# Misunderstanding Work-energy theorem and center of mass properties

• physicsissohard
physicsissohard
Homework Statement
A rod is hinged at one of its ends, and released from rest when it is held parallel to the ground. The question is to find the angular velocity after the rod makes 60 degrees with the horizontal.
Relevant Equations
MgL/2=Ialpha a=r*alpha
To do this apparently, you need to use the work-energy theorem. You can calculate work done by gravity easily. However it was said that work done by the reaction forces from the hinge is zero, I don't get why.

Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass and contribute to the displacement of the center of mass, so how is work done by reaction forces zero?

For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4. So there is a reaction force perpendicular to the rod, at all times so work is not perpendicular so work done is not zero

Last edited by a moderator:
physicsissohard said:
However it was said that work done by the reaction forces from the hinge is zero, I don't get why.
Work is force exerted through a distance. If the friction is zero at the hinge, what is the distance that a force is exerted through?

MatinSAR
physicsissohard said:
Homework Statement: A rod is hinged at one of its ends, and released from rest when it is held parallel to the ground. The question is to find the angular velocity after the rod makes 60 degrees with the horizontal.
Relevant Equations: MgL/2=Ialpha a=r*alpha

To do this apparently, you need to use the work-energy theorem. You can calculate work done by gravity easily. However it was said that work done by the reaction forces from the hinge is zero, I don't get why.

Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass and contribute to the displacement of the center of mass, so how is work done by reaction forces zero?

For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4. So there is a reaction force perpendicular to the rod, at all times so work is not perpendicular so work done is not zero
If you don't believe the work energy theorem, then calculate the angular velocity using torque about the hinge. Then check against the work energy theorem and see how much work is done by the force at the hinge. The change in gravitational potential energy should be straightforward to calculate.

MatinSAR
PeroK said:
If you don't believe the work energy theorem, then calculate the angular velocity using torque about the hinge. Then check against the work energy theorem and see how much work is done by the force at the hinge. The change in gravitational potential energy should be straightforward to calculate.
Seriously, What do you mean I don't believe work energy theorem? I just doubt if the application of WET as shown is correct or not? and there is no way you can calculate angular velocity using torque alone, well you can but that requires you to solve an unsolvable differential equation with elliptic integral and stuff like that.

physicsissohard said:
For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4. So there is a reaction force perpendicular to the rod, at all times so work is not perpendicular so work done is not zero
Sure, there is a non-zero reaction force at the hinge. As pointed out, that force does no real work since it has no displacement.

You could use Newton's 2nd law to calculate the velocity of the center of mass as the rod falls to the 60-degree position. But that will involve some tricky integration. (Note that the reaction force is not constant as the rod falls.)

Much easier to use conservation of energy. The only force acting on the rod that does work is gravity, so you only need to consider changes in gravitational energy.

berkeman and Lnewqban
physicsissohard said:
To do this apparently, you need to use the work-energy theorem.
Using the work-energy theorem is not essential - but it is the simplest approach here.

physicsissohard said:
You can calculate work done by gravity easily. However it was said that work done by the reaction forces from the hinge is zero, I don't get why.
The reaction force (from the hinge on the rod) acts on (say) point A on one end of the rod. Point A has zero displacement - so what can you say about the work done by the reaction force?

physicsissohard said:
Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass
Not sure what 'act on the center of mass' means.

A force does not have to do work on an object to affect the motion of the object. E.g. consider a mass moving in a circle at contant speed; the centripetal force (e.g. tension in a rope) causes acceleration but does no work.

Here the reaction force affects the rod's motion - but that doesn't mean the reaction force does work.

physicsissohard said:
For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4.
Yes. The reaction force exists (and is initially Mg/4). But it doesn't do any work (see above).

physicsissohard said:
So there is a reaction force perpendicular to the rod, at all times
No. For example if the rod eventually reaches it's lowest point, the reaction force acts upwards (parallel to rod).

physicsissohard said:
so work is not perpendicular so work done is not zero
Don't know what that means!

Not, if the reaction force did work here, it would have to have a source of energy - but it doesn't!

Last edited:
MatinSAR and berkeman
physicsissohard said:
⁹Seriously, What do you mean I don't believe work energy theorem? I just doubt if the application of WET as shown is correct or not? and there is no way you can calculate angular velocity using torque alone, well you can but that requires you to solve an unsolvable differential equation with elliptic integral and stuff like that.
You're right. I should have said if you don't believe the force at the hinge does no work.

You don't have to solve an elliptic integral. You could look at the rate of change of KE using torque and rate of change of GPE and check the two are equal and opposite:
$$\tau = \frac{mgl}{2}\cos \theta$$$$\ddot \theta = \frac{\tau}{I} = \frac{mgl}{2I}\cos \theta$$
$$KE = \frac 1 2 I\dot \theta^2$$$$\frac {dKE}{dt} = I\dot \theta \ddot \theta = \frac{mgl\dot \theta}{2}\cos \theta$$
$$PE = -\frac{mgl}{2}\sin \theta$$$$\frac{dPE}{dt} = -\frac{mgl}{2}(\cos \theta)\dot \theta$$Hence the change in KE caused by the torque is entirely accounted for by the change in gravitational PE. The force at the hinge, therefore, does no work.

MatinSAR and TSny
physicsissohard said:
Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass and contribute to the displacement of the center of mass, ##\dots##
That's a serious misconception that needs to be remedied. Yes, the force from the hinge is an external force. Yes, it is part of the net force acting on the rod and contributes to the acceleration of the center of mass. No, it does not act on the center of mass. It acts at the tip of the rod where it is attached to the hinge.

MatinSAR, Doc Al, PeroK and 1 other person

## What is the Work-Energy Theorem and how is it commonly misunderstood?

The Work-Energy Theorem states that the work done by all forces acting on a particle equals the change in its kinetic energy. A common misunderstanding is that it applies only to conservative forces or that it includes potential energy changes directly. In reality, it applies to all forces and only considers kinetic energy changes; potential energy changes are considered separately in the context of conservative forces.

## How does the Work-Energy Theorem relate to non-conservative forces?

Non-conservative forces, such as friction, perform work that converts mechanical energy into other forms like heat. The Work-Energy Theorem still holds, but the work done by non-conservative forces results in a net change in kinetic energy that accounts for energy losses or gains due to these forces, rather than being stored as potential energy.

## What are the common misconceptions about the center of mass in a system of particles?

A frequent misconception is that the center of mass always lies within the material boundaries of an object. In reality, the center of mass is a point that represents the average position of the mass distribution and can lie outside the physical object, especially in systems with irregular shapes or mass distributions.

## How does the motion of the center of mass relate to external forces?

The motion of the center of mass of a system of particles is governed by the net external force acting on the system. A common misunderstanding is thinking that internal forces within the system can affect the motion of the center of mass. However, only external forces can change the velocity of the center of mass according to Newton's second law.

## Why is the concept of the center of mass important in understanding collisions?

In collisions, analyzing the motion of the center of mass helps simplify the problem. A common misunderstanding is neglecting the center of mass when considering momentum conservation. The total momentum of a system of particles is conserved in the absence of external forces, and this conservation can be analyzed more easily by considering the center of mass, which moves as if all the mass and external forces were concentrated at that point.

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