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Phys101 Lectures 14, 15, 16 Rotational

Key points: • Rotational   I • Rotational ; and of 1 • Rotational Kinetic K  I 2 R 2 • Angular and Its Conservation L  I

Ref: 8-1,2,3,4,5,6,8,9. dL   dt

Page 1 Angular Quantities

In purely rotational motion, all points on the object move in circles around the axis of (―O‖). The radius of the circle is R. All points on a straight line drawn through the axis move through the same in the same . The angle θ in is defined: l , R where l is the arc .

2 rad  360 1 rev Sign convention: + CCW     - CW  rad  180 Angular Quantities Angular :

The average angular is defined as the total divided by time:

The instantaneous : Angular Quantities The angular is the at which the angular velocity changes with time:

The instantaneous acceleration: Angular Quantities Every point on a rotating body has an angular velocity ω and a linear velocity v. They are related: dl d  R dl  R d dl d  R dt dt v  R

Only if we use radians! Angular Quantities

Objects farther from the axis of rotation will move faster. Angular Quantities If the angular velocity of a rotating object changes, it has a tangential acceleration:

Even if the angular velocity is constant, each point on the object has a centripetal acceleration: Angular Quantities

Here is the correspondence between linear and rotational quantities:

x represents arc length Example: Angular and linear and . A carousel is initially at rest. At t = 0 it is given a constant α = 0.060 rad/s2, which increases its angular velocity for 8.0 s. At t = 8.0 s, determine the of the following quantities: (a) the angular velocity of the carousel; (b) the linear velocity of a child located 2.5 m from the center; (c) the tangential (linear) acceleration of that child; (d) the centripetal acceleration of the child; and (e) the total linear acceleration of the child.

(a)   0  t  0.0608.0  0.48 rad / s (b) v  R  0.482.5 1.2 m / s 2 (c) atan R  0.0602.5  0.15 m / s v2 1.22 (d) a    0.58m / s2 R R 2.5 2 2 2 2 2 (e) a  aR  at  0.58  0.15  0.60 m / s Angular Quantities

The is the number of complete revolutions per :

Frequencies are measured in :

The period is the time one revolution takes: Constant Angular Acceleration

The for constant angular acceleration are the same as those for , with the substitution of the angular quantities for the linear ones. Example: acceleration. A centrifuge rotor is accelerated from rest to 20,000 rpm in 30 s. (a) What is its average angular acceleration? (b) Through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration?

(a) rev 2 rad 1min   20000  2100 rad / s min 1rev 60 s   2100   0   70 rad / s2 t 30

(b) 1 1 1    t   t 2   t 2  ( 70 ) 302  31500 rad 0 0 2 2 2 1rev   31500 rad  5000 rev 2 rad Torque To make an object start rotating, a is needed; both the and direction of the force matter. The from the axis of rotation to the line along which the force acts is called the arm.

Axis of rotation In a FBD, don’t shift the sideways! Torque

A longer lever arm is very helpful in rotating objects. Torque

Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown. Torque

The torque is defined as:

R  R sin   RF sin

Also:   R F That is, F  F sin     is the angle between F and R.   R F sin Example: Torque on a compound . Two thin disk-shaped , of radii

RA = 30 cm and RB = 50 cm, are attached to each other on an that passes through the center of each, as shown. Calculate the net torque on this compound wheel due to the two forces shown, each of magnitude 50 N. [Solution]

  FARA  FB RB sin60  50( 0.30 )50( 0.50 )sin60  6.7 m N

CCW: + CW: - Rotational Dynamics; Torque and Rotational Inertia Tangential component of ’s law leads to

  FR  maR  mR2  I ( a  R I  mR2 )

This is for a single point ; what about an R extended object? As the angular acceleration is the same for the whole object, we can write: Newton's Law for rotation   I where I  mR2    I – . Rotational Dynamics; Torque and Rotational Inertia The quantity is called the rotational inertia (moment of inertia) of an object. The distribution of mass matters here—these two objects have the same mass, but the one on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation. Rotational Dynamics; Torque and Rotational Inertia The rotational inertia of an object depends not only on its but also the location of the axis of rotation—compare (f) and (g), for example. I  mR2 

Newton's Law for rotation   I Example: Atwood’s .

An consists of two , mA and mB, which are connected by a cord of negligible mass that passes over a . If the pulley has radius R0 and moment of inertia I about its axle, determine the acceleration of the masses mA and mB. [Solution] FBD for each object • Don’t use a point to represent a rotating object; • Don’t shift forces sideways.    T FP A TB x

mA g TA  mAax

TB  mB g  mBax

TAR0 TB R0  I   a  R  x m g  x 0 mA g  P  m g B 4 unknowns: a ,,T ,T . TA TB x A B mA g TA  mAax (1)

TB  mB g  mBax (2)

TAR0 TB R0  I (3)

ax  R0 (4)

Iax Iax subs (4) into (3) : R0(TA TB )   TA TB  2 (5) R0 R0

(1)  (2): ( mA  mB )g TA TB  ( mA  mB )ax (6)

