A uniform rod of mass M= 2 kg and length L = 3 m is suspended at 35 degrees belo
ID: 2216605 • Letter: A
Question
A uniform rod of mass M= 2 kg and length L = 3 m is suspended at 35 degrees below the horizontal ceiling . The supporting massless string at its right end is perpendicular to the rod, and the pivot at O is frictionless. What is the magnitude of the torque due to the weight of the rod about an axis perpendicular to the screen and passing through O? The string is cut and the rod falls from rest. Immediately after the string is cut what is the angular acceleration a of the rod in units of radians/ sec^s where torque is due to the mass of the rod about the O in N-m as in the question above?
A uniform rod of mass M= 2 kg and length L = 3 m is suspended at 35 degrees below the horizontal ceiling . The supporting massless string at its right end is perpendicular to the rod, and the pivot at O is frictionless. What is the magnitude of the torque due to the weight of the rod about an axis perpendicular to the screen and passing through O? The string is cut and the rod falls from rest. Immediately after the string is cut what is the angular acceleration a of the rod in units of radians/ sec^s where torque is due to the mass of the rod about the O in N-m as in the question above?Explanation / Answer
torque = rfsin
f = mg, r = distance between point O and point of action of force mg(at center of rod)
=> r = l/2, = 35.
=> = 16.86 kgm^2/s^2 about axis passing through O
the angular acceleration of the rod, mmediately after the string is cut = /I
we have
I = ml^2/3 = 6
=> angular acceleration = 2.81 radians/ sec^s
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