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PhMichael
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1. Given data
Ball: I (moment of inertia), m (mass), R (Radius), V0 (Velocity, as shown ), [tex]\omega_{0}[/tex] (angular velocity, as shown).
The ball collides instantaneously with the wall and immediately after this collision, the ball ceases to rotate and only moves vertically upwards parallel to the wall, as shown in the figure. The question is to find this velocity.
2. The attempt at a solution
I think that this problem relates to conservation of angular momentum since the word "instantaneously" is explicitly written. Now, angular momentum is conserved about the instantaneous point between the ball and the wall, and there I can write that the total angular momentum about this point, after the collision is:
[tex] \vec{L}_{after} = -R \hat{x} \times mV \hat{y} = -mVR \hat{z} [/tex]
Now, how can I express the linear momentum before the collision?! I mean, about the ball's center of mass, it is obviously [tex] I \omega_{0} [/tex], however, is it the right expression to be used in the conservation equation? OR, should I used the parallel axis theorem and write [tex] (I + mR^{2}) \omega_{0} [/tex] instead?!
One more question: Is mechanical energy conserved in this problem?!
In an inelastic collision, energy is never conserved, however, here I'm dealing with one object hitting and infinitely heavy object so is it treated differently or in any case of inelastic collisions, I mustn't use the conservation of energy?!
Uploaded with ImageShack.us
1. Given data
Ball: I (moment of inertia), m (mass), R (Radius), V0 (Velocity, as shown ), [tex]\omega_{0}[/tex] (angular velocity, as shown).
The ball collides instantaneously with the wall and immediately after this collision, the ball ceases to rotate and only moves vertically upwards parallel to the wall, as shown in the figure. The question is to find this velocity.
2. The attempt at a solution
I think that this problem relates to conservation of angular momentum since the word "instantaneously" is explicitly written. Now, angular momentum is conserved about the instantaneous point between the ball and the wall, and there I can write that the total angular momentum about this point, after the collision is:
[tex] \vec{L}_{after} = -R \hat{x} \times mV \hat{y} = -mVR \hat{z} [/tex]
Now, how can I express the linear momentum before the collision?! I mean, about the ball's center of mass, it is obviously [tex] I \omega_{0} [/tex], however, is it the right expression to be used in the conservation equation? OR, should I used the parallel axis theorem and write [tex] (I + mR^{2}) \omega_{0} [/tex] instead?!
One more question: Is mechanical energy conserved in this problem?!
In an inelastic collision, energy is never conserved, however, here I'm dealing with one object hitting and infinitely heavy object so is it treated differently or in any case of inelastic collisions, I mustn't use the conservation of energy?!
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