Chapter 8: Torque and Angular Momentum

• Concept. Questions: 2, 4.

• Problems: 5,13,18, 27, 39, 43, 55, 69, 73.

• Rotational Kinetic Energy

• Torque & Angular Acceleration

• Torque & Angular Momentum

• (Vector Nature of)

2

Rotational Dynamics: Newton’s 2nd Law for Rotation

Inertia Rotational

TorqueNet Accel.Angular

Clockwise (CW)

Counter-clockwise (CCW)

221 mvKE

2mrI2

21 IKE

)( 2221 rm

2221 mr

Rotational Inertia & Energy

Central Axis

4

2222 mamamamaI

24maI

Axis on End

5

2222 )2()2(00 amammmI

28maI

6

Calculated Rot. Inertias

• rotational inertias of solid objects can be calculated

• The calculated values are listed in your textbook on p.263

• /

Ex: Rotational Inertia: A 0.3kg meter stick is held horizontally from one end. Its rotational inertia about one end is:

7

231 MLI

2

231

10.0

)0.1)(3.0(

mkg

mkg

Torque = lever-arm x force

meter-newton

ft-lb

Inertia Rotational

TorqueNet Accel.Angular

9

torque

• lever-arm is the shortest distance from axis to line of the force

• Torque (giam7-11)

rFFr

FFr

rF

)sin(

)sin(

sin

Ex: Zero and Non-Zero Torque

• Zero Torque

• Large Torque

Ex: Torque due to Gravity: A 0.3kg meter stick is held horizontally from one end. The torque due to gravity about the end is:

12

cmgrav Mgx

Nm

mkgNkg

47.1

)5.0)(/8.9)(3.0(

13

Newton’s 2nd Law (Rotation)

INet

Inertia Rotational

TorqueNet Accel.Angular

Ex: Angular Acceleration: A 0.3kg meter stick is held horizontally from one end. Its angular acceleration when released is:

14

221 MLI

gravnet

ssrad

mkg

Nm

//7.14

1.0

47.12

Ex: A merry-go-round has a rotational inertia of 100kgm^2 and a radius of 1.0 meter. A force of 250 N is applied tangentially at its edge. The angular acceleration is:

15

I

F

Inet

ssrad

mkg

Nm

//25.0

100

)250)(0.1(2

16

Equilibrium Problems

• Equilibrium is state when: Net force = 0 & Net torque = 0

• You can choose the axis anywhere, so we choose it where an unknown force acts.

• 1st Step: torque-ccw = torque-cw

• 2nd Step: force-up = force-down

• /

17

Ex: The drawing shows a person whose weight is 584N. Calculate the net force with which the floor pushes on each end of his body.

18

Rotational Kinetic Energy

• Rotational K = ½(I)2.

• Example: Constant Power Source has 100kg, 20cm radius, solid disk rotating at 7000 rad/s.

• I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2.

• Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ

19

Rotational Work-Energy Theorem

• (Work)rot = .

• Example: torque of 50 mN is applied for one revolution.

• rotational work = (50Nm)(2rad) = 314 J

• (Rotational Work)net = Krot.

• /

20

Angular Momentum (L)

• analog of translational momentum

• L = I [kgm2/s]

• Example: Disk R = 1m, M = 1kg, = 10/s

• I = ½MR2 = ½(1)(1)2 = 0.5 kgm2

• L = I = (0.5kgm2)(10/s) = 5kgm2/s

21

Conservation of Angular Momentum

• For an isolated system

• (I)before = (I)after

• Example: Stationary disk M,R is dropped on rotating disk M, R, i.

• (I)before = (I)after

• (½MR2)(i) = (½MR2 + ½MR2)(f)

• (f) = ½ (i)

22

8 Summary

• Mass Rotational Inertia

• Force Torque

• Rotational KE

• Angular Acceleration

• Work and Energy

• Angular Momentum

23

Concept Review

• Torque: rotational action

• Rotational Inertia: resistance to change in rotational motion.

• Torque = force x lever-arm

24

Mass-Distribution

Larger radius

Larger Speed

Larger Effort

Rotational Inertia ~ R2

25

Torque () [m·N]

F

F

F

= lever-arm

26

Rotational Inertia ( I )

222

211 rmrmI

Example

22 )2)(5()3)(4( mkgmkgI

kg(m)2

2562036 mkg

27

28

Problem 33

• Pivot at left joint, Fj = ?, but torque = 0.

