How Does Firing a Cannonball in Space Affect an Astronaut's Rotation?

[Satvik Pandey]

Homework Statement

An astronaut in space weights 100 kg and is holding a cannon which can shoot a 10kg cannonball. The astronaut is moving at 10 m/s. The astronaut wishes to fire the cannonball such that he turns the maximum possible angle. Find this angle in degrees.

Homework Equations

The Attempt at a Solution

If we consider astronaut and cannon ball to be a system then on that system no external force and torque acts on it so we can use equations for conservation of momentum and angular momentum and energy.

I have a confusion, when astronaut fires the cannon an impulsive force(due to recoiling) acts on it, due to which the astronaut rotates. Also as no other torque acts on it afterwards so I think it will go on rotating forever.

However I proceeded. I think for turning through maximum angle I think the initial kinetic energy of should get converted to the rotational kinetic energy. For finding rotational energy I need to calculate moment of inertia of astronaut first. But how? Also we can not consider as a point mass. Can we?

 

[Orodruin]

I believe what the problem intends to ask is the angle for which the direction of travel after the firing deviates the most from the direction of travel before (naturally in the frame where the astronaut is traveling at 10 m/s). Assuming the astronaut fires the cannon in such a way that the force is directed through the centre of mass, the astronaut will not actually gain any angular momentum.

 

[gneill]

Homework Statement

An astronaut in space weights 100 kg and is holding a cannon which can shoot a 10kg cannonball. The astronaut is moving at 10 m/s. The astronaut wishes to fire the cannonball such that he turns the maximum possible angle. Find this angle in degrees.

Homework Equations

The Attempt at a Solution

If we consider astronaut and cannon ball to be a system then on that system no external force and torque acts on it so we can use equations for conservation of momentum and angular momentum and energy.

I agree with the conservation of momenta, but there's another repository of energy in the system that would have to be taken into account here if you plan to use conservation of energy. Think about what makes a canon fire.

However, I think you might be able to get away without having to apply conservation of energy here.

I have a confusion, when astronaut fires the cannon an impulsive force(due to recoiling) acts on it, due to which the astronaut rotates. Also as no other torque acts on it afterwards so I think it will go on rotating forever.

However I proceeded. I think for turning through maximum angle I think the initial kinetic energy of should get converted to the rotational kinetic energy. For finding rotational energy I need to calculate moment of inertia of astronaut first. But how? Also we can not consider as a point mass. Can we?

I think the "rotation" here is meant to be the angle through which the trajectory changes, not a rotation of the spacecraft itself.

<<EDIT: Ah! Orodruin beat me to it!>>

 

[Orodruin]

Ah! Orodruin beat me to it!

Only because I did not take the time to draw a picture ...

 

[gneill]

By the way Satvik, were you given information about the speed for the canon ball after firing?

 

[Satvik Pandey]

By the way Satvik, were you given information about the speed for the canon ball after firing?

Thanks for replying.
No, velocity of canon is not given.

 

[Satvik Pandey]

In order to use conservation of momentum I should know the direction and magnitude of the speed of the canon. Should I assume that canon is fired at angle ##\theta## (please refer to figure) with the horizontal.

 

[gneill]

No, velocity of canon is not given.

Well that makes things easier Suppose the speed of the canon ball after firing is such that its momentum is exactly equal to the initial momentum of the astronaut plus canon ball before firing. What does that tell you about the new velocity of the astronaut? What if the speed of the canon ball were even larger? What then is the range of possible turning angles?

 

[gneill]

In order to use conservation of momentum I should know the direction and magnitude of the speed of the canon. Should I assume that canon is fired at angle ##\theta## (please refer to figure) with the horizontal.
View attachment 74777

The recoil angles of the two objects will not necessarily be equal. You can say though that by conservation of momentum the net momentum in the X and Y directions are separately conserved.

 

[Satvik Pandey]

Well that makes things easier :) Suppose the speed of the canon ball after firing is such that its momentum is exactly equal to the initial momentum of the astronaut plus canon ball before firing. What does that tell you about the new velocity of the astronaut?

