Simple Harmonic Motion of a Rigid Body

In summary, the period of oscillation for a suspended meter stick with a light, rigid rod attached to one end is determined by the equation omega = (Mgd/I)^1/2. The "d" in the numerator represents the distance between the pivot and the center of mass of the rod. The moment of inertia of the rod can be determined using the parallel axis theorem, taking into account the additional mass of the rod (md^2) when calculating the moment of inertia about the pivot. In this problem, the correct answer was obtained by using the moment of inertia about the center of mass and setting d as the distance between the pivot and the center of mass.
  • #1
Cintdrix
4
0

Homework Statement



A very light, rigid rod with a length of 0.570 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Determine the period of oscillation.


Homework Equations



omega = (Mgd/I)^1/2
omega = 2pi/period.

The Attempt at a Solution



I solved this question, but I was just looking for clarification as to why it was right. The moment of inertia of the rod would be 1/3 ML^2 and using the parallel axis theorem I would have to add Md^2 on top of that.

The masses cancel in the first equation, but I was wondering about the "d" in the numerator. What does it represent? I originally thought it was the distance from the centre of mass to the pivot but it wasn't the case for this problem. Any insight is appreciated.

I when i subbed d=0.57 into the equation i got the correct answer of 2.15s, but I don't understand why.
 
Physics news on Phys.org
  • #2
Cintdrix said:

The Attempt at a Solution



I solved this question, but I was just looking for clarification as to why it was right. The moment of inertia of the rod would be 1/3 ML^2 and using the parallel axis theorem I would have to add Md^2 on top of that.

The parallel axes theorem says that the moment of inertia about distance d from the center of mass is given by

I = ICM + md2, where ICM = (1/12)mL2 for a uniform rod not (1/3)mL2. If you got the right answer, you must have made two errors that canceled each other.
 
  • #3
True. I was using the moment of inertia about one end of the stick, which is indeed (1/3)mL2. In my case i guess the D was still the distance from the old axis to the new axis. In using the moment of inertia about the CM, and making d the distance between the pivot and the CM, I was able to again get the correct answer.

Thanks for your help
 

Related to Simple Harmonic Motion of a Rigid Body

1. What is Simple Harmonic Motion of a Rigid Body?

Simple Harmonic Motion (SHM) of a rigid body is a type of motion where the object moves back and forth in a periodic manner, about a fixed equilibrium position. This type of motion is characterized by a restoring force that is directly proportional to the displacement from the equilibrium position, and acts in the opposite direction of the displacement.

2. What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x(t) = A*cos(ωt + φ), where A is the amplitude (maximum displacement from equilibrium), ω is the angular frequency (2π divided by the period of motion), and φ is the phase angle (related to the starting position of the object).

3. What is the difference between Simple Harmonic Motion and Uniform Circular Motion?

Simple Harmonic Motion involves a back and forth motion of an object, while Uniform Circular Motion involves a circular motion of an object at a constant speed. However, both types of motion can be described using similar equations and involve a restoring force that acts perpendicular to the direction of motion.

4. What is the significance of the restoring force in Simple Harmonic Motion?

The restoring force in Simple Harmonic Motion is responsible for bringing the object back to its equilibrium position whenever it is displaced from it. This force is directly proportional to the displacement and acts in the opposite direction, leading to the periodic motion of the object.

5. What are some real-life examples of Simple Harmonic Motion of a Rigid Body?

Some common examples of Simple Harmonic Motion of a Rigid Body include the motion of a pendulum, the motion of a mass-spring system, and the vibration of guitar strings. These systems exhibit SHM because they have a restoring force (gravity, spring force, tension) that is proportional to the displacement from equilibrium.

Similar threads

  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
958
  • Introductory Physics Homework Help
Replies
10
Views
314
Replies
13
Views
417
  • Introductory Physics Homework Help
Replies
16
Views
500
  • Introductory Physics Homework Help
Replies
6
Views
3K
Replies
21
Views
243
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top