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A uniform rod of mass M and length L stands vertically on the edge of a frictionless table. A blow of impulse J is applied horizontally at the lower end of the rod.

a) What are the initial horizontal velocity of the CM of the rod relative to the table, and the velocity of the lower end of the rod relative to the table?

b) How far does the CM of the rod move horizontally in the time the rod rotates one revolution?

w = angular velocty of rod.

I = moment of inertia of rod

Vcm = velocity of CM of rod

Vtan = velocity of end of the rod

OK, I place the origin at the bottom of the rod so that angular momentum is conserved.

Iw = (L/2)MVcm

[(1/12)ML^2]w = (L/s)MVcm

Impulse in x direction is equal to the change of momentum in the x direction so: Vx = Vcm = J/M

w = 6J/(ML) so Vtan = 3J/M => My textbook says it's 4J/M

Plug w back into the equation above and you get.

Vcm = J/(2M) => Textbook says it's J/M

b) Using the above I get piL/6

I've thought about the problem pretty carefully and can't see why my answers don't agree with the book.

Can someone please check them.

Thanks.

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# Rotating Rod - Please check answer.

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