# Kinetic energy transfer from a rotating body in an inelastic collision

• KataruZ98
So we have linear momentum and angular... what about kinetic energy?The cylinder has a moment of inertia of ~1.67kg*m² and rotational KE of 2.058J. At the point of impact also, assuming the body hits the sphere at a 90deg angle after traversing 90deg of displacement, it should(?) exert a force of 1.31N - enough to give an acceleration of 1.31m/s².In summary, the cylinder in question would have a moment of inertia of ~1.67kg*m² and rotational KE of 2.058J. At the point of impact also, assuming the body hits the sphere at af

#### KataruZ98

Homework Statement
A cylindrical body pivoting around a base strikes with the other a small metal sphere laying still on the ground. The cylinder is 1m long and 0.1m wide, with mass of 5kg - whereas the sphere is 1m across and 1kg heavy. Supposing the cylinder was rotating at 90deg/s, this translates to a tangential velocity of 1.57m/s where the impact happened.
Relevant Equations
How can I find the kinetic energy transferred in the form of rotational KE to the sphere from the cylinder? What about the translational KE of the sphere post-collision?
The cylinder in question would have a moment of inertia of ~1.67kg*m² and rotational KE of 2.058J. At the point of impact also, assuming the body hits the sphere at a 90deg angle after traversing 90deg of displacement, it should(?) exert a force of 1.31N - enough to give an acceleration of 1.31m/s².

How should I move from this point onward to get the desired data?

it should(?) exert a force of 1.31N - enough to give an acceleration of 1.31m/s².
Why?

Why?

I thought that’d be the case, considering the body’s torque of 1.31N*m and distance of point of impact to the pivot point. Guess I was wrong though lol.

I thought that’d be the case, considering the body’s torque of 1.31N*m and distance of point of impact to the pivot point. Guess I was wrong though lol.
I do not see how you arrived at a "torque" of 1.31 N*m either.

It is an impulsive collision. The force and associated torque will be infinite. Or at least very large... and irrelevant.

KataruZ98
I do not see how you arrived at a "torque" of 1.31 N*m either.

It is an impulsive collision. The force and associated torque will be infinite.

Well isn’t torque of the rotating body equal to its KE divided by the angular displacement (90 degrees for the question’s purpose)?

Well isn’t torque of the rotating body equal to its KE divided by the angular displacement (90 degrees for the question’s purpose)?
No.

A rotating body does not have a torque. It has a kinetic energy. And it has an angular momentum. A torque is a rate of transfer of angular momentum through an interface. A rotating body does not have such a thing.

KataruZ98
No.

I see. Apologies, but would you guide me through my errors to see what went wrong in my assumptions?

EDIT: just saw what you’ve add. Could you explain more in depth then the mechanism of KE transfer in this situation?

I see. Apologies, but would you guide me through my errors to see what went wrong in my assumptions?
Let us start with a description of the problem. A problem well described is half solved.

We have this cylinder. It is tall (1 meter) and narrow (0.1 meter). It is standing upright on one of its circular ends. It is rotating but is otherwise nearly motionless. It is 5 kg mass.

We have this sphere. It is 1 meter in diameter. It is just sitting there. It is 1 kg mass.

We are told that these two objects collide. We are told (per the thread title) that the collision is inelastic. This suggests that we have not been given the correct and complete problem statement.

Since we have been told that a collision has occurred but have not been given an initial linear velocity for the rotating cylinder, we will assume that it has somehow slid sideways into the stationary sphere at a speed so low as to be negligible for our purposes.

We are told that the collision is inelastic. So we assume that the cylinder and the sphere have somehow welded themselves together as a result of the collision. The pair is now moving or wobbling as a consolidated lump.

You are asking for the rotational and linear kinetic energy in the motion of the sphere as it gyrates as part of the post-collision lump.

For me, the place to start is with conservation laws. What quantities are conserved across the collision event?

BvU, KataruZ98 and Steve4Physics
I see, that’d be momentum and energy if IIRC

I see, that’d be momentum and energy if IIRC
Yes, momentum is always conserved in a collision (barring impulsive external forces associated with the collision).

Is mechanical energy conserved in an inelastic collision?

KataruZ98
Is mechanical energy conserved in an inelastic collision?

As far as I know, momentum is always conserved - be it angular or linear. However mechanical energy isn’t in an inelastic collision.

As far as I know, momentum is always conserved - be it angular or linear. However mechanical energy isn’t in an inelastic collision.
You are correct. In the absence of external forces, both linear momentum and angular momentum are always conserved.

In the case of a collision, both are conserved even if external forces are present, so long as those external forces are not impulsive. In the case at hand, the only external forces are gravity and the supporting normal force from the ground. Neither is impulsive.

We can wave our hands a bit here. By inspection, the collision impulse is entirely in the horizontal plane, so there is no vertical impulse. Nor are we particularly worried about the cylinder tipping slightly since the collision acts at its vertical midpoint. So we need not envision the cylinder tipping slightly, catching on the ground and launching itself skyward due to an impulsive interaction with the ground. We can pretend that both cylinder and sphere are sliding on an icy surface.

