# Shortening the length of pendulum

## Homework Statement ## The Attempt at a Solution

I tried with the conservation laws. Angular momentum conservation won't work. To use energy conservation, I need the force pulling up the string but I don't have it. The force in the given case is tension but the tension continuously changes and I am not sure if the tension force is the conservative one. I don't have any idea about how to start making the equations. :(

Any help is appreciated. Thanks!

#### Attachments

• 30.4 KB Views: 1,949
• 1 person

Related Introductory Physics Homework Help News on Phys.org
Assuming that the string remains tight throughout the process, the tension is always perpendicular to the direction of motion and thus will do no work on the mass 'system'. So according to me, the total mechanical energy of the mass will be conserved (kinetic and gravitational potential energy).

Assuming that the string remains tight throughout the process, the tension is always perpendicular to the direction of motion and thus will do no work on the mass 'system'. So according to me, the total mechanical energy of the mass will be conserved (kinetic and gravitational potential energy).
I am not sure how it is always perpendicular to motion. The point mass moves in the vertical direction and there is some component of tension in that direction.

Hi Pranav...

I guess by slowly moving the mass ,it means that there is no acceleration of the mass along the string i.e we may assume that the velocity of the mass is always tangential to the string .Now the work done by tension results in the increase in mechanical energy by mgL/2 .

What do you think ?

Hi Pranav...

I guess by slowly moving the mass ,it means that there is no acceleration of the mass along the string i.e we may assume that the velocity of the mass is always tangential to the string .Now the work done by tension results in the increase in mechanical energy by mgL/2 .

What do you think ?
Yes but then how do I use the conservation of energy here? I mean at what two instants should I use? Even if I use the maximum velocities during the oscillation for conservation, I need the amplitude.

And btw, the given answer is 1.68.

The work of tension is $$\int \vec T \cdot \mathrm d \vec r = \int T v_r \mathrm d t$$ where $v_r$ is the radial velocity. Now we have been told that $v_r$ is very small. Does that mean that the work is negligible?

The work of tension is $$\int \vec T \cdot \mathrm d \vec r = \int T v_r \mathrm d t$$ where $v_r$ is the radial velocity. Now we have been told that $v_r$ is very small. Does that mean that the work is negligible?
No but how do I find the work? The tension continuously changes. I can express T in terms of angle with the vertical but I have no idea for $v_r$. I do not think you can find the work without knowing $v_r$.

I do not think you can find the work without knowing $v_r$.
So...is the question incomplete?

TSny
Homework Helper
Gold Member
So...is the question incomplete?
No, the question is ok. It's actually a fairly well-known example in mechanics that is used to illustrate the concept of "adiabatic invariant".

The idea is that the string is pulled so slowly that the pendulum can be thought of as making many oscillations during the time that the string is shortened an infinitesimal distance. So the infinitesimal work done on the system by the tension force can be calculated using the average value of the tension over an oscillation.

No, the question is ok. It's actually a fairly well-known example in mechanics that is used to illustrate the concept of "adiabatic invariant".

The idea is that the string is pulled so slowly that the pendulum can be thought of as making many oscillations during the time that the string is shortened an infinitesimal distance. So the infinitesimal work done on the system by the tension force can be calculated using the average value of the tension over an oscillation.
Tension at any time t is given by
$$T=mg\cos\theta(t)+m\omega^2L$$
where $\theta(t)=\theta_m\sin(\sqrt{g/L}t)$ and $\omega$ is the angular velocity of object about the pivot.

How do I find the average of cos(sin(something))? Should I take the small angle approximation and then find the average?

The detail in the question informs "that the angle of oscillation is very small" so I assume you can make the simplifying angle tends to zero assumption.

• 1 person
TSny
Homework Helper
Gold Member
Should I take the small angle approximation and then find the average?
Yes, good. You'll need to expand $\cos\theta$ to second order in $\theta$. Also, write out ω = $\dot{θ}$ explicitly as a function of time.

