One end of a thin, uniform rod is connected to a frictionless hinge as shown in

Question

One end of a thin, uniform rod is connected to a frictionless hinge as shown in (Figure 1) . The rod has a length of 0.9 m and a mass of 2 kg. It is held up in the horizontal position (=90) and then released.

Calculate the angular velocity of the rod at =90. rad/s

Calculate the angular acceleration of the rod at =90. rad/s2

Calculate the angular velocity of the rod at =60. rad/s

Calculate the angular acceleration of the rod at =60. rad/s2

Calculate the angular velocity of the rod at =0. rad/s

Calculate the angular acceleration of the rod at =0. rad/s2

Explanation / Answer

at @ = 90 deg ,

angular velocity = 0

using torque = I x alpha

(0.9 /2) x 2 x 9.81 = ( 2 x 0.9^2 / 3) alpha

alpha = 16.35 rad/s^2

--------------------------------------------

height of centre of mass comes lower,

h = L/2 ( 1 -cos60) = 0.225 m

using energy conservation,

mgh = I w^2 /2

2 x 9.81 x 0.225 = ( 2 x 0.9^2 / 3) w^2 /2

w = 4.04 rad/s

using torque = I x alpha

(0.9 /2) x 2 x 9.81 x sin60 = ( 2 x 0.9^2 / 3) alpha

alpha = 14.16 rad/s^2

----------------------------

height of centre of mass comes lower,

h = L/2 = 0.45 m

using energy conservation,

mgh = I w^2 /2

2 x 9.81 x 0.45 = ( 2 x 0.9^2 / 3) w^2 /2

w = 5.72 rad/s

using torque = I x alpha

(0.9 /2) x 2 x 9.81 x sin0 = ( 2 x 0.9^2 / 3) alpha

alpha = 0

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