm completely lost. Please help!
 
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  • #2
So I take it no one else can figure this one out either?
 
  • #3
burianek said:

Homework Statement


A pendulum consists of a uniform disk with radius r=0.100m and mass 0.500 kg attached to the end of a uniform rod with length L=0.500 m and m 0.250 kg. It pivots at the other end of the rod. a) Calculate the rotational inertia of the pendulum about the pivot point. b) What is the distance between the pivot point and the center of mass of the pendulum? c)Calculate the period of oscillation


Homework Equations


I=Icom+mh^2
T=2pi(sqrt(I/(mgh)))


The Attempt at a Solution



I got the first part using the parallel axis theorum for both the disk and the rod and adding them together. I=0.205 kg*m^2 (checked, correct).
I can't figure out how to get the distance between the pivot point without having the period. I tried setting I=mL^2 and solving for L, but this was incorrect. The correct answer is 0.477 m. I can get part c) once I have some clues to b. I'm completely lost. Please help!

Where is the center of mass of the disk? Where is it for the rod? Once you have those two, you can use the standard center of mass formula to find the center of mass of the entire object.
 

What is the definition of center of mass?

The center of mass of an object is the point where the entire mass of the object can be considered to be concentrated, and through which all external forces act as if the object were a point mass.

How is the center of mass of a physical pendulum different from other objects?

A physical pendulum is a rigid body that oscillates around an axis of rotation, as opposed to a simple pendulum which is a point mass attached to a string. The center of mass of a physical pendulum is not at the pivot point, but rather at a point along the pendulum's length where the mass is evenly distributed.

How is the center of mass of a physical pendulum calculated?

The center of mass of a physical pendulum can be calculated using the formula: x = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where m is the mass at a given point and x is the distance from the pivot point.

Why is the center of mass important in understanding the motion of a physical pendulum?

The center of mass is important because it is the point about which the pendulum swings and experiences rotational motion. It is also used to calculate the pendulum's period and frequency.

How does the center of mass affect the stability of a physical pendulum?

The lower the center of mass, the more stable the pendulum will be, as it will require a greater force to cause it to topple over. This is why objects with a low center of mass, such as a pyramid, are difficult to tip over.

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