Rotation of Rigid Bodies:
Let us consider a couple of questions involving rigid body rotation.
(1) A meter scale is held vertically with one end on the floor and is allowed to fall. Assuming that the end on the floor does not slip, what will be the linear velocity of the other end when it strikes the floor?
(a) 2.7m/s (b) 3.1m/s (c) 5.4m/s (d) 9.8m/s (e) 11.2m/s
When the meter scale is allowed to fall, its gravitational potential energy gets converted into rotational kinetic energy so that we have
mgl/2 = ½ I ω 2 where ‘l’ is the length(1metre for a metre scale), ‘m’ is the mass, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the scale. l/2 appears in the potential energy expression since the centre of gravity of the scale is initially at a height l/2. You should note that the moment of inertia of the scale is about the end in contact with the floor and is equal to ml2/3.
From the above equation, ω =√(3g/l).
The linear velocity of the feree end of the scale = ωl = √(3gl). On substituting for l(=1) and g(=9.8) the linear velocity is 5.4m/s [Option (c)].
(2) An impulsive force F acting for a short time interval ∆t is applied at one end of a thin uniform bar of mass M and length L, in a direction perpendicular to the lengthof the bar. The angular velocity with which the bar will rotate is
(a) F∆t/4ML (b) F∆t/2ML (c)2F∆t/ML (d) 4F∆t/ML (e) 6F∆t/ML
The impulse received by the bar is F∆t which is equal to the linear momentum supplied. The bar will rotate about its centre of mass. The ‘lever arm’ for the angular momentum is L/2 so that we have, (L/2)F∆t = Iω = ML2ω /12. So, ω =6F∆t/ML
Let us consider a couple of questions involving rigid body rotation.
(1) A meter scale is held vertically with one end on the floor and is allowed to fall. Assuming that the end on the floor does not slip, what will be the linear velocity of the other end when it strikes the floor?
(a) 2.7m/s (b) 3.1m/s (c) 5.4m/s (d) 9.8m/s (e) 11.2m/s
When the meter scale is allowed to fall, its gravitational potential energy gets converted into rotational kinetic energy so that we have
mgl/2 = ½ I ω 2 where ‘l’ is the length(1metre for a metre scale), ‘m’ is the mass, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the scale. l/2 appears in the potential energy expression since the centre of gravity of the scale is initially at a height l/2. You should note that the moment of inertia of the scale is about the end in contact with the floor and is equal to ml2/3.
From the above equation, ω =√(3g/l).
The linear velocity of the feree end of the scale = ωl = √(3gl). On substituting for l(=1) and g(=9.8) the linear velocity is 5.4m/s [Option (c)].
(2) An impulsive force F acting for a short time interval ∆t is applied at one end of a thin uniform bar of mass M and length L, in a direction perpendicular to the lengthof the bar. The angular velocity with which the bar will rotate is
(a) F∆t/4ML (b) F∆t/2ML (c)2F∆t/ML (d) 4F∆t/ML (e) 6F∆t/ML
The impulse received by the bar is F∆t which is equal to the linear momentum supplied. The bar will rotate about its centre of mass. The ‘lever arm’ for the angular momentum is L/2 so that we have, (L/2)F∆t = Iω = ML2ω /12. So, ω =6F∆t/ML
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