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When edge of the pencil hits the floor, normal and friction forces at the hitting point (point A) will be very large, so we can ignore the other forces.
Because these two forces only works on the edge, we can conclude that angular momentum of the pencil relative to point A is constant.

Assume before it hits the floor, its CM velocity is v, and v' after hitting the floor. See the picture below.Because mass of the pencil is concentrated at its center, it has zero moment inertia.
Angular momentum relative to point A before and after hitting the floor:





Hence,



Now, we have to calculate the x and y-component of the velocity before and after hitting the floor.






Impulse from the floor change the velocity from v to v', for x and y-components.
For y-component, there is a normal force from the floor, so,





And for x-component, there is a friction force from the floor toward negative x-axis, so,





Because , we can also say that:


Hence,





If you have any mistake, suggestion, or comment, just leave a comment below.

Solution - Rolling PencilSocialTwist Tell-a-Friend

4 comments

Anonymous said... @ February 4, 2009 at 5:57 PM

I doubt if u can conserve angular momentum abt point of contact. because there are two different points of contact .before hitting the floor normal rxn is not passing through A . so net torque will not b zero.

Correct me if m wrong .

Copycat91 said... @ February 4, 2009 at 9:56 PM

I c.
When it hits the floor, the force exerted on point A is very large, because it needs a great force to stop point A at a short time.
So we can ignore other forces.

Anonymous said... @ February 6, 2009 at 2:08 AM

hmmmm got it . Thnx a lot for ur suggestion.

Anonymous said... @ November 14, 2009 at 12:41 AM

What about the weight? it should be integral of (N-mg)dt = m(vy' - vy)

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