(C) Let $m = 0.1 \ kg$ be the mass of the sleeve,$R = 5 \ m$ be the radius,and $\omega = 2 \ rad/s$ be the angular velocity.For the sleeve to be in eq

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(C) Let $m = 0.1 \ kg$ be the mass of the sleeve,$R = 5 \ m$ be the radius,and $\omega = 2 \ rad/s$ be the angular velocity.For the sleeve to be in equilibrium relative to the rotating frame,the forces acting on it are gravity ($mg$ downwards) and the normal force $(N)$ from the wire.The horizontal component of the normal force provides the centripetal acceleration: $N \sin \theta = m \omega^2 (R \sin \theta)$.Since $\sin \theta \neq 0$ (not at top or bottom),we can cancel $\sin \theta$ to get $N = m \omega^2 R$.Substituting the values: $N = (0.1 \ kg) \times (2 \ rad/s)^2 \times (5 \ m) = 0.1 \times 4 \times 5 = 2.0 \ N$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

Medium

$A$ small $100 \ g$ sleeve $B$ can slide on a smooth,circular and rigid wire frame $A$ of radius $5 \ m$ placed in a vertical plane. The wire frame is rotating about its vertical diameter at $2 \ rad/s$. When the sleeve is brought to a particular angular position $\theta$ (measured from the vertical downward axis) other than the bottom and the top of the ring,the sleeve will not slide on the wire frame. Calculate the normal force $N$ (in Newtons) of interaction between the sleeve and the wire frame at this position.

A
$0.5$
B
$1.0$
C
$2.0$
D
$4.0$

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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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