Sir please solve question no 5
Asked by shobhit
23rd April 2018,
10:40 PM
Answered by Expert
Answer:
when the loop roates about a vertical axis passing through its diameter, the bead moves in a circular motion with its plane of motion perpendicular to the axis of rotation. Hence the bead experiences two forces (1) due to its own weight mg and (2) due to circular motion, centri petal force mω2r, where r is the radius of circular motion of the bead as shown in figure.
when mg > mω2r the bead comes to the lowest point of the loop. when the angular frequency is increased and the condition mg = mω2R is satisfied, then the bead stays at the diametrical position of the loop as shown in figure.
hence the condition, mω2R > mg or ω > √(g/R) for the bead to stay at the diametrical position of the loop otherwise it will fall down to the lowest point of loop.
when the condition ω > √(g/R) is satisfied the radius vector makes an angle 90° at centre of the loop with the vertical axis.
Answered by Expert
24th April 2018,
6:13 AM
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