End A of a rod AB is being pulled on the floor with a constant velocity v0 as shown. Taking the length of the rod as l, at an instant when the rod makes an angle 37∘ with the horizontal, calculate
The velocity of the CM of the rod
A
57v0 at tan−143 below horizontal
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B
57v0 at tan−134 below horizontal
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C
56v0 at tan−134 below horizontal
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D
45v0 at tan−134 below horizontal
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Solution
The correct option is D45v0 at tan−134 below horizontal The center of mass of the rod will be at a distance of l2.
where, ICR = insterntenious center of rotation
Velocities at point A:
Along the length Vcm=V0cosθ (θ=37o) Vcm=V0×45 Vcm=45V0