A thin, uniform ring of radius π4m and mass 83 kg is fixed with two rods of length π4m each. The system is pivoted about the centre of the ring P and made to rotate in the vertical plane. When the ring is in its lowest position (i.e. the position shown in the figure), the angular velocity of the system is 4 rad/s. At this instant, the force exerted by one of the rods on the ring is _______(in N) [Assume that the force on the ring by a rod is always purely along the length of the rod, take g=10 m/s2 ]

Question

A thin, uniform ring of radius  π4m and mass 83 kg  is fixed with two rods of length  π4m each. The system is pivoted about the centre of the ring P  and made to rotate in the vertical plane. When the ring is in its lowest position (i.e. the position shown in the figure), the angular velocity of the system is 4 rad/s. At this instant, the force exerted by one of the rods on the ring is _______(in N) [Assume that the force on the ring by a rod is always purely along the length of the rod, take g=10  m/s2  ]

Accepted Answer

Let the mass of the ring be m, let its radius be r, and let the angular velocity of the system when the ring is at its lowest position be $ω$
Let the force exerted by any one rod be T
Due to symmetry at the instant when the rod is at the lowest position, the force exerted by the other rod is also T
Since the centre of mass of the ring is moving in circular path of radius $\frac{2 R}{π}$
$2 T \cos 30 ° - m g = m ω^{2} (\frac{2 R}{π}); T = 144 N$

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