Rotational MotionHard
Question
A 2m long rod of negligible mass is free to rotate about its centre. An object of mass 5 kg is threaded into the rod at a distance of 50 cm from its end in such a way that the object can move without friction. The rod is then released from its horizontal position. The speed of the rod′s end in the rod′s vertical position is (in m/s)
Options
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Solution
Since friction and the rod′s mass is negligible, the only force acting on the object is gravitational force, therefore the object undergoes free-fall.
The object-moves a distance of h =
m until it drops off from the rod.
Its velocity at this moment v =
The object′s velocity perpendicular to the rod equals to the velocity of the rod′s end at the moment when the object leaves the rod. After this moment the rod′s end maintains its speed, so in vertical position its
speed = v cos60o =
m / s
The object-moves a distance of h =
m until it drops off from the rod.Its velocity at this moment v =
The object′s velocity perpendicular to the rod equals to the velocity of the rod′s end at the moment when the object leaves the rod. After this moment the rod′s end maintains its speed, so in vertical position its
speed = v cos60o =
m / sCreate a free account to view solution
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