A uniform rod of mass M = 245.0 g and length L = 56.0 cm stands vertically on a

Question

A uniform rod of mass M = 245.0 g and length L = 56.0 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 = 45.6° with respect to the vertical.
angular speed ?
vertical acceleration ?
normal force ?

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.

acceleration?

Explanation / Answer

mg sin43.6
moment of center of the rod= F*l/2=mg*sin43.6*0.49/2
moment of inertia of the rod=1/3ml^2
angular acceleration of the rod=moment/moment of inertia
=(mg*sin45.6*l/2)/(1/3*m*l^2)
=(3*g*sin45.6)/(2*l)
=(3*9.81*sin45.6)/(2*0.56)
=18.8rad/s^2

linear acceleration of the end=18.8*0.56=10.5m/s^2

vertical acceleration= 10.1*sin45.6=7.5m/s^2


the force pointing towards the contact point= mg cos43.6
the vertical component of this force(the normal force)=mg (cos45.6)
=0.245*9.81*cos45.6
=1.68N

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