Goals:

1
Lecture 18

• Goals

• Define and analyze torque
• Introduce the cross product
• Relate rotational dynamics to torque
• Discuss work and work energy theorem with
  respect to rotational motion
• Specify rolling motion (center of mass velocity
  to angular velocity
• So what causes rotation?

2
So what forces make things rotate?

• Which of these scenarios will cause the bar to
  spin?

Fixed rotation axis.
A
F
B
C
F
D
F
F
F
E
3
Net external torques cause objects to spin

• An external force (or forces) properly placed
  induced changes in the angular velocity. This
  action is defined to be a torque
• A force vector or a component of a force vector
  whose line of action passes through the axis of
  rotation provides no torque
• A force vector or the component of a force vector
  whose line of action does NOT pass through the
  axis of rotation provides torque
• The exact position where the force is applied
  matters. Always make sure the force vectors
  line of action contacts the point at which the
  force is applied.

4
Net external torques cause objects to spin

• Torque increases proportionately with increasing
  force
• Only components perpendicular to r vector yields
  torque
• Torque increases proportionately with increasing
  distance from the axis of rotation.

• This is the magnitude of the torque
• q is the angle between the radius and the force
  vector
• Use Right Hand Rule for sign

5
Force vector line of action must pass through
contact point

• Force vector cannot be moved anywhere
• Just along line of action

q
F
r
r
line of action
q
F
6
Resolving force vector into components is also
valid

• Key point Vector line of action must pass
  through contact point (point to which force is
  applied)

r
line of action
q
Fr
F
7
Exercise Torque

• In which of the cases shown below is the torque
  provided by the applied force about the rotation
  axis biggest? In both cases the magnitude and
  direction of the applied force is the same.
• Remember torque requires F, r and sin q
• or the tangential force component times
  perpendicular distance

1. Case 1
2. Case 2
3. Same

8
Torque is constant along the line of action

• Even though case 2 has a much larger radius
  vector the torque remains constant.
• Case 1 t L F
• Case 2 t r F sin q F r sin q
• Notice that the sin q sin (p-q) - cos p sin
  q sin q
• Case 2 t r F sin q F r sin q F r
  sin (p-q) F L
• Here q 135 and r 2½ L so r sin q
  2-1/2 (2½ L ) L

9
Torque, like w, is a vector quantity

• Magnitude is given by (1) r F sin q
• (2) Ftangential r
• (3) F rperpendicular to line of
  action
• Direction is parallel to the axis of rotation
  with respect to the right hand rule

• Torque is the rotational equivalent of force
• Torque has units of kg m2/s2 (kg m/s2) m N m

10
Torque can also be calculated with the vector
cross product

• The vector cross product is just a definition

11
Torque and Newtons 2nd Law

• Applying Newtons second law

12
Example Wheel And Rope

• A solid 16.0 kg wheel with radius r 0.50 m
  rotates freely about a fixed axle. There is a
  rope wound around the wheel.
• Starting from rest, the rope is pulled such
  that it has a constant tangential force of F 8
  N.
• How many revolutions has the wheel made after
  10 seconds?

F
r
13
Example Wheel And Rope

• m 16.0 kg radius r 0.50 m
• F 8 N for 10 seconds
• Constant F ? Constant t ? constant a
• Isolid disk ½ mr2 2 kg m2
• I a t r F ? a 4 Nm/ 2 kg m2 2 rad/s2
• q q0 w0 Dt ½ a Dt2 ? q - q0 w0 Dt ½ a
  Dt2
• Rev (q - q0) / 2p ( 0 ½ a Dt2 )/ 2p
• Rev (0.5 x 2 x 100) / 6.28 16

14
Work

• Consider the work done by a force F acting on an
  object constrained to move around a fixed axis.
  For an angular displacement d? then ds r d?
• ?dW FTangential dr Ft ds
• dW (Ft r) d?
• ?dW ? d? (and with a constant torque)
• We can integrate this to find W ? ? t
  (qf - qi)

15
Rotation Kinetic Energy...

• The kinetic energy of a rotating system looks
  similar to that of a point particle
• Point Particle Rotating System

v is linear velocity m is the mass.
? is angular velocity I is the moment of
inertia about the rotation axis.
16
Work Kinetic Energy

• Recall the Work Kinetic-Energy Theorem
• ?K WNET or WEXT
• This is true in general, and hence applies to
  rotational motion as well as linear motion.
• So for an object that rotates about a fixed axis

17
Example Wheel And Rope

• A solid 16 kg wheel with radius r 0.50 m
  rotates freely about a fixed axle. There is a
  rope wound around the wheel.
• Starting from rest, the rope is pulled such
  that it has a constant tangential force of F 8
  N.
• What is the angular velocity after 16
  revolutions ?

