# Rotational energy of an object.

## Homework Statement

An object, consisting of two point masses with mass m are connected by a rigid shaft with negligible mass. They are lying horizontally at rest. One of the point masses is instantly given a speed of v. Calculate the total energy of the system once the system has stopped rising in the gravitational field.

## The Attempt at a Solution

Not really sure if it can be solved by simple methods. I assume I need to calculate the rotational energy of the system and also the kinetic energy of any velocity perpendicular to g-acceleration, since once it stops rising, velocity parallel to the gravitational acceleration will be gone. One mass starts moving up, the others inertia makes it move in a curved path, causing centripetal acceleration, perhaps sideways motion of the center of mass etc? Any tips to get me on the right path and if it can be solved without calculus are welcomed.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

An object, consisting of two point masses with mass m are connected by a rigid shaft with negligible mass. They are lying horizontally at rest. One of the point masses is instantly given a speed of v. Calculate the total energy of the system once the system has stopped rising in the gravitational field.

## The Attempt at a Solution

Not really sure if it can be solved by simple methods. I assume I need to calculate the rotational energy of the system and also the kinetic energy of any velocity perpendicular to g-acceleration, since once it stops rising, velocity parallel to the gravitational acceleration will be gone. One mass starts moving up, the others inertia makes it move in a curved path, causing centripetal acceleration, perhaps sideways motion of the center of mass etc? Any tips to get me on the right path and if it can be solved without calculus are welcomed.
Describe the motion of the center of mass of the system.

Describe the motion of the system with respect to the center of mass.

I'm not really sure how the two objects would behave in this situation. Is it as simple as such that the center of mass starts moving upwards with a velocity of v/2 and therefore the rest of the energy goes into rotational kinetic energy?

Energy of one mass at the beginning E=mv^2/2

Energy of the center of mass: E=2m(v/2)^2/2=mv^2/4

Rotational energy: E1-E2=mv^2/4

But wouldn't the object start moving sideways since one mass is moving faster and causes more centripetal acceleration to the other mass?

SammyS
Staff Emeritus
Homework Helper
Gold Member
I'm not really sure how the two objects would behave in this situation. Is it as simple as such that the center of mass starts moving upwards with a velocity of v/2 and therefore the rest of the energy goes into rotational kinetic energy?

Energy of one mass at the beginning E=mv^2/2

Energy of the center of mass: E=2m(v/2)^2/2=mv^2/4

Rotational energy: E1-E2=mv^2/4

But wouldn't the object start moving sideways since one mass is moving faster and causes more centripetal acceleration to the other mass?
The center of mass will only move vertically. Both point masses will begin to move sideways, toward the the vertical line passing through the center of mass, as the object rotates.

What is the initial angular velocity (rotational speed) of the object ?

If the distance from the center to the masses is r, then as E=1/2*I*w^2
and I=2*m*r^2
(mv^2)/4=1/2*2*m*r^2*w^2
w=((mv^2/4)/(m*r^2))^0,5=(v^2/(4*r^2))^0,5=v/(2r)

SammyS
Staff Emeritus
Homework Helper
Gold Member
If the distance from the center to the masses is r, then as E=1/2*I*w^2
and I=2*m*r^2
(mv^2)/4=1/2*2*m*r^2*w^2
w=((mv^2/4)/(m*r^2))^0,5=(v^2/(4*r^2))^0,5=v/(2r)
Put all of that together to find E in terms of ω , m , and r .

Filip Larsen
Gold Member
If you are to find the total mechanical energy, i.e the sum of potential energy from gravity (from some arbitrary reference level) and rotational and linear kinetic energy, you can utilize that this is constant over time when there is no friction. This means that the total energy at the "top" is equal to the total energy at "start", which is very easy to find since only one mass is moving.

Of course, it is good practice to calculate the linear and rotational energy of the system, as suggested by SammyS, and check that their sum is equal to the total kinetic energy at the start.

Do you mean that E=m*r^2*w^2=1/4m*v^2 ?
But since I don't know the w and I have to calculate it and it seems to give the same answer as the kinetic energy only method, I guess that way is easier?

If I don't know the height at the beginning nor at the end nor the relationship between them, can I still apply the total mechanical energy method?

Filip Larsen
Gold Member
I am not sure what you really ask. The intention of my post was in part to hint that you can calculate the instantaneous kinetic energy of the system in two ways and both should end up giving same result.

One way (the "particle approach") would be to simply add the (linear) kinetic energy of each of the two masses, which at the time of start is easy for this problem since we are given that one mass is at rest and the other is moving with speed v. However, note that since this model does not capture how the two masses interact we are not able to directly calculate the total kinetic energy at any later time.

Another way to calculate the total kinetic energy (the "system approach") would be to model the two masses as a rigid system with center of mass and linear and rotational speed and then calculate the linear and rotational kinetic energy of this system. At time of start it is fairly easy to calculate the initial linear kinetic energy and rotational kinetic energy of this system using information from the problem description.

You can then either do one or the other calculation or you can do both and check that you get the same total kinetic energy in the two cases. Or, knowing that the two energies must be equal, you can use the first way to calculate the rotational kinetic energy of the system without having to establish a rotational model directly.

Why is the "system approach" at all interesting? The problem text seems to ask about the total (mechanical) energy of the system, which can be calculated directly using the "particle approach" and noting that the total mechanical energy must be constant over the time interval in question (from start to top) since there is no modeled friction, and in this case the rotational kinetic energy is not really interesting.

However, if you wanted to calculate the how high the system gets in the gravitational field you can use the fact that the linear motion of the systems center of mass is decoupled from the rotational state, and, assuming a uniform gravitational field, the torque on the system from gravity is zero which means the the rotational momentum and hence rotational energy is constant. Thus, the increase in potential energy (the gain in height) of the system only comes from a reduction in linear kinetic energy of the system (since total mechanical energy is conserved without friction). For this particular setup then, knowing how much of the initial total kinetic energy that goes into linear kinetic energy of the system instead of rotational energy allows you to find height as a function of initial speed.

Last edited: