Answer in units of rad / s 2 . A uniform 4 kg rod with length 20 m has a frictio

Question

Answer in units of rad/s2.

A uniform 4 kg rod with length 20 m has a frictionless pivot at one end. The rod is a frictionliess pivot at one encd. Ihe rod 1s released from rest at an angle of 29° beneath the horizontal. Note: The moment of inertia of a rod about ass is 12 m 12, where m is the the center of mass is mI2, where m is the mass of the rod and L is the length of the rod The mo ment of inertia of a rod about either end is m L2 3 The acceleration of gravity is 9.8 m/s 29° 20 Note: The rod is initially at rest at 29 below the horizontal.

Explanation / Answer

We know that the angular acceleration for any body is given as:

Angular acceleration = Torque / Inertia

Therefore we need to determine the inertia of the rod about the pivot and the torque so as to determine the angular acceleration of the rod.

Now, as mentioned in the question, the inertia of the rod about the pivot is ML2 / 3

Putting in the values for M and L, we get the inertia as:

I = 4 x 400 / 3 = 1600 / 3 kg-m2

Also, torque about the pivot at the given moment would be: MgCos29 x L/2

Torque = 4 x 9.8 x Cos29 x 10 = 342.851 N-m

Therefore, angular acceleration = 1600 / 3 x 342.851 = 1.556 rad/s

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