1301W.600 Lecture 16. November 6, PDF Free Download

1301W.600 Lecture 16 November 6, 2017

You are Cordially Invited to the Physics Open House Friday, November 17 th, 2017 4:30-8:00 PM Tate Hall, Room B20 Time to apply for a major? Consider Physics!! Program Information! Research Opportunities! Alumni Career Speakers! Lab Tours & Current Research Discussions! Faculty Speaker, Professor Vuk Mandic, will be discussing his work with LIGO, the most precise measuring tape in the world, which won the Nobel Prize in Physics this year! A free pizza dinner will be provided for participants!

iclicker questions

iclicker question 1 To increase the moment of inertia of a body about an axis, you must A: increase the angular acceleration. B: increase the angular velocity. C: make the body occupy less space. D: place part of the body farther from the axis.

iclicker question 2 What are the moments of inertia of the 4-mass system in these two cases? Treat the masses as equal point masses and assume they are equidistant from the axis in the left case. A: 4ma 2 and 4ma 2 B: 4am 2 and 4am 2 C: 0 and 8ma 2 D: 4ma 2 and 8ma 2

iclicker question 3 A turntable rotates through 10 radians in 4 seconds. The turntable experiences uniform acceleration. If the turntable started from rest, what is its angular velocity at the end of the 4 seconds? A: 2.5 rad/s B: 5 rad/s C: 4 rad/s D: 40 rad/s

iclicker question 4 A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? A: 1 B: 2 C: 3 D: 4 E: 5

iclicker question 4 A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? A: 1 The only forces acting on the rider are the gravitational force and the forces applied by the wall. The wall can only apply a normal force, which is normal to the wall by definition, and a frictional force which is parallel to the wall. The answer is A and not E, because the forces shown in diagram E would not result in circular motion about the indicated axis. (It would also result in the rider somehow accelerating through the wall, which sounds rather unpleasant.)

Resistance to rotation Moment of Inertia Unlike Mass, the Moment of Inertia depends on the axis we choose. That is the resistance to rotation will depend on which axis we are trying to rotate the object about. I = ML2 3 I = ML2 12 So, there is no unique Moment of Inertia!

However, in general, the moment of inertia depends on the mass distribution and the location and orientation of the rotation axis. For a discrete mass distribution: I = X i m i r 2 i For a continuous mass distribution: Z Z I = r 2 dv = r 2 dm where r is the radial distance from the axis of rotation to the mass element dm

Example Compute the moment of inertia of a uniform hollow-core cylinder of inner radius R1 and outer radius R2, length L and mass M about the axis parallel to the cylinder s length and passing through its center. L

The inertia per unit volume is density: = M/V L = M (R 2 2 R 2 1 )L dm = dv The moment of inertia: Z Z I = r 2 dm = r 2 dv

R1 R2 L r dr We divide the cylinder into shells of the width dr. Each shell has a surface area 2 rl Thus, the volume element is and thickness dr. dv =2 rldr Now we compute the moment of inertial for the whole cylinder: I = Z r 2 dv = Z r 2 2 rldr =2 L Z r 3 dr = L 2 r4 R 2 R 1

L R1 r R2 dr I = Z r 2 dv = Z r 2 2 rldr =2 L Z r 3 dr = L 2 r4 R 2 R 1 I = L 2 (R 2 4 R 1 4 )= L 2 M L(R 2 2 R 1 2 ) (R 2 4 R 1 4 )= M 2 (R 2 2 + R 1 2 )

Parallel axis theorem It is usually easier to calculate moment of inertia through an object s center of mass and then use this to find out moment of inertia through another, parallel axis. I = I CM + Mh 2 Note that the moment of inertia is minimal when the rotation axis passes through an object s center of mass and then increases as the rotation axis is moved further from the CM.

Example using the parallel axis theorem We have seen last lecture that the moment of inertia for a rod about an end is I = ML2 3 The parallel theorem tells us I end = I CM + Mh 2 I CM = I end Mh 2 = ML2 3 M( L 2 )2 = ML2 12 We have recovered the result of the last lecture!

Perpendicular axis theorem For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object.