So we have linear momentum and angular momentum to work with. What is the combined total linear momentum before the collision? What is the combined total angular momentum before the collision?

KataruZ98
You are correct. In the absence of external forces, both linear momentum and angular momentum are always conserved.

In the case of a collision, both are conserved even if external forces are present, so long as those external forces are not impulsive. In the case at hand, the only external forces are gravity and the supporting normal force from the ground. Neither is impulsive.

We can wave our hands a bit here -- by inspection, the collision impulse is entirely in the horizontal plane, so there is no vertical impulse. Nor are we particularly worried about the cylinder tipping slightly since the collision acts at its vertical midpoint. So we need not envision the cylinder tipping slightly, catching on the ground and launching itself skyward due to an impulsive interaction with the ground. We can pretend that both cylinder and sphere are sliding on an icy surface.

So we have linear momentum and angular momentum to work with. What is the combined total linear momentum before the collision? What is the combined total angular momentum before the collision?

Alright, the total linear momentum would be:

Mass(cylinder)*Initial Velocity(cylinder)+Mass(sphere)*Initial Velocity(sphere)

I didn’t give an initial linear velocity for the cylinder I realize, so if we call that X the formula would be:

5X+0

Alright, the total linear momentum would be:

Mass(cylinder)*Initial Velocity(cylinder)+Mass(sphere)*Initial Velocity(sphere)

I didn’t give an initial linear velocity for the cylinder I realize, so if we call that X the formula would be:

5X+0
I agree. Strictly speaking, we should have units on that. If it were me, I'd probably have used different variable names. Maybe ##p_0## for starting linear momentum, ##m_c## for the mass of the cylinder and ##v_c## for the initial velocity of the cylinder. So we'd have:$$p_0 = m_cv_c$$There is a complication. We will see that when we get to the angular momentum calculation. Is this a head-on impact? A glancing impact? Something in between? We will need another parameter for that. Assuming a negligible linear velocity for the cylinder would make this concern go away.

KataruZ98
I see. Say, Vc is 1m/s, would that suffice?

I see. Say, Vc is 1m/s, would that suffice?
That is fine. Let us see the angular momentum calculation now.

KataruZ98
Alright, that’d be:

I(c)*w(c)+I(s)*w(s)

Or:

1.67*1.57+0.17*0

Alright, that’d be:

I(c)*w(c)+I(s)*w(s)

Or:

1.67*1.57+0.17*0
There is a complication. But let me unwrap your calculation first.

You have the angular momentum of the cylinder about its vertical axis. That is the moment of inertia of the cylinder times the angular velocity of the cylinder: ##I_c \omega_c##. You have the same for the sphere: ##I_s \omega_s##.

Where did that 1.67 come from? It is not equal to the moment of inertia of the cylinder.
Where did that 1.57 come from? It is not equal to the angular velocity of the cylinder.
Where did the 0.17 come from? It is not equal to the moment of inertia of the sphere.

[I am pretty sure that I know how you got the 0.17 -- it is for a hollow sphere, not a solid sphere. And I think that I figured out the 1.67 -- it is for a solid rod rotating around one end instead of about its long axis. I know where you got the 1.57 -- that's tangential velocity, which is not the same as angular velocity]

I will hold off on the other complication until we get the moments of inertia and the angular velocities straightened out.

Hint: Angular velocity (conventionally denoted by the greek letter ##\omega##) is a rotation rate in radians per second.

Hint: What is the radius of the cylinder about its vertical axis?

Last edited:
Where did that 1.67 come from? It is not equal to the moment of inertia of the cylinder.
I mistakenly used the formula for the moment of inertia of a cylinder about an axis perpendicular and passing through the center of mass, then used the parallel axis theorem to get the moment of inertia by an axis perpendicular to the cylinder and passing through a base.
This actually reminds me, I initially wanted to ask for the impact between the sphere and the cylinder with the latter moving like a pendulum - hitting the sphere with the extremity opposed to the pivot point.

Where did that 1.57 come from? It is not equal to the angular velocity of the cylinder.
Shouldn’t that be the velocity I’ve given of 90deg/s translated into rad/s?
[I am pretty sure that I know how you got the 0.17 -- it is for a hollow sphere]
I got it from this Wikipedia article:

https://en.wikipedia.org/wiki/List_of_moments_of_inertia

Supposedly it is for a solid sphere, or am I wrong?

Where did that 1.67 come from? It is not equal to the moment of inertia of the cylinder.
I mistakenly used the formula for the moment of inertia of a cylinder about an axis perpendicular and passing through the center of mass, then used the parallel axis theorem to get the moment of inertia by an axis perpendicular to the cylinder and passing through a base.
The 1.67 would be the moment of inertia for a 5 kg thin rod with length 1 meter rotating around one end.

I got my list of moments of inertia from the same Wiki page that you used. For an axis at one end, the moment of inertia is ##\frac{1}{3}mL^2## read directly from the page.

The approach of using the parallel axis theorem is also valid you start with ##\frac{1}{12}mL^2## and then add an adjustment of ##\frac{1}{4}mL^2## for a total of ##\frac{4}{12}mL^2##.

