Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products Definition of vector product and its properties Axis of rotation unit vector normal to the plane defined by and

Vector (cross) products (cont’d) Properties of vector product z : unit vector in x,y,z direction y Consider three vectors: x Then

Vector (cross) products (cont’d) Properties of vector product (cont’d) z y x

Torque Case for 1 point-like object of mass m with 1 force Torque is a quantitative measure of the tendency of a force to cause or change the rotational motion. Case for 1 point-like object of mass m with 1 force r is constant only tangential component of causes rotation y x massless rigid rod torque moment of inertia unit Nm

Torque Case for 1 point-like object of mass m with 1 force Define y Define x is aligned with the rotation axis lever arm

Torque (cont’d) Case for 2 point-like objects with 2 forces increases y increases decreases Example: A see-saw in balance x M m R r mg Mg

Work & energy (I) W=DK A massive body on massless rigid rod Work done by the force: Also x W=DK

Work & energy (I) (cont’d) Power in rotational motion Work done by the force: Power :

Correspondence between linear & angular quantities displacement velocity acceleration mass force Newton’s law kinetic energy work

Work & energy (II) A massive body in rotational & translational motion Kinetic energy: Now where is the velocity with respect to the center of mass (COM) and the velocity of COM. Then com

Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) kinetic energy due to translational motion kinetic energy due to rotational motion about a rotation axis through COM

Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) Example: A rolling cylinder (without sliding due to friction) O O s P s P View point 1: + translation = rotation O O O This view point was used in the last few slides

Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) Example: A rolling cylinder (without sliding) (cont’d) View point 2: Any point in the cylinder rotates around P From the parallel axis theorem P rotation axis rotational translational

Angular momentum & torque Angular momentum of a particle y Define angular momentum as: angular momentum x Since and lever arm net torque

Angular momentum & torque Angular momentum of a multi-particle system As before let’s break down vectors into two components velocity of com Then position vector for com ang. mom. about COM + ang. mom. of COM

Conservation of angular momentum If the net torque is zero, the total angular momentum of the system is conserved. Conservation of angular momentum

Angular momentum & torque Angular momentum vector and axis of rotation z plane A plane B x plane A plane B In general, because of non-zero x/y component of angular momentum unless the object is symmetric about the axis of rotation. If the object is symmetric about the axis of rotation, then

Gyroscope Gyroscopic motion remove support pivot  mg precession pivot

Gyroscope Principle of gyroscopic z y pivot x 

Gyroscope y Assumption: The angular momentum vector is associated only with the spin of the flywheel and is purely vertical. The procession is much slower than the rotation, x Precession angular speed:

Gyroscope

Gyroscope

Spinning Top

Example A disk of mass M and radius R rotates around the axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f. 1) wf = wi 2) wf = ½ wi 3) wf = ¼ wi z f z i

First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! f z i z 2 1

Example You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

What happens to the angular momentum as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT L1 L2

What happens to your angular velocity as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT w1 w2 I2 I1 L

What happens to your kinetic energy as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT (using L = I ) w1 w2 I2 I1 L

Example A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens? 1. She starts to spin CCW. 2. She starts to spin CW. 3. Nothing CORRECT

Since there is no net external torque acting on the student stool system, angular momentum is conserved. Remember, L has a direction as well as a magnitude! Initially: LINI = LW,I Finally: LFIN = LW,F + LS LS LW,I LW,I = LW,F + LS LW,F

Example A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase? (a) 2 (b) 4 (c) 8 w

Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved. L1 = I1w1 = mR2w1 w1 m R mR2w1 = m R2w2 w1 = w2 w2 = 4w1

Example A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2. What is the angular speed wF of the stick after the collision? (Ignore gravity) M F D m D/4 v1 v2 initial final

Set Li = Lf using M F D m D/4 initial final v1 v2

Exercises Problem 1 Find the acceleration of an object of mass m. n y Solution R For the object, Newton’s 2nd law gives: ay>0 in –y direction (1) T For the cylinder, the total torque is: M (2) Mg T m The tangential acceleration of the cylinder is equal to that of the object: mg (3) h (2)+(3): x (4)

