A uniform rod of mass M = 225.0 g and length L = 47.0 cm stands vertically on a

Question

A uniform rod of mass M = 225.0 g and length L = 47.0 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.)

(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?=49.2?with respect to the vertical.

(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 withg.

=

angular speed vertical acceleration normal force A uniform rod of mass M = 225.0 g and length L = 47.0 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?=49.2?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

A uniform rod of mass M = 225.0 g and length L = 49.0 cm stands vertically on a horizontal table. It is released from rest to fall. 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 = 43.6° with respect to the vertical. 1,vertical acceleration ? 2, normal force ?

F=mg sin43.6moment of center of the rod= F*l/2=mg*sin43.6*0.49/2moment of inertia of the rod=1/3ml^2angular acceleration of the rod=moment/moment of inertia=(mg*sin43.6*l/2)/(1/3*m*l^2)=(3*g*sin43.6)/(2*l)=(3*9.81*sin43.6)/(2*0.49)=20.7rad/s^2linear acceleration of the end=20.7*0.49=10.1m/s^2vertical acceleration= 10.1*sin43.6=7.0m/s^2the force pointing towards the contact point= mg cos43.6the vertical component of this force(the normal force)=mg (cos43.6)^2=0.225*9.81*(0.724)^2=1.16N

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