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Considering a rod AB lying on a frictionless horizontal surface, it's mass being M and length being L. A mass m strikes the rod at end A with a velocity v and perpendicular to AB and the mass comes to rest immediately after collision.

Now, here I have two questions.

  1. I know the center of percussion of a hinged rod is the point at which if a horizontal impulsive force is applied, there is no reaction at the pivot. But, in my case, there is no pivot (the rod is unhinged). About which axis does the rod start rotating just after collision? Is the about the axis passing through the center of mass perpendicular to the place of the horizontal surface, or the center of percussion, perpendicular to the plane of the surface?

Is the center of percussion at rest and the axis of rotation immediately after the collision?

  1. Would the axis of rotation, or the distance of the center of percussion of rod (2L/3) change if the mass m stuck to the rod?
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  • $\begingroup$ Are you familiar with the laws of conservation of linear momentum and conservation of angular momentum? If so, have you applied them before? $\endgroup$ Jun 25, 2021 at 16:26
  • $\begingroup$ @Chemomechanics Yes I have. But here angular momentum can be conserved about any axis as there is no net torque. I know how to solve these questions, I only have the doubts regarding the center of percussion $\endgroup$
    – Techie5879
    Jun 25, 2021 at 16:27
  • $\begingroup$ As no pivot exists in this problem, I'm not sure why the center of percussion is relevant. $\endgroup$ Jun 25, 2021 at 16:36

1 Answer 1

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An object rotates about its center of mass for any unhinged collision, and its center of mass translates in the direction of of the impulse vector. If the point mass sticks to the rod, it is the pointmass-rod system that rotates, so the center of mass is the center of the pointmass-rod system.

If the collision is known to be elastic, conserve kinetic energy, shared by the pointmass, rod (translation) and rod (rotation). If the collision is inelastic or unknown, you can't conserve kinetic energy, and will need to calculate the final energy of the system after conserving momentum and angular momentum.

Conserve momentum, shared by the pointmass, rod (translation).

Conserve angular momentum shared by the pointmass (equal to mvL/2 for right-angle collision), rod (rotation). The rod (translation) has zero angular momentum because the rod's center of mass is, of course, zero distance from the rod's center of rotation.

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