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 the fixed point. The tension in the string is

For the conical pendulum: ω = π 2 rev / s = 4 rad / s . cos θ = ω 2 l g = 16 × 1 10 = 8 5 . Vertical balance: T cos θ = m g , so T = cos θ m g = 8 5 0.1 × 10 = 5 8 N .

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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 the fixed point. The tension in the string is

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Solution:

For the conical pendulum: $ω = \frac{2}{π} rev / s$
$= 4 rad / s$ .
$cos θ = \frac{g}{ω ^{2} l} = \frac{10}{16 \times 1} = \frac{5}{8} .$
Vertical balance: $T cos θ = m g$ ,
$so T = \frac{m g}{cos θ} = \frac{0.1 \times 10}{\frac{5}{8}} = \frac{8}{5} N .$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

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