Question
Explain your answer
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6.00 m long, weighs 12.0 N, and rotates at 250 rev/min. (a) Calculate the rotational inertia of the rod about the axis of rotation. 14.69 kg m (b) hat is the magnitude of the angular momentum of the rod about that axis W x kg m /s the rotation axis is shifted to one end? How is the about its center. How do you calculate the rotational inertia when Table gives the rotational inertia of a rod 10-2 angular momentum of a rigid body rotating about a fixed axis related to the rotational inertia and the angular speed?
Explanation / Answer
a)inertia= (ml^2)/3=((12/9.8)*6*6)/3=14.69 kg-m^2
b) angular momentum = I*w= (14.69)*(250*2*3.14)/60=384.38 kg-m^2/s