Exercises Problem 1 (cont’d) Solution (cont’d) n (1)+(4): y R (5) T (4)+(5): M Mg T The final velocity of the object when it was at rest initially: m mg h x

Exercises Problem 2 Solution n (a) The normal force on the cylinder is: y R T (b) Compare n with (m+M)g: M As the suspended mass is accelerating down the tension is less than mg. Therefore n is less than the total weight (m+M)g. Mg T m (c) If the cylinder is initially rotating clockwise and so that the object gets initial velocity upward, what effect does this have on T and n? As long as the cable remains taut, T and n remain the same. mg h x

Exercises pulley Problem 3 m1 T1 M A glider of mass m1 slides without friction on a horizontal air track. It is attached to an object of mass m2 by a massless string. The pulley is a thin cylindrical shell with mass M and radius R, and the string turns the pulley without slipping or stretching. Find the acceleration of each body, the angular acceleration of the pulley, and the tension in each part of the string. R T2 glider m2 hanging object Solution y x x y glider: (1) + x (2) n1 object: T1 T1 y T2 pulley: m1 m2 T2 m1g m2g (3) Mg no stretching and slipping: n2 (4) glider hanging obj. pulley

Exercises Problem 3 (cont’d) Solution cont’d (1)-(4):

Exercises Problem 4 A primitive yo-yo is made by wrapping a string several times around a solid cylinder with mass M and radius R. You hold the end of the string stationary while releasing the cylinder with no initial motion. The string unwinds but does not slip or stretch as the cylinder drops and rotates. Find the speed vcm of the center of mass of the solid cylinder after it has dropped a distance h. 1 h 2 Solution Energy conservation

Exercises Problem 5 (Atwood’s machine) R I Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. C mA mB A B Solution The accelerations of blocks A and B will have the same magnitude . As the cord does not slip, the angular acceleration of the pulley will be . If we denote the tensions in cord as and , the equations of motion are:

Exercises Problem 6 A solid uniform spherical boulder starts from rest and rolls down a 50.0 m high hill. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill? rough h=50.0 m smooth Solution 1st half (rough): 2nd half (smooth):

Exercises Problem 7 Occasionally, a rotating neutron star undergoes a sudden and unexpected speedup called a glitch. One explanation is that a glitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed w0=70.4 rad/s underwent such a glitch in October 1975 that increased its angular speed to w=w0+Dw, where Dw/w0=2.01x10-6. If the radius of the neutron star before the glitch was 11 km, by now how much did its radius decrease in the starquake? Assume that the neutron star is a uniform sphere. Solution Conservation of angular momentum:

Exercises Problem 8 A small block with mass 0.250 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks. Solution The tension on the string: The radius at which the string breaks is obtained from:

Exercises Problem 9 pivot A ball catcher whose mass is M and moment of inertia is I is hung by a frictionless pivot. A ball with a speed v and mass m is caught by the catcher. The distance between the pivot point and the ball is r which is much greater that the radius of the ball. r v Find the angular speed of the catcher right after it catches the ball. (b) After the ball is caught, the catcher – ball system swings up as high as h. Find the angular speed at the maximum height h. pivot catcher M,I

Exercises Problem 10 pivot Solution The initial angular momentum with respect to the pivot is: The final total moment of inertia is r v h Since pivot catcher M,I (b) The kinetic energy after the collision is: potential energy at height h

Exercises d= 42.0 cm Problem 11 A 42.0 cm diameter wheel, consisting of a rim and six spokes, is constructed from a thin rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 58.0 m high. How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled? l= 25.0 g/cm No sliding h= 58.0 m

Exercises d= 42.0 cm Problem 11 (cont’d) l= 25.0 g/cm Solution (a) Conservation of energy: No sliding h= 58.0 m (b) Doubling the density would have no effect! As , doubling the diameter would reduce the angular velocity by half. But would be unchanged.