Practice free Rotational Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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 ….
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 $
A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is
$ 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 $
A binary star consist of two stars A (2.2 Ms) and B(mass 11Ms) where Ms is the mass of sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B. about the centre of mass is
Here C.M wrt A $ = \left( { 11M_5 \over 11M_5 + 2.2 M_5 } \right) d = {5d \over 6 } $ $ L_A = 2.2 M_5 w \left( { 5d \over 6} \right) ^2 = 55M_5 w { d^2 \over 36 } $ $L_B = 11 M_5 \omega \left( { d \over 6} \right)^2 = 11 M_5 \omega { d^2 \over 36 } $
A particle of mass m slides down on inclined plane and reaches the bottom with
linear velocity V. If the same mass is in the form of ring and rolls
without slipping down the same inclined plane. Its velocity will be .
$ Here in first case { 1 \over 2} mg^2 = mvh $ $ \therefore v = \sqrt { 2 gh } $ $ as for the ring K = R v' = \sqrt {gh} $
Match list I with list II and select the correct answer. List - I System (x) A ring about it axis (y) A uniform circular disc about it axis (z) A solid sphere about any diameter (w) A solid sphere about any tangent
List - II (1) $ { M R^2 \over 2 } $ (2) $ { 2\over 5 } MR^2 $ (3) $ { 7 \over5 } MR^2 $ (4) $ MR^2 $ (5) $ { 9 \over 5 } MR^2 $
The moments of inertia for different systems are as follows:
Statement -1 — The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about any point on the circumference of the circle. Statement -2— If no net torque outs, the angular momentum of a system is conserved
The correct choice is [D] since the centrifetal force is radial. Forque is zero so L = constant.
Statement -1— A sphere and a cylinder slide without rolling from rest from the top of an inclined plane. They will reach the bottom with the same speed.
Statement -2 — Bodies of all shapes, masses and sides slide down a plane with the same acceleration.
The correct choice is [A] If a body slides down an indined plane its acaleration is $ a = g (sin \theta – \mu cos \theta )$ which depends only on g, $ \theta $ and $ \mu $ .
Statement -1— Friction is necessary for a body to roll on surface Statement -2— Friction provides the necessary tangential force and torque
Statement - 1 is correct because friction is necessary for rolling motion to occur as it prevents slipping. Statement - 2 is also correct because friction provides the necessary tangential force and torque needed for rolling. Thus, statement - 2 is the correct explanation for statement - 1.
A Solid sphere of mass M and radius R is released from rest at the top of a frictionless inclined plane of length 'd' and inclinatSolid spion ?. In case (a) it rolls down the plane without slipping and in case (b) it slides down the plane The ratio of the velocity of the spheres when it reaches the bottom of the plane in case (a) to that in case (b) is
$ using V^2 = 2 ad we can find $ $ { v_1^2 \over v_2^2 } = { a_1 \over a_2 } = \sqrt { 5 \over 7 } $
A Solid sphere of mass M and radius R is released from rest at the top of a frictionless inclined plane of length 'd' and inclinatSolid spion ?. In case (a) it rolls down the plane without slipping and in case (b) it slides down the plane The ratio of the velocity of the spheres when it reaches the bottom of the plane in case (a) to that in case (b) is
$ from d = { 1 \over 2 } a t ^2 , we find that { d_1 \over d_2 }= \sqrt{ 7 \over 5 } $
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