F
r
18
Example Wheel And Rope

• Mass 16 kg radius r 0.50 m Isolid disk½mr2
  2 kg m2
• Constant tangential force of F 8 N.
• Angular velocity after you pull for 32p rad?
• W F (xf - xi) r F Dq 0.5 x 8.0 x 32p J
  402 J
• DK (Kf - Ki) Kf ½ I w2 402 J
• w 20 rad/s

F
r
19
Exercise Work Energy

• Strings are wrapped around the circumference of
  two solid disks and pulled with identical forces
  for the same linear distance.
• Disk 1, on the left, has a bigger radius, but
  both have the same mass. Both disks rotate
  freely around axes though their centers, and
  start at rest.
• Which disk has the biggest angular velocity
  after the pull?

w2
w1
W ? ? F d ½ I w2 Smaller I bigger w (A)
Disk 1 (B) Disk 2 (C) Same
F
F
start
d
finish
20
Home Example Rotating Rod

• A uniform rod of length L0.5 m and mass m1 kg
  is free to rotate on a frictionless pin passing
  through one end as in the Figure. The rod is
  released from rest in the horizontal position.
  What is
• (A) its angular speed when it reaches the lowest
  point ?
• (B) its initial angular acceleration ?
• (C) initial linear acceleration of its free end
  ?

21
Example Rotating Rod

• A uniform rod of length L0.5 m and mass m1 kg
  is free to rotate on a frictionless hinge passing
  through one end as shown. The rod is released
  from rest in the horizontal position. What is
• (B) its initial angular acceleration ?
• 1. For forces you need to locate the Center of
  Mass
• CM is at L/2 ( halfway ) and put in the Force on
  a FBD
• 2. The hinge changes everything!

S F 0 occurs only at the hinge
but tz I az r F sin 90 at the center of
mass and (ICM m(L/2)2) az (L/2) mg and solve
for az
mg
22
Example Rotating Rod

• A uniform rod of length L0.5 m and mass m1 kg
  is free to rotate on a frictionless hinge passing
  through one end as shown. The rod is released
  from rest in the horizontal position. What is
• (C) initial linear acceleration of its free end
  ?
• 1. For forces you need to locate the Center of
  Mass
• CM is at L/2 ( halfway ) and put in the Force on
  a FBD
• 2. The hinge changes everything!

a a L
mg
23
Example Rotating Rod

• A uniform rod of length L0.5 m and mass m1 kg
  is free to rotate on a frictionless hinge passing
  through one end as shown. The rod is released
  from rest in the horizontal position. What is
• (A) its angular speed when it reaches the lowest
  point ?
• 1. For forces you need to locate the Center of
  Mass
• CM is at L/2 ( halfway ) and use the Work-Energy
  Theorem
• 2. The hinge changes everything!

L
W mgh ½ I w2
m
W mgL/2 ½ (ICM m (L/2)2) w2 and solve for
w
mg
L/2
mg
24
Connection with CM motion

• If an object of mass M is moving linearly at
  velocity VCM without rotating then its kinetic
  energy is

• If an object of moment of inertia ICM is rotating
  in place about its center of mass at angular
  velocity w then its kinetic energy is

• What if the object is both moving linearly and
  rotating?

25
Connection with CM motion...

• So for a solid object which rotates about its
  center of mass and whose CM is moving

VCM
?
26
Rolling Motion

• Now consider a cylinder rolling at a constant
  speed.

VCM
CM
The cylinder is rotating about CM and its CM is
moving at constant speed (VCM). Thus its total
kinetic energy is given by
27
Rolling Motion

• Again consider a cylinder rolling at a constant
  speed.

2VCM
CM
VCM
28
Rolling Motion

• Now consider a cylinder rolling at a constant
  speed.

VCM
CM
The cylinder is rotating about CM and its CM is
moving at constant speed (VCM). Thus its total
kinetic energy is given by
29
Motion

• Again consider a cylinder rolling at a constant
  speed.

30
For Tuesday

• Read though 11.3