• 2 people
The detail in the question informs "that the angle of oscillation is very small" so I assume you can make the simplifying angle tends to zero assumption.
Yes, good. You'll need to expand $\cos\theta$ to second order in $\theta$. Also, write out ω = $\dot{θ}$ explicitly as a function of time.
Thanks! :)

This is what I did:
$$T=mg\left(1-\frac{\theta^2}{2}\right)+m\dot{\theta}^2L$$
The average of $\theta^2=\theta_m^2 \sin^2(\sqrt{g/L}\,t)$ is $\theta_m^2/2$ and the average of $\dot{\theta}^2=\theta_m^2g/L \cos^2(\sqrt{g/L}\,t)$ is $\theta_m^2g/(2L)$. Hence, I have
$$T_{avg}=mg\left(1-\frac{\theta_m^2}{4}\right)+m\frac{\theta_m^2g}{2L}L$$
$$\Rightarrow T_{avg}=mg\left(1+\frac{\theta_m^2}{4}\right)$$
I think the next step is energy conservation but I am not sure how to write down the equation. If the length of string changes by dL, then the change in gravitational potential energy is mgdL and work done by tension is TdL. How do I write down the change in kinetic energy? TSny
Homework Helper
Gold Member
Not only will there be an increase in the gravitational PE due to raising the pendulum, there will also be a change in the SHM energy.

Edit: Your expression for the average tension looks good.

Not only will there be an increase in the gravitational PE due to raising the pendulum, there will also be a change in the SHM energy.

Edit: Your expression for the average tension looks good.
Isn't SHM energy simply the sum of kinetic energy and potential energy? TSny
Homework Helper
Gold Member
Yes. You should be able to write the total SHM energy in terms of just the amplitude of the motion.

The total SHM energy includes potential energy of gravity associated with the oscillation, but it does not include the increase in gravitational potential energy of the system due to shortening the length of the string.

• 1 person
Yes. You should be able to write the total SHM energy in terms of just the amplitude of the motion.

The total SHM energy includes potential energy of gravity associated with the oscillation, but it does not include the increase in gravitational potential energy of the system due to shortening the length of the string.
I write the oscillation energy of pendulum.

$$E=-mgL\cos\theta(t)+\frac{1}{2}m\dot{\theta}^2L^2$$
where I have set the potential energy to be zero at the horizontal line passing through pivot.

Honestly, I am not sure what I am doing here, do I take the average again? TSny
Homework Helper
Gold Member
In the small angle approximation, the energy of the system is E = ESHM + mgy where y is the height of the lowest point of the swing above the floor. Like any SHM, you should be able to express ESHM in terms of the amplitude of the motion. Then consider dE for a small change in length dL of the pendulum and relate it to the work done by the tension.

In the small angle approximation, the energy of the system is E = ESHM + mgy where y is the height of the lowest point of the swing above the floor. Like any SHM, you should be able to express ESHM in terms of the amplitude of the motion. Then consider dE for a small change in length dL of the pendulum and relate it to the work done by the tension.
Is
$$E_{SHM}=\frac{1}{2}mgL\theta_m^2$$?

TSny
Homework Helper
Gold Member
That looks good to me.

That looks good to me.
Thanks! :)

I use conservation of energy now.
$$mgy+E=mg(y+dL)+(E+dE)+TdL \Rightarrow -TdL=mgdL+dE$$
Here,
$$dE=\frac{1}{2}mg\theta_m^2\,dL+mgL\theta_m\,d\theta_m$$
and
$$T=mg\left(1+\frac{\theta_m^2}{4}\right)\,dL$$
Substituting and solving it further doesn't seem to give me the right answer. I believe I have made a sign error but I am unable to find it. :(

TSny
Homework Helper
Gold Member
Do dy and dL have the same sign?

Do dy and dL have the same sign?
I have no dy but if you ask about y and dL, then I think no as the length decreases. So does that mean the final potential energy is mg(y-dL)?

TSny
Homework Helper
Gold Member
So does that mean the final potential energy is mg(y-dL)?
Sounds right. Final PE = mg(y+dy).

• 1 person