A uniform rod of mass M = 225.0 g and length L = 55.0 cm stands vertically on a

Question

A uniform rod of mass M = 225.0 g and length L = 55.0 cm stands vertically on a horizontal table. It is released from rest to fall.

(a) What forces are acting on the rod? (Select all that apply.)

force of gravitynormal forcefriction force

(b) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle = 41.6° with respect to the vertical.

angular speed =

vertical acceleration=

normal force=

(c) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table.
a =  

Compare it with g.

Explanation / Answer

Examine the rod at the point it has rotated through an angle of

the forces acting on the rod are gravity at the center of mass of the rod and vertically downwards
and centripetal radially from the base to the tip which is m*r*^2
and the reaction force at the point of contact with the table which will have a horizontal component which is friction and as long as it is not overcome (no slipping) react to the horizontal component of the centripetal force cos with a normal force in the vertical equal to the weight of the rod minus the vertical component of the centripetal. sin

look at energy
m*g*L/2 *( 1-cos) is the loss on PE
and the gain in KE is
.5*I*^2+.5*m*v^2
and *L/2=v
and I=m*L^2/3
substituting
=(12*g*(1-cos)/(7*L)^.5
use the chain rule to find
d/d

which is equal to
d/dt * dt/d

or
/
Therefore
=*{[(12*g*(1-cos)/(7*L)^.5]d}
and a=*r

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