If we are dealing with a solid cylinder spinning around its long axis then ##\frac{1}{2}mr^2## is appropriate. The radius is, of course, 0.05 in that direction.

As you've already acknowledged. So we are good here now.
This actually reminds me, I initially wanted to ask for the impact between the sphere and the cylinder with the latter moving like a pendulum - hitting the sphere with the extremity opposed to the pivot point.
We could switch to that problem. The scale seems a bit off though. A 1 meter long pendulum striking a 1 meter high sphere? Maybe we would want a 10 meter long, 1 meter radius pendulum?
Shouldn’t that be the velocity I’ve given of 90deg/s translated into rad/s?
Well, let us see. I took your original post at face value. There you called off 1.57 m/s as a tangential velocity. Maybe that one is instead correct for angular velocity.

90 degrees per second times ##\frac{2 \pi \text{ radians}} {360 \text{ degrees}} = 1.57 \text{ radians per second}##

You are correct. The 1.57 radians per second is the angular velocity.

Which means that the "tangential velocity" in the original post was incorrect. The tangential velocity is the radius (0.05 meters) times the angular velocity (1.57 radians per second) = 0.0785 meters per second.

[Tangential velocity and angular velocity will be numerically equal only if the radius is 1 unit. Or if both are zero]
I got it from this Wikipedia article:

https://en.wikipedia.org/wiki/List_of_moments_of_inertia

Supposedly it is for a solid sphere, or am I wrong?
For a solid sphere that page lists ##\frac{2}{5}mr^2##. With a radius of 0.5, that yields a moment of inertia of 0.1.

For a hollow sphere, that page lists ##\frac{2}{3}mr^2##. With a radius of 0.5, that yields a moment of inertia of 0.167

KataruZ98
If you don’t mind switching up, could we focus on the case of the cylinder hitting like a pendulum the sphere - as it was my original query? Anyway though I must thank you for the explanation, and I apologize for having sidetracked.

EDIT: we can also treat it as a 10m long, 2m wide cylinder if it’s seems more appropriate. I’d keep though an angular velocity of 1.57rad/s.

If you don’t mind switching up, could we focus on the case of the cylinder hitting like a pendulum the sphere - as it was my original query? Anyway though I must thank you for the explanation, and I apologize for having sidetracked.
OK. We will switch.

Now we have a vertical cylinder that is hinged at the top end. It strikes a sphere.

Does it strike the sphere exactly with the end of the rod? So that the end of the rid lines up with the equator line on the sphere? I assume so.

I assume that we still have a collision that results in the sphere being firmly attached to the pendulum.

If the pendulum/rod has length 1 meter then indeed the tangential velocity and angular velocity will be numerically equal at 1.57 meters per second and 1.57 radians per second respectively.

The moment of inertia of the rod about its top end will be properly assessed as 1.67 kg m^2 as you had it.

The moment of inertia of the sphere about its center line will be 0.1 kg m^2. But since its angular velocity is zero, that fact is not currently relevant.

We can calculate the pre-collision angular momentum about an axis at the top of the rod as ##I_c \omega_c = 1.67 \times 1.57 = 2.6219##.

Recall that I mentioned a "complication" a few posts back. We are computing total angular momentum here by adding the angular momenta of the two parts: ##L_c + L_s = I_c \omega_c + I_s \omega_s##.

But you can only add angular momenta if they are taken about the same reference axis.

Fortunately, the linear momentum of the sphere is zero. This means that its angular momentum is the same when taken about any other parallel axis. So we can shift to an axis at the top of the pendulum rod and still have a zero angular momentum for the sphere. So we dodged that bullet.

We had previously decided that angular momentum is conserved across the collision. However, there was that proviso about no impulsive external torques.

In this case we have the potential for an impulsive external torque.

We have that hinge at the top of the pendulum. When the bottom of the pendulum slams into the sphere, that retarding force on the bottom end will tend to force the top end of the pendulum into the hinge. The result is an impulsive force from hinge on pendulum. However, that force is exerted exactly at the point of our chosen reference axis. The moment arm is zero. So the resulting torque is zero. So we dodged that bullet as well.

So angular momentum about the pendulum's top end is conserved. So it should still be 2.6219 kg m^2/s.

How does that translate to motion in the fused lump -- pendulum plus sphere? What is the moment of inertia of the fused lump? What then is the angular velocity of the fused lump?

I leave those questions to you.

KataruZ98
fused lump
I can't see that stated in any of the posts. Might it not be wholly or partly elastic?
@KataruZ98, are you sure about the sphere's size? It says "small sphere", but the sizes you quote make it much larger in volume than the cylinder.

I can't see that stated in any of the posts. Might it not be wholly or partly elastic?
@KataruZ98, are you sure about the sphere's size? It says "small sphere", but the sizes you quote make it much larger in volume than the cylinder.
The thread title says "inelastic collision". I am assuming "completely inelastic" which would entail the pieces mergine and moving together as a result of the collision.

Yes, you are correct that partly elastic is not ruled out by anything in the OP.

My suspicion is that this is a made-up problem rather than something crafted in a textbook.

I also agree that the size of the target relative to the pendulum is troubling.