A string of length 1 m is fixed at one end with a bob of mass 100 g and the string makes π 2 rev/s around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

For the conical pendulum: tan θ = g ω 2 r , with r = l sin θ , so cos θ = ω 2 l g . ω = π 2 rev / s = 4 rad / s , so cos θ = 16 × 1 10 = 8 5 and θ = cos − 1 ( 8 5 ) .

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PhysicsCircular MotionDynamicsMedium2 minQB

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A string of length 1 m is fixed at one end with a bob of mass 100 g and the string makes $\frac{2}{π}$ rev/s around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

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For the conical pendulum: $tan θ = \frac{ω ^{2} r}{g}$ , with $r = l sin θ$ ,
so $cos θ = \frac{g}{ω ^{2} l}$ .
$ω = \frac{2}{π} rev / s$
$= 4 rad / s$ ,
so $cos θ = \frac{10}{16 \times 1}$
$= \frac{5}{8}$
and $θ = cos^{- 1} (\frac{5}{8})$ .

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

⏱ 2mℹ️ Source: QB

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