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Work and Rotational Kinetic EnergyAs we discussed in Chapter 7, when a force
For motion confined to an This reduces to The rate at which the work is done is the power, which we can find with Eqs. 7-43
Now let us consider a rotational situation that is similar. When a torque accelerates a rigid body in rotation about a fixed axis, it does work
Here,
where
The rate at which the work is done is the power, which we can find with the rotational equivalent of Eq. 11-43,
Proof of Eqs. 11-44 through 11-47 Let us again consider the situation of Fig. 11-16, in which force
Using
(11- From Eq. 11-26, the rotational inertia for this one-particle body is
which is Eq. 11-44. We derived it for a rigid body with one particle, but it holds for any rigid body rotated about a fixed axis. We next relate the work
From Eq. 11-32, we see that the product
The work done during a finite angular displacement from which is Eq. 11-45. It holds for any rigid body rotating about a fixed axis. Equation 11-46 comes directly from Eq. 11-45. We can find the power P for rotational motion from Eq. 11-51: which is Eq. 11-47 Sample Problem 11-10 A rigid sculpture consists of a thin hoop (of mass (a) In terms of
A Key Idea here is that we can separately find the rotational inertias of the hoop and the rod and then add the results to get the sculpture's total rotational inertia where we have used the fact that (b) Starting from rest, the sculpture rotates around the rotation axis from the initial upright orientation of Fig. 11-19. What is its angular speed SOLUTION: Three Key Ideas are required here: 1.We can relate the sculpture's speed 2.We can relate 3.For the gravitational potential energy we can treat the rigid sculpture as a particle located at the center of mass, with the total mass We can write the conservation of mechanical energy ( As the sculpture rotates from its initial position at rest to its inverted position, when the angular speed is From Eq. 8-7 where To find When the sculpture is inverted, the center of mass is this same distance Now let's pull these results together. Substituting Eqs. 11-55 and 11-56 into 11-54 gives us Substituting
Q. 2.01. Do the internal forces affect the motion of a system under the effect of some external force ? Ans. No. Torques acting on the system due to internal forces cancel out. Q. 2.02. What do you mean by a rigid body? Ans. A body, whose constituent particles remain at their respective positions, when a body is in translational or rotational motion is called a rigid body.
Q. 2.01. Torque and work are both defined as force times distance. Explain, how do they differ. Ans. (i) Whereas work is a scalar quantity, torque is a vector quantity. (ii) Work done is measured as the product of the applied force and the distance, which the body covers along the direction of the force. On the other hand, torque is measured as the product of the force and its perpendicular distance from the axis of rotation. Q.2.02. Why is a ladder more apt to slip, when you are high up on it than when you just begin to climb? Ans. When a person is high up on a ladder, then torque produced due to his weight about the point of contact between the ladder and the floor becomes quite large. On the other hand, when he starts climbing up, the torque is small. Due to this reason, the ladder is more apt to slip, when one is high up on it. Q. 2.03. A planet moves around the sun under the effect of gravitational force exerted by the sun. Why is the torque on the planet due to the gravitational force zero? Ans. The torque on the planet due to the sun,
where Ans. An object will not acquire angular momentum, if no external torque acts on it. During its flight, a torque acts on the projectile due to gravity and hence it acquires angular momentum. Q. 2.06. A planet revolves around a massive star in a
1. Starting from Newton's second law of motion, derive the equation of motion of a particle (capable of rotation about an axis) on which a torque 2 Derive expression for torque on a system in cartesian coordinates. 3, Derive expression for the torque acting on a system of 5. Derive expression for the angular momentum of a system in cartesian co-ordinates. 6. Prove that the time rate of change of the angular momentum of a particle is equal to the torque acting on it. 7. Derive a relation between angular momentum and torque. 8. Derive the relation between the torque and the angular momentum. Hence obtain and state the law of conservation of angular momentum. 9. State and prove the principle of conservation of angular momentum.
B. On Torque 9. A disc of radius 0.5 m is rotating about an axis passing through its centre and perpendicular to its plane.A tangential force of 2,000 N is applied to bring the disk to rest in 2 s. Calculate its angular momentum. [Ans. 2,000 kg m2/s 10. A torque of 20 N m is applied on a wheel initially at rest. Calculate the angular momentum of the wheel after 3 s. [Ans. 60 kg m2/s Type C. On Angular momentum 11. In hydrogen atom, electron revolves in a circular orbit of radius 0.53 A with a velocity of 2.2 x 106 m/s. If the mass of the electron is 9.0 x 10-31 kg, find its angular momentum. [Ans. 1.05 x 10-34 kg m2/s 12. Find the angular momentum of Neptune about the sun. Given distance of Neptune from the sun is 5 x 1012 m. period of revolution about the sun = 5 x 109s and mass of Neptune = 1027 kg. [Ans. 3.14 x 1043 kg m2I
Date: 2015-01-12; view: 2338
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