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NANDHU001

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mass of rod= x ,mass of ball=y,angular velocity of rod=v,arm distance from center where collision takes place=d.

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- Thread starter NANDHU001
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In summary, a rotating rod can have no linear momentum before impact, but gains linear momentum after impact if a ball of the right mass and velocity is hit at the same time.

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NANDHU001

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mass of rod= x ,mass of ball=y,angular velocity of rod=v,arm distance from center where collision takes place=d.

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- #2

mfb

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The rod has to be attached to something in some way to do that, otherwise linear momentum is not conserved.At an instant a rigid ball is introduced near the rod so that it collides with the rod and the stick comes to rest(rotationally and translationally).

No, as some momentum has to be exchanged with the environment.Then can the ball be said to posses the equivalent linear momentum of the rotational(angular) momentum possessed by the rod initially.

Probably with conservation of angular momentum, assuming the rod is fixed to rotate around its center of mass.Also if the parameters are as below please tell me how to calculate the final linear momentum of the ball.

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sophiecentaur

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The rod could be spinning in space with no linear momentum (in the observer's frame). A situation can arise where a ball of the right mass and velocity (linear momentum but with angular momentum about the axis of spin of the rod, on impact) can strike the end of the rod and leave the rod not rotating but moving off in one direction, with the mass bouncing back - or sticking to the rod, depending on the details. Linear momentum and angular momentum would be preserved.

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AlephZero

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sophiecentaur said:A situation can arise where a ball of the right mass and velocity (linear momentum but with angular momentum about the axis of spin of the rod, on impact) can strike the end of the rod and leave the rod not rotating but moving off in one direction, with the mass bouncing back - or sticking to the rod, depending on the details. Linear momentum and angular momentum would be preserved.

The OP's question says the rod has no linear momentum before and after the impact.

I don't think that is possible unless the rod is rotating about a fixed pivot. The rod has no linear momentum before the impact. If it has no linear momentum after impact, the total impulse on the rod must be zero. The ball hitting the rod gives a non-zero impulse, so you another impulse on the rod to balance it - for example the reaction at the pivot.

So the only solution is the "trick answer" that the rotation speed of the rod before the impact was zero, and you can put the ball anywhere you like.

- #5

NANDHU001

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A rod is spinning around its center of mass with no linear velocity with respect to the observer.

Is it possible for a ball (placed so that it collides with the rod) to make the linear as well as angular momentum of the rod ZERO( The collision is neither elastic nor the ball sticks to the rod). If so how can the linear momentum of the ball be calculated. And can the final linear momentum of the ball be said to be the equivalent linear momentum of the angular momentum possessed by the rod.

- #6

DrZoidberg

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It's possible if the rod collides with two balls at the same time.

Say the rod is floating in space and rotating around it's center but otherwise isn't moving relative to the observer.

If both ends of the rod collide with a ball at the same moment and the two balls have the right mass, the rod will stop moving alltogether. It's angular momentum will be 0 and the two balls will each have gained a linear momentum.

However - the angular momentum is still conserved. Remember the definition of angular momentum. L = r x p. An object doesn't need to rotate to have an angular momentum. r is the position of the object and p is it's linear momentum. It doesn't matter relative to what point in the universe you measure r, as long as you use the same point for all measurements the total angular momentum of all particles combined will stay the same, as will the linear momentum. And yes, the magnitude of the linear momentum of both balls combined will be equal to the magnitude of the angular momentum the rod originally had.

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sophiecentaur

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NANDHU001 said:

A rod is spinning around its center of mass with no linear velocity with respect to the observer.

Is it possible for a ball (placed so that it collides with the rod) to make the linear as well as angular momentum of the rod ZERO( The collision is neither elastic nor the ball sticks to the rod). If so how can the linear momentum of the ball be calculated. And can the final linear momentum of the ball be said to be the equivalent linear momentum of the angular momentum possessed by the rod.

If linear and angular momentum are to be conserved then the answer has to be NO. The ball starts with some linear momentum and it can hardly interact with the rod without transferring some of that.

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NANDHU001

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Please note the word '

- #9

jtbell

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If the rod's axis of rotation (center of mass) is not held effectively stationary by being fastened to the Earth or some other very massive object, the answer is "no."

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.

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sophiecentaur

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jtbell said:

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.

I agree. The problem seems to keep shifting in subtle ways, I think but the basic conservation laws must apply. Changing the model slightly, it would be easy to bring a rotating bat to a halt by striking an appropriate ball. This was what I thought we were heading for aamof.

- #11

NANDHU001

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jtbell said:

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.

Thanks, will there be any change in the rotational velocity of the rod after the event.

If so, tell me how to calculate it.

- #12

sophiecentaur

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NANDHU001 said:Thanks, will there be any change in the rotational velocity of the rod after the event.

If so, tell me how to calculate it.

I'd suggest you write out two equations, one for linear and one for angular momentum. I would imagine that there will be a suitable ratio of masses that will give you whatever outcome you specify. The variables are the position of the impact and the ratio of masses, which is two unknowns, which two equations should enable you to find.

Angular momentum is a measure of the rotational motion of an object, while linear momentum is a measure of the translational motion of an object.

Angular momentum can be converted to linear momentum through the application of an external torque, which causes the object to rotate and therefore have both angular and linear momentum.

The conversion of angular momentum to linear momentum is affected by the mass and velocity of the object, as well as the direction and magnitude of the external torque.

No, the conversion of angular momentum to linear momentum is only possible when there is an external torque acting on the object. In the absence of torque, angular momentum remains constant.

The conversion of angular momentum to linear momentum is seen in various everyday activities such as throwing a frisbee, riding a bicycle, and swinging a bat. It is also important in understanding the movement of planets and other celestial bodies in space.

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