# Is the Angular Momentum of a Pendulum Conserved?

• brotherbobby
In summary, the content discusses a pendulum swinging in a vertical plane and addresses questions about its angular momentum, torque, and rate of change of angular momentum. The direction of the angular momentum changes as the pendulum swings, and the torque is caused by the tension in the rope. The magnitude of the rate of change of angular momentum is also discussed.
brotherbobby
Statement of the problem :

A ball shown in the figure is allowed to swing in a vertical plane like a simple pendulum. Answer the following :

(a) Is the angular momentum of the ball conserved?

No, the angular momentum ##L = mvl##, where m is the mass of the ball and v is its speed at an instant. Note, this is an example of (non-uniform) circular motion whereby vector v ##\perp## radius vector r at all instants. Clearly, v changes, reaching its maximum value when ##\theta = 0## and hence the angular momentum L also changes, becoming maximum at the mean position.

(b) Calculate the direction of L at some time. Does it change?
No. The motion takes place in a plane. From the definition of L = r ##\times## p, using either the right hand cork screw rule or taking the convenient mean position of the pendulum to evaluate (r ##\rightarrow - \hat y##, p ##\rightarrow \hat x##, ##-\hat y \times \hat x = \hat z##), we find that the direction of L is along the +z axis, out of the page.

(c) What force acting on the pendulum gives a zero torque about an axis perpendicular to the motion plane and through the point of support?

Tension T. By the definition of torque, ##\tau = r \times F \Rightarrow \tau = l \hat l \times -T \hat l = 0##, where ##\hat l## is the unit vector along the rope out from the point of support.(d) Calculate the torque due to the weight of the ball about this axis at an angle ##\theta##.

From definition, ##\tau = r \times F##. At the angle ##\theta##, the only contributing force to the torque is the "vertical" component of weight, ##mg \sin \theta## (The tension and the "horizontal" component of weight ##mg \cos \theta## both lie along the rope). Hence the magnitude of torque ##\mathbb{\tau = mg \sin \theta l}##. (e) Calculate the magnitude of the rate of change of angular momentum of the pendulum bob at ##\theta##.

If ##L = mvl## [see (a)], then ##\frac {dL} {dt} = m \frac {dv} {dt} l = m g \sin \theta l = mgl \sin \theta##, as the rate of change of v, ##\frac {dv}{dt} = g \sin \theta##, in a direction perpendicular to the rope and along the motion of the bob.

Thank you for you interest.

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I do not see a question here.

I would quibble a bit with your response to b. Is the angular momentum always going to be out of the page? Can it ever be into the page instead?

The questions are listed from (a) - (e) above, in italics. You will see my responses in normal font.

Hmm, no the angular momentum will be "into" the page when the pendulum swings back from the right extreme position. When it reaches the mean position, we get for the direction of L (r ##\times## p) = ##-\hat y \times -\hat x = -\hat z##, into the page in the direction of the -z axis.

Thank you for the correction.

brotherbobby said:
The questions are listed from (a) - (e) above, in italics.
jbriggs meant, what question are you asking the forum? Presumably it was "is this right?"

haruspex said:
jbriggs meant, what question are you asking the forum? Presumably it was "is this right?"

Yes, I'd be glad if you could let me know if my solutions were correct. I have figured out that (b) above was not.

brotherbobby said:
Yes, I'd be glad if you could let me know if my solutions were correct. I have figured out that (b) above was not.

After your correction to that, all looks good.

## 1. What is angular momentum?

Angular momentum is a measurement of the rotational motion of an object around an axis. It is a vector quantity and is calculated by multiplying the moment of inertia (a measure of an object's resistance to rotation) by the angular velocity (the rate at which the object is rotating).

## 2. How does a pendulum have angular momentum?

A pendulum has angular momentum because it is constantly rotating around a fixed point (the point of suspension). As the pendulum swings back and forth, its velocity and direction are constantly changing, resulting in a continuous change in angular momentum.

## 3. What factors affect the angular momentum of a pendulum?

The factors that affect the angular momentum of a pendulum include the length of the string, the mass of the bob (the object at the end of the string), and the initial angle of release. Additionally, external forces such as air resistance can also affect the angular momentum of a pendulum.

## 4. How does the angular momentum of a pendulum change over time?

The angular momentum of a pendulum remains constant if there are no external forces acting upon it. This is known as the conservation of angular momentum. However, if external forces are present, the angular momentum of the pendulum may change over time.

## 5. How is the angular momentum of a pendulum related to its period?

The period of a pendulum (the time it takes to complete one full swing) is directly related to its length and the acceleration due to gravity. The angular momentum of a pendulum also depends on these factors, and as a result, the period and angular momentum of a pendulum are inversely proportional.

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