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A molecule consist of two atoms each of mass 'm' and separated by a distance
of 'd' If 'K' is the average rotational K.E. of the molecule at particular temperature then its angular frequency is….

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Explanation

The M.I of the molecule = $ M \left( { d \over 2 } \right) + m \left( { d \over 2} \right)^2 $ $ I = 2m \left( { d^2 \over 4 } \right) = { md^2 \over 2 } $ The Rotational K.E. of the moldule $ (K) = { 1 \over 2} I \omega^2 $ $ \omega = \sqrt { 2K \over I } $

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A car is moving with a constant speed the wheels of the car make 120 rotations per minute the breaks are applied and the car comes to rest in 8 sec how many rotation are completed by the wheels before the car is brought to rest.

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Explanation

$ \omega_0 = { 2 \pi \times 120 \over 60} = 4 \pi rad /sec $ Now $ \omega = \omega_0 + \alpha t $ Total angle descrited in 8 second is $ \theta = w_0 t + { 1 \over 2} \alpha t^2 $

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The angular momentum of a wheel changes from 2L to 5L in 3 seconds what is the magnitudes of torque acting on it?

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Explanation

$ \tau = { dL \over dt} = { 5L - 2L \over 3 } ={ 3L \over 3 } + L $

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A uniform disc of mass 500kg and radius 2 m is rotating at the rate of 600 r.p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of 100 sec ?

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Explanation

$ For disc I = { MR^2 \over 2} = { 500 \times 4 \over 2 } $ $ angular speed = \omega = { 2 pi \times 600 \over 60 } $ so angular momentum $ L = I \omega$ final angular momentum in opposite direction $ = -1000 \times 20 \pi kgm^2 / sec $ So change in angular momentum = $ \triangle = 2 \times 1000 \times 20 \pi kg m^2 / sec $ $ \tau = { dL \over dt } = { 2 times 1000 \times 20 \pi \over 100 } = 400 \pi N.m$

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The moment of inertia of a meter scale of mass 0.6kg about an axis perpendicular to the scale and passing through 30 cm position on the scale is given by (Breath of scale is negligible).

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Explanation

By Parallel axis therom solved problem

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How much constant force should be applied tangential to equator of the earth to stop its rotation in one day ?

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Explanation

$ \omega_1 = 2 \pi rad/day and \omega_2 = 0 , t =1 day $ $ \alpha = { w_2 - w_1 \over t } $ Torque required to stop the earth = T = $ I \alpha $ = F.R $ F = { I.\alpha \over R } $

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A constant torque of 1500 Nm turns a wheel of moment of inertia 300 kg m2 about an axis passing through its centre the angular velocity of the wheel after 3 sec will be….. rad/sec

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Explanation

$ \tau = I \alpha = I { d \omega \over dt} $

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A mass m is moving with a constant velocity along the line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin

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Explanation

$ L = Momentum \times perpenlicular distance between point of rotation and line of action $ = m.V.y all remain constant L = remaing constant

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A body is rolling down an incline plane. If the rotational K.E. of the body is 40% of its translational K.E. then the body is ….

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Explanation

R.K.E = 40 /100 T.K.E $ {1 \over 2} I \omega^2 = { 4 \over 10 } \times { 1 \over 2} mv^2 = { 1 \over 5} mv^2 $ $ \therefore { 1 \over 2} mk^2 \times {V^2 \over r^2 } = { 2 \over 5} mv^2 $

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A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is

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Explanation

$ Total energy E = { 1 \over 2} I \omega^2 + {1 \over 2} mv^2 = { 1 \over 2} \times { 2 \over 5 } mr^2 \omega^2 + { 1 \over 2} mr^2 \omega^2 $

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