A rod of with mass .25 kg and a length of .5 m stands vertically on a horizontal

Question

A rod of with mass .25 kg and a length of .5 m stands vertically on a horizontal table. It is released from rest and falls. If the rod falls onto the table and no slipping is involved, find the linear acceleration of the end point of the rod when it makes contact with the surface of the table.

I know that linear acc. = sqrt((centrip acc)^2 + (tangential acc)^2), or a = sqrt(ac^2 + at^2), but for some reason I cannot get the correct answer. It is supposed to be 25.487 m/s^2. Could someone please explain how this is done.

Explanation / Answer

there are two component of acceleration here. the tangential one. a/L=mg*L/2 /(mL^2/3) so a=g*3/2. ------- conservation of energy. mg*L/2=w^2*(mL^2/3*2) so that w^2=3g/L. so that w^2*L=3g. so that a=sqrt(a^2+(w^2L)^2)=

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