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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.

a/g =

Explanation / Answer

a) there are three force forces acting on rod :

gravitational force, frictional force and normal force due to table.

b) the @ = 41.6 deg

then centre of mass of rod comes to h height down.

h = L/2 ( 1 - cos41.6) = 0.2522L/2 = 0.0694 m

using energy conservation,

mgh = Iw^2 /2

m x 9.81 x 0.0694 = ( m L^2 /3) w^2 /2

4.08 = (0.55^2)w^2

w = 3.67 rad/s ...........Ans

using torque = I x alpha

(L/2 x mgsin41.6) = (m L^2 /3) alpha

alpha = (3 x 9.81 x sin41.6) / ( 2 x 0.55) = 17.76 rad/s

acc a = alpha* L = 17.76 x 0.55 =9.77 m/s^2 ......Ans

for centre of mass,

a = alpha * L/2 = 4.885 m/s^2

using mg - N = ma

0.225 x9.81   - N = 0.225 x 4.885

N = 1.11 N

c) at that moment ,

torque = I x alpha

L/2 x mg = ( m L^2 / 3) alpha

alpha = 26.75 rad/s^2

a = alpha*L = 26.75 x 0.55 = 14.71 m/s^2

a/g = 14.71/9.81 =1.50

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