Perpendicular axis theorem The perpendicular axis theorem for planar objects can be demonstrated by looking at the contribution to the three axis moments of inertia from an arbitrary mass element. From the point mass moment, the contributions to each of the axis moments of inertia are

Rotational inertia of a ring about an axis through its center I = Z r 2 dm = R 2 Z dm = R 2 M

Rotational inertia of a disc about an axis through its center I = Z r 2 dm = Z r 2 dv = Z R 0 r 2 M V 2 rldr = Z R 0 r 3 M R 2 l 2 ldr I = 2M R 2 Z R 0 r 3 dr = 2M R 2 R 4 4 R 0 = MR2 2

Rotational inertia of a sphere about an axis through its center I = 2 5 MR2

Demo: Ring, Disc, and Sphere A ring, disk, and sphere, each with the same diameter, are allowed to roll down an adjustable incline. In which order they reach the bottom?

Demo: All Disks Roll the Same

5 min break

Angular momentum vs linear momentum Momentum for translational motion: ~p = m~v Momentum for rotational motion about given axis: ~L = I~! Angular momentum of a rigid body is defined as a product of the moment of inertial and the angular velocity

Angular momentum of a rigid body ~L = I~! The direction of angular momentum is given by L and ω are perpendicular to the plane of rotation. Using the right hand rule, the direction of both angular velocity and angular momentum is defined as the direction in which the thumb of your right hand points when you curl your fingers in the direction of rotation.

Angular momentum of a point-like particle ~L = I~! I = mr 2! = v R L = mr 2 v R = mvr

Angular momentum is the cross product ~L = ~r ~p Angular momentum for discrete mass distribution ~L = X i ~r i ~p i

L = r? mv

Choice of coordinate system is crucial when calculating angular momentum because both the magnitude and the direction depend on the choice. y r 2 p r 1 A particle will have a non-zero angular momentum about some origin if its position vector measured from this point appears to rotate about this point. If the position vector only increases or decreases in magnitude, the particle is moving along a line passing through the origin, the particle will have zero angular momentum. x

You must define an origin before calculating an angular momentum! y 2 r p r 1 x ~L = rp sin ẑ

Conservation of angular momentum In the absence of tangential forces (i.e. no torque), angular momentum is conserved: ~L = const Consequences: Because angular momentum is conserved, internal redistributions of mass lead to change in the angular velocity, ω. In other words, if the moment of inertia changes, the angular speed must also change to ensure L = constant. I 1 ~! 1 = I 2 ~! 2

Divers increase their spin by tucking in their arms and legs and reducing their rotational inertia.

Suppose the outstretched body of a diver rotates at 1.2 revolutions per second before he pulls his arms and knees into his chest, reducing his rotational inertia from 9.4 kg m 2 to 3.1 kg m 2. What is his rotational velocity after he tucks in his arms and legs?

SOLUTION Once the diver is off the board, the only force exerted on his is the gravitational force exerted by Earth. This force does not affect the angular momentum of the diver (dropped objects do not spontaneously start to rotate), and so his angular momentum must remain constant.

From the conservation of the angular momentum we have: L f = L i I f! f = I i! i! f = I i I f! i = 9.4kgm2 3.1kgm 2 (1.2s 1 )=3.6s 1

Squeezatron Two heavy spheres are allowed to rotate freely. When the handle on the apparatus is squeezed, the spheres are drawn toward the center. They rotation rate increases.

iclicker question 5 Two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different distances from the axis of rotation. If you release the hanging blocks simultaneously from rest, and if the ropes do not slip, which block lands first? A: the block on the left B: the block on the right C: at the same time

Whirlybird The rod with adjustable masses is placed perpendicular to a pulley. A string is wrapped around the pulley. The system rotates due to a falling mass attached to the string.

Rotational dynamics: Torque Chapter 12 Chapter Goal: To study the causes of the changes in rotational motion using the concepts of torque and angular momentum.

Rotational dynamics To spin a disk, we need a tangential acceleration. This means we must apply a tangential force. The ability of a force to rotate an object about an axis is called torque.

To tighten a nut, you need to apply a torque Does the magnitude of the force F matter? Does the length, r, of the wrench matter? Does the angle, θ, at which you push on the wrench matter?

To push a seesaw to lift a child seated on the opposite side as shown in the figure, it is best to push as far as possible from the pivot, and in a direction that is perpendicular to the seesaw.

Torque torque =(lever arm) (force) = r? F We can see that applying the force as far as possible from the pivot point increases the torque.

Torque equation = r? F = rf sin So, only the component of the force perpendicular to the wrench generates a torque!

Torque as a cross product ~ = ~r ~ F Magnitude of torque: = r? F = rf sin Direction of torque: Right-Hand rule

thank you

1301W.600 Lecture 16. November 6, 2017