Finding Angular Velocity and Center of Mass Velocity for a Free Rotating Rod

In summary, the homework statement states that a rod of mass m and length l will rotate around its center with a given impulse J. Find the angular velocity ω and the velocity of the center of mass v0.
  • #1
physicsisgreat
10
1

Homework Statement


I have a rod of mass m and length l on a table without any kind of friction. I give it an impulse J in any point of distance d from the center of the rod, parallel to the table and perpendicular to the rod.
Find the angular velocity ω and the velocity of the center of mass v0.

Homework Equations


Moment of inertia of the rod rotating around its center: I = m l2 / 12
L = I · ω

The Attempt at a Solution


From the impulse theorem:
J = ΔP = P'
I can calculate ω from the angular momentum relations:
L = d x J = I · ω
ω = d J / I = 12 d J / (m l2),
which is 0 if I hit the rod on its center and max if I hit it on d = l/2.
Now I fail to calculate v0 :P

Thank you in advance :)
 
Last edited:
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  • #2
Hello P.i.G., welcome to PF :)

Your relevant equation needs one or two colleagues: currently v0 doesn't feature there, ##\omega## doesn't, etc.

Your relevant equation also needs improvement. There is a factor missing.
 
  • #3
Should you not apply a couple of equal torques (one clock and one anti-clock) to stop any motion of the centre of mass?
 
  • #4
dean barry said:
Should you not apply a couple of equal torques (one clock and one anti-clock) to stop any motion of the centre of mass?
Not a good idea. 1. It doesn't work. 2. There is nothing that can be considered a cause for these ##\tau##. 3. The center of mass is not "stopped"
 
  • #5
Hi, I edited my post adding the factor missing and an equation about ω (which I actually used later in my attempt at a solution).
I cannot give another equation about v0 though, as it is exactly what I am looking for! :)
 
Last edited:
  • #6
Mmh, following the impulse-momentum theorem
J = ΔP
I would say that
v0 = J / m.

At the same time I wonder: can this be true? Can the velocity of the center of mass of the rod be independent from the point in which the impulse is applied?
 
  • #7
physicsisgreat said:
Mmh, following the impulse-momentum theorem
J = ΔP
I would say that
v0 = J / m.

At the same time I wonder: can this be true? Can the velocity of the center of mass of the rod be independent from the point in which the impulse is applied?
Yes, it's correct. It feels wrong, until you realize that to impart the same impulse further from the centre of mass of the rod you have to 'work' harder. The rod tends to swing out of the way, so takes less momentum off the impacting object. To achieve the same imparted momentum the impacting object has to start with more momentum.
 
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  • #8
Thank you very much haruspex! :)
 

Related to Finding Angular Velocity and Center of Mass Velocity for a Free Rotating Rod

What is impulse on a free rod?

Impulse on a free rod refers to the change in momentum of a rod that is free to move and is subjected to an external force for a certain period of time.

What factors affect the impulse on a free rod?

The impulse on a free rod is affected by the magnitude of the external force applied, the duration of the force, and the mass and velocity of the rod.

How is impulse related to the motion of a free rod?

According to Newton's Second Law of Motion, the impulse acting on a free rod will result in a change in its momentum, which in turn will cause the rod to either accelerate or decelerate depending on the direction of the force.

What is the formula for calculating impulse on a free rod?

The formula for impulse is given by the product of the external force and the time for which it acts, given by the equation I = F x Δt. This can also be represented by the area under the force-time graph.

How is impulse different from force?

Force refers to the push or pull exerted on an object, while impulse refers to the change in momentum caused by the force. In other words, force acts for a short duration of time, while impulse takes into account the duration of the force.

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