• ccw (Fm)sin15(18) = mg(26) = cw

• ccw (Fm)sin15(18) = (3)g(26) = cw

• (Fm) = (3)g(26)/sin15(18) = 160N

• Note: any point of arm can be considered the pivot (since arm is at rest)

29

If ball rolls w/o slipping at 4.0m/s, how large is the height h in the drawing?

rolling w/o slipping

30

#39

• Left force = mg = 30g, Right = 25g

• mg = 30g + 25g m = 55kg

• ccw mg(xcg) = cw 30g(1.6)

• (55)g(xcg) = 30g(1.6)

• (55)(xcg) = 30(1.6)

• Xcg = (30/55)(1.6)

31

#60, z-axis

• Each mass has r2 = 1.52 + 2.52.

• I = sum mr2 = (2+3+1+4)(1.52 + 2.52)

32

#65

• First with no frictional torque, then with frictional torque as specified in problem.

• M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8

33

#83

• Pulley M, R. what torque causes it to reach ang. Speed. 25/s in 3rev?

• Alpha: use v-squared analog eqn.

• Torque = I = (½MR2)(

34

#89, uniform sphere part

• Rolling at v = 5m/s, M = 2kg, R = 0.1m

• K-total = ½mv2 + ½I2.

• = ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2.

• = 25 + 10 = 35J

• Roll w/o slipping, no heat created, mech energy is conserved, goes all to Mgh.

• 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m

35

#111

• Ice skater, approximate isolated system

• Therefore:

• (I)before = (I)after

• (100)(i) = (92.5)(f)

• (f) = (100/92.5)(i)

• K-rot increases by this factor squared times new rot. Inertia x ½.

36

Example: Thin rod formulas.

37

Angular Momentum• Symbol: L Unit: kg·m2/s

• L = mvr = m(r)r = mr2 = I.

• v is perpendicular to axis

• r is perpendicular distance from axis to line containing v.

38

Angular Momentum• Symbol: L Unit: kg·m2/s

• L = mvr = m(r)r = mr2 = I.

• v is perpendicular to axis

• r is perpendicular distance from axis to line containing v.

39

13) Consider a bus designed to obtain its motive power from a large rotating flywheel (1400. kg of diameter 1.5 m) that is periodically brought up to its maximum speed of 3600. rpm by an electric motor at the terminal. If the bus requires an average power of 12. kilowatts, how long will it operate between recharges?Answer: 39. minutesDiff: 2 Var: 1 Page Ref: Sec. 8.4

40

6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end.(a) What is the total force provided by both supports?(b) With what force does the support, closest to the painter, push upward?

41

28) A 4.0 kg mass is hung from a string which is wrapped around a cylindrical pulley (a cylindrical shell). If the mass accelerates downward at 4.90 m/s2, what is the mass of the pulley?A) 10.0 kgB) 4.0 kgC) 8.0 kgD) 2.0 kgE) 6.0 kg

42

19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and sticks to the disk at a distance d = 80. cm from the axis of rotation.(a) What was the moment of inertia before the gum fell?(b) What was the moment of inertia after the gum stuck?(c) What is the angular velocity after the gum fell onto the disk?

(a) 2.0 kg-m2(b) 2.3 kg-m2(c) 31. rpm

43

1. A pair of forces with equal magnitudes and opposite directions is acts as shown. Calculate the torque on the wrench.

44

3. The drawing shows the top view of two doors. The doors are uniform and identical. The mass of each door is M and width as shown below is L. How do their rotational accelerations compare?

45

A Ring, a Solid-Disk, and a Solid-Sphere are released from rest from the top of an incline. Each has the same mass and radius. Which will reach the bottom first?

46

5. The device shown below is spinning with rotational rate i when the movable rods are out. Each moveable rod has length L and mass M. The central rod is length 2L and mass 2M.

Calculate the factor by which the angular velocity is increased by pulling up the arms as

shown.

47

Rotational Review

rs

(angles in radians)

tavg

rvt

rat tavg

2rac

+ 4 kinematic equations

tc aaa

48

Angular Momentum Calculation

L = I

Example: Solid Disk M = 2kg R = 25cm

Spins about its center-of-mass at 35 rev/s

skgm

revradsrevmkgIL

/7.13

)/2)(/35()25)(.2(2

221

49

4. A one-meter-stick has a mass of 480grams. a) Calculate its rotational inertia about an axis perpendicular to the stick and through one of its ends.b) Calculate its rotational inertia about an axis perpendicular to the stick and through its center-of-mass.c) Calculate its angular momentum if spinning on axis (b) at a rate of 57rad/s.

50

Conservation of Angular Momentum

• Example: 50 grams of putty shot at 3m/s at end of 200 gram thin 80cm long rod free to rotate about its center.

• Li = mvr = (0.050kg)(3m/s)(0.4m)

• Lf = I = {(1/12)(0.200kg)(0.8m)2 + (0.050kg)(0.4m)2}()

• final rotational speed of rod&putty =

51

tavg

tavg

52