Iax (5)  (6) : ( mA  mB )g  ( mA  mB )ax  2 R0 ( m  m )g Solve for a : a  A B x x I ( mA  mB ) 2 R0

,TA ,and TB can be determined using (4) and then (1) and (2). Rotational The kinetic energy of a rotating object is given by

By substituting the rotational quantities, we find that the rotational kinetic energy can be written as:

 1  1 1 1 K   mv2   mR2 2   2 mR2   2I  2  2 2 2

An object in both translational and rotational motion has both translational and rotational kinetic energy: Example (Old final exam, #19): A pencil, 16 cm long (l=0.16m), is released from a vertical position with the eraser end resting on a table. The eraser does not slip. Treat the pencil like a uniform rod. (a) What is the angular acceleration of the pencil when it makes a 30° angle with the vertical? (b) What is the angular of the pencil when it makes a 30°angle with the vertical? (a) Rotational dynamics (about o)  1 l    I,  I  ml2 ,   mg sin30  3 2  CM  l  1 mg  sin30  ml2  2  3 30 3g sin30 39.80.5     46 (rad/s 2 ) 2l 20.16 o (b) Mechanical energy is conserved because there is no non- conservative ( does not do work here).

E  E l l l i f h   cos 30  (1 cos 30 ) 1 2 2 2 mgh  I 2 1 2 I  ml2 2mgh 3   I y 2mgl(1 cos 30 ) / 2   2 CM ml / 3 h CM 3g(1 cos 30 )  l 30 3( 9.8 )(1 cos 30 )  0.16 4.96 (rad/s) Rotational Work The torque does work as it moves the wheel through an angle θ:

W  Fl  F R    Rotational Plus Translational Motion;

(a) In (a), a wheel is rolling without slipping. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity v . In (b) the same wheel is seen (b) from a reference frame where C is at rest. Now point P is moving with velocity – . The linear speed of the wheel is related to its angular speed: Demo: Which one reaches the bottom first? Why?

Wnc=0, ME is conserved (ignore energy loss due to rolling friction). y For each object, 1 1 mgH  mv2  I 2 2 CM 2 0 and vCM  v  r The hoop: 1 1 v2 1  I   mgH  mv2  I  mv2 1  • mass is distributed away 2 2 r 2 2  mr2  from the axis of rotation (large I). 2gh 1 hoop v  I  • Large I means large I 1/ 2  solid cylinder rotational KE, 1 mr2 2  • which means small mr 2 / 5  solid sphere translational KE (since I The smaller the , the faster the v. KE +KE =mgH=const.), mr2 R T • which means slow v.  Solid sphere 1st, then solid cylinder, and the hoop is the last. —Objects Rotating About a Fixed Axis

The rotational analog of linear momentum is angular momentum, L:

Then the rotational analog of Newton’s second law is:

This form of Newton’s second law is valid even if I is not constant. Conservation of Angular Momentum

In the absence of an external torque, angular momentum is conserved:

dL 0 andLI   constant. dt

The total angular momentum of a system remains constant if the net external torque is zero. This means: Iii  I f  f  constant Therefore, a figure skater can increase her angular speed by reducing her moment of inertia.

Ii  f  i I f

Demo i-clicker question 18-1 The condition for angular momentum to be conserved is (A) It’s a closed system. (B) The net external force is zero. (C) No non-conservative work. (D) The net external torque is zero. (E) The angular momentum is always conserved. Example: Object rotating on a string of changing length. A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v1 = 2.4 m/s in a circle of radius R1 = 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to R2 = 0.48 m. What is the speed, v2, of the mass now? [Solution] Net torque = 0. Angular momentum is conserved.

Ii i  I f  f

2 v1 2 v2 mR1   mR2  R1 R2 Given: v1, R1, R2; Want: v2. R1v1  R2v2

R1v1 0.80 2.4 v2    4.0m / s R2 0.48 Angular Momentum of a Moving Particle You could open a door by throwing a large piece of play- doh at the door.    When p  R,  R p L  pR  mvR  m( R )R  mR2  I   When p is not perpendicular to R,   p only the perpendicular component of p p contributes to L :   L  p R  pR sin p//  R

That is equivalent to R

L  pR  pR sin i.e., L  mvRsin Example: A bullet of mass m moving with velocity v strikes and becomes embedded at the edge of a cylinder of mass M and radius R0. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of the cylinder after this ? Is kinetic energy conserved? [Solution] Net external torque = 0. Angular momentum is conserved.

Li  Lf

 2 1 2  mvR0  mR0  MR0   2  mvR 2mv   0  2 1 2 2m  M R mR  MR   0 Kinetic energy is not 0 2 0 conserved. As the bullet being

1 2 embedded into the cylinder, KE: Ki  mv some kinetic energy is 2 converted into heat. 1 2 1  2 1 2  2 Internal forces can change the K f  I  mR0  MR0  2 2  2  kinetic energy of the system! Vector Nature of Angular Quantities* The angular velocity vector points along the axis of rotation, with the direction given by the right-hand rule. If the direction of the rotation axis does not change, the angular acceleration vector points along it as well.