Torque calculation for a compound pendulum

In summary, @TSny is saying that the force on the bar due to the springs will have a vertical component and a component in the direction of the angle between the bar and the springs.
  • #1
charon_9
7
0
Homework Statement
Newton's second law on a compound pendulum
Relevant Equations
I*α = ΣM, F=-kx, G=mg
sss.png


I need to write an equation for Newton's second law for the above system, where k1=k2 (both springs are the same). The red line represents a bar with m=2kg, l=2m.

I know that I*α = M1 + M2 + M3

If we displace the bar by x, we get the angle of displacement theta.

M1=M2=-k*x

I know that, judging by the picture, tan(Θ)=x/l, thus x=tan(Θ)*l.
However, my workbook states that M1=-k*tan(Θ)*l*l*cos(Θ). Where did the l*cos(Θ) come from?
 
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  • #2
Setting the workbook aside for the moment, how would you write Newton's second law for this arrangement?
 
  • #3
charon_9 said:
Where did the l*cos(Θ) come from?
Strictly speaking, the springs will not remain horizontal. As the rod swings, its end will rise slightly. So is ##x=l\tan(\theta), l\sin(\theta), ## or ##2l\sin(\theta/2)##?
I'm sure we will be doing a small angle approximation, so it is not going to matter. I.e. ##\cos(\theta)\approx 1##.
 
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  • #4
haruspex said:
Strictly speaking, the springs will not remain horizontal. As the rod swings, its end will rise slightly. So is ##x=l\tan(\theta), l\sin(\theta), ## or ##2l\sin(\theta/2)##?
I'm sure we will be doing a small angle approximation, so it is not going to matter. I.e. ##\cos(\theta)\approx 1##.
Yes, we're always using an approximation that $$cos(\theta) = 1$$ for small angles.
Thing is, I can understand that $$M_1 = kx \cdot r$$, where I already figured out that $$x=l\tan{\theta}$$

However, I really can't figure out where they got the $$l\cos{\theta}$$ in

$$M_1=-kl\tan{\theta} \cdot l\cos{\theta}$$
 
  • #5
kuruman said:
Setting the workbook aside for the moment, how would you write Newton's second law for this arrangement?

Well,

$$I\alpha = M_1 + M_2 + M_3$$ where $$M_1 = M_2$$.

$$M_1 = M_2 = -k \cdot l\tan{\theta}$$

Now I get stuck figuring out the radius part. I have no clue where they got the cosine from.
 
  • #6
charon_9 said:
However, I really can't figure out where they got the $$l\cos{\theta}$$
For the figure shown below, how would you express the moment (torque) about O of the force F in terms of F, l, and ##\theta##?
1674941401568.png


(They appear to be ignoring the tilt of the springs pointed out by @haruspex .)
 
  • #7
charon_9 said:
Now I get stuck figuring out the radius part. I have no clue where they got the cosine from.
To add to what @TSny posted, the picture that you posted in #1 is misleading. Strictly speaking, as the rod swings to some angle, the springs will no longer be horizontal. However, in the small angle approximation, you have ##l\cos\!\theta \approx l## and it wouldn't matter that the springs are not horizontal.
 
  • #8
TSny said:
For the figure shown below, how would you express the moment (torque) about O of the force F in terms of F, l, and ##\theta##?
View attachment 321364

(They appear to be ignoring the tilt of the springs pointed out by @haruspex .)
Thank you for taking the time to answer.

Wasn't it $$M=r \times F = r \cdot F \sin{\theta}$$
 
  • #9
charon_9 said:
Wasn't it $$M=r \times F = r \cdot F \sin{\theta}$$
##|\vec r \times \vec F| = rF \sin\phi##, where ##\phi## is the angle between ##\vec r## and ##\vec F##. Is the angle ##\theta## is the diagram the same as the angle between ##\vec r## and ##\vec F##?
 
Last edited:
  • #10
kuruman said:
To add to what @TSny posted, the picture that you posted in #1 is misleading. Strictly speaking, as the rod swings to some angle, the springs will no longer be horizontal. However, in the small angle approximation, you have ##l\cos\!\theta \approx l## and it wouldn't matter that the springs are not horizontal.
Sorry for posting a misleading picture, that's what I have been provided with by my school. I'm once again sorry.
I did some testing so please tell me if I'm right or wrong.
Because the springs are a bit elevated. their force F has a vertical component and a component in the direction of $$\theta$$. We need the component in the direction of theta, so $$G_t = G\cos{\theta}$$.
As for the ##l##, it is the radius, or the distance from the tip of the bar to the point about which it oscillates. Is this correct?
 
  • #11
charon_9 said:
Sorry for posting a misleading picture
No, that's ok, we know it is not your picture. I think @kuruman was just making sure you were not being misled by it.

charon_9 said:
Because the springs are a bit elevated. their force F has a vertical component and a component in the direction of $$\theta$$. We need the component in the direction of theta, so $$G_t = G\cos{\theta}$$.
G? Gt? What has that to do with the tension in the springs?
The elevation of the springs will not be to angle ##\theta##. It will depend on the spring length, so taking them to be extremely long gets rid of that problem.
charon_9 said:
As for the l, it is the radius, or the distance from the tip of the bar to the point about which it oscillates. Is this correct?
Yes.
 