Then its speed should be zero.

What if the speed of the canon ball were even larger?

Then the velocity of astronaut should be opposite to the velocity of the canon.

What then is the range of possible turning angles?

It is ##0 to 180##?

 

[Satvik Pandey]

The recoil angles of the two objects will not necessarily be equal. You can say though that by conservation of momentum the net momentum in the X and Y directions are separately conserved.

I was too thinking that. But if we assume that the velocity vector of canon makes angle ##\phi## with the horizontal and velocity of astronaut makes ##\theta## with the horizontal then by using conservation of momentum in X and Y direction and by conservation of energy I can make three equations but there are 4 variables.

 

[gneill]

It is ##0 to 180##?

Right.

If the canon ball's velocity is not limited then the turning angle can be anything you want, up to and including 180°.

 

[Satvik Pandey]

Right.

If the canon ball's velocity is not limited then the turn angle can be anything you want, up to and including 180°.

So the answer is 180?

 

[gneill]

I was too thinking that. But if we assume that the velocity vector of canon makes angle ##\phi## with the horizontal and velocity of astronaut makes ##\theta## with the horizontal then by using conservation of momentum in X and Y direction and by conservation of energy I can make three equations but there are 4 variables.

Conservation of energy is a dubious thing to use here since the firing of the canon is analogous to an inelastic collision happening in reverse. Kinetic energy is not conserved across an inelastic collision.

If the canon ball's speed is not specified then perhaps a better approach is to simply assume that the firing of the canon imparts a change of momentum ΔP of some magnitude and direction. The canon ball carries off momentum ΔP and the astronaut suffers a change of -ΔP (since momentum is conserved).

Without knowing what the magnitude of ΔP can be you can't really say anything more specific.

 

[gneill]

So the answer is 180?

What do you think?

If the complete problem statement is as you've presented it then you can't determine a particular angle other than by considering extremes of the unspecified variables.

You could come up with a set of expressions for different cases where the |ΔP| is less than, equal to, or greater than that of the initial momentum. But that seems like a lot of work for such a loosely phrased problem. How does this problem's complexity compare to others in the same problem set?

 

[Satvik Pandey]

Answer of this question is 27.
You can find this problem on (https://brilliant.org/community-problem/astronaut-in-space/?group=hjkNbxhQsI1z&ref_id=455645 )

 

[gneill]

Answer of this question is 27.
You can find this problem on (https://brilliant.org/community-problem/astronaut-in-space/?group=hjkNbxhQsI1z&ref_id=455645 )

The problem statement does not contain enough information to arrive at a specific answer. The problem is flawed.

 

[Satvik Pandey]

Thank you gneill for helping. Sorry I posted incomplete question.:s

 

[Orodruin]

Don't be sorry, it is the fault of brilliant.org (or rather people posting problems there). I generally do not trust those problems as it seems nobody is checking them. I still have a problem with some ants on a tetrahedron in vivid memory ... Solvable, but not with the alleged answer of the poster, and definitely not using maths as simple as the poster imagined.

 

[Satvik Pandey]

Don't be sorry, it is the fault of brilliant.org (or rather people posting problems there). I generally do not trust those problems as it seems nobody is checking them. I still have a problem with some ants on a tetrahedron in vivid memory ... Solvable, but not with the alleged answer of the poster, and definitely not using maths as simple as the poster imagined.

Yes I remember the problem 'Ants on tetrahedron'.
Thank you Orodruin and gneill for helping.
Congratulation Orodruin. You are a staff now.:w

 

[haruspex]

Thank you gneill for helping. Sorry I posted incomplete question.:s

The question looks wrong straight away. You have three given quantities which, dimensionally, are M, M, LT^-1. The desired answer is dimensionless. There is no way to involve the speed and get a dimensionless answer, so it must be irrelevant. (There remains the possibility that the two masses alone suffice, but as has been shown in this thread that does not work either.)