Pendulum & Torque HW: Calc Torque at 4.6 & 15.6 Degrees

  • Thread starter Thread starter kiwikahuna
  • Start date Start date
  • Tags Tags
    Pendulum Torque
Click For Summary

Homework Help Overview

The problem involves calculating the torque exerted by a simple pendulum at specific angles with respect to the vertical. The pendulum consists of a point mass and a string, and the torque is influenced by the gravitational force acting on the mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate use of the torque equation and the correct interpretation of the angle theta in relation to the vertical. There are inquiries about the derivation of the torque equation and the identification of vectors involved in the torque calculation.

Discussion Status

Participants are actively engaging with the concepts of torque and vector relationships. Some have provided insights into the cross product and the significance of the angle between vectors. There is an ongoing exploration of how to apply these concepts to the specific problem at hand.

Contextual Notes

There is some uncertainty regarding the definition of the angle used in the torque calculation, as well as the role of the tension in the string. Participants are also navigating their understanding of vector operations, particularly the cross product.

kiwikahuna
Messages
60
Reaction score
0

Homework Statement



A) A simple pendulum consits of a 2.1 kg point mass hanging at the end of a 4.3long light string that is connected to a pivot point.
a. Calculate the magnitude of the torque (due to the force of gravity) around this pivot point when the string makes a 4.6 degree angle with the vertical.
B) Repeat this calculation for an angle of 15.6 degrees.

Homework Equations


torque = mglsin(theta)


The Attempt at a Solution


I think the above equation is the right equation to use in this problem but I'm not sure what to plug in for theta. Is it the angle that's given or do I have to add 90 degrees to the given angle because the string makes the angle with the vertical? I have the same question for Part B as well. Thank you in advance for any help/advice.
 
Physics news on Phys.org
first draw a force diagram. next think about how one may derive the equation torque = mgl.sin(theta), ... ok... mg is a force (which is a vector). now what does mgl.sin(theta) looks like? yes.. the magnitude of a cross product of some sort eh... so which are your two vectors? I have already identify one for you.
remember in general torque is given by
[tex]\vec \tau= \vec R \times \vec F[/tex]
so what is your R in this case?

once you have identified your two vectors... then use your knowledge of the cross product which tells you that the angle (theta) is the (smaller) angle between the two vectors.
 
Last edited:
Would my R be the tension from the string [lsin(theta)]?

Could you explain to me more about what you mean when you say "the smaller angle?"

I'm not very familiar with the cross product although we did learn about the dot product.
 
kiwikahuna said:
Would my R be the tension from the string [lsin(theta)]?
[itex]\vec{R}[/itex] is the position vector of the pendulum bob; it is parallel to the string and has a magnitude equal to the length of the string. The angle you need is that between the position vector (the string) and the force (gravity, which is vertical). So the theta needed is the given angle that the string makes with the vertical.

Could you explain to me more about what you mean when you say "the smaller angle?"
Any two vectors (A & B) in a plane make some angle with respect to each other. You can describe the angle as A to B or B to A. One of those angles will be less than 180 degrees, the other greater. (But the sine of either angle will have the same magnitude.)
 
kiwikahuna said:
Would my R be the tension from the string [lsin(theta)]?

Could you explain to me more about what you mean when you say "the smaller angle?"

I'm not very familiar with the cross product although we did learn about the dot product.

if [tex]\vec A, \vec B[/tex] are two vectors, then the magnitude of the cross product between A and B is given by
[tex]|\vec A \times \vec B| = |\vec A||\vec B|\;\sin \;\theta[/tex]
where theta is the angle between the two vectors.

I was trying to point out to you where the sin theta actually comes from and hopefully from that you can deduce which angle to use.
 
Thank you to both Doc Al and mjsd. I think I finally got it!
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 22 ·
Replies
22
Views
4K