  • #12
haruspex said:
No, that's ok, we know it is not your picture. I think @kuruman was just making sure you were not being misled by it.
Yes.
 
  • #13
haruspex said:
No, that's ok, we know it is not your picture. I think @kuruman was just making sure you were not being misled by it.G? Gt? What has that to do with the tension in the springs?
The elevation of the springs will not be to angle ##\theta##. It will depend on the spring length, so taking them to be extremely long gets rid of that problem.

Yes.
Sorry. I was quick to reply so I mistook ##G## for ##F##, the elastic force.

This is my guess. Let's say the springs tilt up for a small angle. This means that the elastic force has a vertical component to it (in the negative direction of the y axis in an XY plane). We need the other component - the one that "pulls" to the left.

IMG_20230128_234706.jpg


I didn't draw the springs in order to not make the sketch a mess. I just named the components as such, where ##t## would be tangential and ##n## would be perpendicular.
 
  • #14
charon_9 said:
Let's say the springs tilt up for a small angle. This means that the elastic force has a vertical component to it (in the negative direction of the y axis in an XY plane). We need the other component - the one that "pulls" to the left.

View attachment 321374
As I posted, the angle the springs make to the horizontal will not be the same angle the rod makes to the vertical. To see that, try including the springs in the diagram, making them much longer than the rod.
Because we don’t know the lengths of the springs, we cannot know the angle they make.
 
  • #15
haruspex said:
As I posted, the angle the springs make to the horizontal will not be the same angle the rod makes to the vertical. To see that, try including the springs in the diagram, making them much longer than the rod.
Because we don’t know the lengths of the springs, we cannot know the angle they make.
Okay, thanks. I have one more question, though. I decided not to open a new thread, but please tell me if I need to.

I took this photo from a very old textbook, where it asked us to write the same equation for this system, where the mass of the disc is ##M=2##, and ##k=37.5## for both springs. The solution says that
$$I\alpha = kx \cdot 2R$$

$$x = 2R \cdot \tan{\theta}$$

$$I \alpha = -4kR^2\tan{\theta}$$

I now have two questions:

1. Why did they only include the torque for one spring? Shouldn't it be ##2 \cdot -4kR^2\tan{\theta}##, like in the first example?

2. Why isn't the torque for the gravitational force included, also like in the first example?
1674948786729.png
 
  • #16
charon_9 said:
Okay, thanks. I have one more question, though. I decided not to open a new thread, but please tell me if I need to.

I took this photo from a very old textbook, where it asked us to write the same equation for this system, where the mass of the disc is ##M=2##, and ##k=37.5## for both springs. The solution says that
$$I\alpha = kx \cdot 2R$$

$$x = 2R \cdot \tan{\theta}$$

$$I \alpha = -4kR^2\tan{\theta}$$

I now have two questions:

1. Why did they only include the torque for one spring? Shouldn't it be ##2 \cdot -4kR^2\tan{\theta}##, like in the first example?

2. Why isn't the torque for the gravitational force included, also like in the first example?
View attachment 321377
No idea. Both look like mistakes to me. Leaving out gravity would have made sense if the disc were rolling back and forth on a flat surface, but it is clearly mounted on a pivot.
 

1. What is torque and why is it important in a compound pendulum?

Torque is a measure of the force that causes an object to rotate around an axis. In a compound pendulum, torque is important because it determines the motion of the pendulum and affects its period and amplitude.

2. How is torque calculated for a compound pendulum?

Torque can be calculated by multiplying the force applied to the pendulum by the distance from the axis of rotation to the point where the force is applied. This distance is known as the moment arm or lever arm. The formula for torque is T = Fd, where T is torque, F is force, and d is the moment arm.

3. What factors affect the torque in a compound pendulum?

The factors that affect torque in a compound pendulum include the mass of the pendulum, the length of the pendulum, the angle at which the force is applied, and the distance between the axis of rotation and the point where the force is applied. These factors can all impact the motion and behavior of the pendulum.

4. How does torque affect the period of a compound pendulum?

Torque is directly related to the period of a compound pendulum. A larger torque will result in a larger angular acceleration, causing the pendulum to swing with a larger amplitude and shorter period. On the other hand, a smaller torque will result in a smaller angular acceleration, causing the pendulum to swing with a smaller amplitude and longer period.

5. What is the difference between torque and moment of inertia in a compound pendulum?

While torque is a measure of the force that causes rotation, moment of inertia is a measure of an object's resistance to rotation. In a compound pendulum, the moment of inertia is affected by the mass distribution and shape of the pendulum, while torque is affected by external forces applied to the pendulum. Both torque and moment of inertia play important roles in determining the motion of a compound pendulum.

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