Transcript Physics 106P: Lecture 1 Notes

Example
A 40-kg mass placed 1.25 m on the
opposite side of the support point
balances a mass of 25 kg, placed (x) m
from the support point of a uniform
beam. What is the value of the
unknown distance x?
A large seed initially at rest explodes
into two pieces which move off. Which
of these could be possible paths the
two pieces would take?
(II)
(I)
(III)
A ball is projected straight up. Which
graph shows the total energy of the
ball as a function of time?
(B)
(A)
t
(E)
t
t
(F)
t
(D)
(C)
t
t
Chapter 8
Torque ()
and
Angular Momentum (L)
Rotational Inertia (I)
Recall: From Newton’s 2nd law of linear motion:
Fnet = ma = m(v-vo)/t
mass (m) = measure of inertia of an object.
= measure of how difficult it is to
change linear velocity (v) of an object.
The larger the mass, the more difficult it is to
change its velocity v.
For rotation, moment of inertial or rotational
inertia (I) = measure of how difficult it is to
change angular velocity () of an object.
Rotational Inertia (I)
Rotational inertia (I) = measure of how difficult
it is to change angular velocity () of an object.
Rotational inertia (I) – Also called moment of
inertia:
• Depends on mass, m. (I  m).
• Depends on square of radius of rotation (I  r2)
• Depends on how mass m is distributed.
Same mass.
Different radius.
Which one easier to
get to start
rotating?
Rotational Inertia
For a point object, I = mr2
Point object = one whose
size is small c.f. radius r
m
r
Units of I = kg-m2
Rotational Inertia
For a point object, I = mr2
m1
r1
For many discrete point
objects with different
shapes rotated about the
same axis,
Total rotational inertia
= sum of I for each object.
Itotal = miri2
Rotational Inertia
 For
 For
a point object, I = mr2
objects of comparable size to
radius r, the moment of in inertia
depends on distribution of mass.
Rotational Inertia Table
Rotational Inertia (I)
Mass of earth = 5.975 x 1024 kg
Mean radius = 6.37 x 106 m
Earth-sun distance = 1.5 1011 m
What is the rotational inertia of
the earth’s spin about its axis?
Example
m1 = 3 kg
m2 = 5 kg
Two solid spherical balls are joined together
by a 4-meter long steel rod. If they are
spun about a vertical axis passing through
their center of mass, find the total
rotational inertia. [Hint, treat the spinning
balls as point objects with I = mR2].
Rotational Inertia (I)
Linear
Displacement (x)
Velocity (v)
Acceleration (a)
Mass (m)
Ktran = ½ mv2
Angular (Rotational)
Angle () in radians
Angular Velocity ()
Angular Acceleration ()
Rotational Inertia (I)
Krot = ½ I 2
Rotational Kinematics
Linear
Rotational
a = constant
v = vo + at
x = xo + vot + ½at2
 = constant
 = o + t
 = o + ot + ½ t2
v2 = vo2 + 2a(x - xo)
2 = o2 + 2 ( - o)
Comment on axes and sign
(i.e. what is positive and negative)
Whenever we talk about rotation, it is implied
that there is a rotation “axis”.
This is usually called the “z” axis (we usually
omit the z subscript for simplicity).
Counter-clockwise (increasing ) is usually
called positive.
Clockwise (decreasing ) is usually
called negative.
z
+
Example
You and a friend are playing on a merry-goround. You stand at the outer edge of the
merry-go-round and your friend stands
halfway between the outer edge and the center.
Assume the rotation rate of the merry-goround is constant. Who has the greatest
angular velocity?
1. You do
2. Your friend does
3. Same
Example
Who has the greatest tangential velocity (v)?
1. You do
2. Your friend does
3. Same
Rotational Kinetic Energy
v
r
m
Mass m in rotational motion.
Its rotational inertia, I = mr2
Since it is moving, it has kinetic energy.
K = ½ mv2
From v = r,
K = ½ mr22 = ½ (mr2) 2 = ½I2
Kinetic Energy of Rotating Disk
Consider a disk with radius R and mass M,
spinning with angular frequency 
Each “piece” of disk has speed vi=ri
Each “piece” has kinetic energy
»Ki = ½ mi v2
mi
» = ½ mi 2 ri2
ri
Combine all the pieces
» SKi = S ½ mi 2 ri2
»
= ½ (S mi ri2) 2
»
= ½ I 2
Rotational Kinetic Energy (Krot)
A rigid object spinning about a fixed
axis (pure rotational motion) has
rotational kinetic energy Krot = ½I2
[eg, spinning wheel]
If the rigid object is moving (sliding)
with velocity v without any spin (ie pure
translational motion), it has only
translational kinetic energy K = ½ mv2
[eg, skidding wheel]
Rotational Kinetic Energy (Krot)
If the rigid object is spinning with angular
velocity  while its center of mass moves
linearly with velocity vcm, it has both
translational and rotational kinetic energy.
Its total kinetic energy will be
Ktot = Ktran + Krot
½ mvcm2 + ½ I2
[eg, wheel rolling normally on the ground, ball
rolling on the ground]
 Ktran = ½ m v2 Linear Motion
 Krot = ½ I 2
Rotational Motion
Example
Who has the greater kinetic energy?
(Assume masses are equal)
1. You do
2. Your friend does
3. Same
Example
A 10-kg hollow cylindrical shell rolls on the
ground at a linear velocity of 5 m/s. Find
its total kinetic energy.
50 cm
50 cm
Torque ()
 Force
(F) is responsible for
change of linear velocity. Net
force results in linear
acceleration.
 Torque
() is responsible for
change in angular velocity. Net
torque results in angular
acceleration.
Torque
r

r
F
 = rF
= (rsin )F = rFsin
r = rsin = lever arm (moment arm)
Torque = lever arm x force
Unit: m-N
r
 = rF = rsinF

F
Larger lever arm r, larger torque  if F stays
unchanged.
2. Larger force F, larger torque  if lever arm
stays unchanged.
3. If r = 0,  = 0.
4. If  = 0,  = 0.
5. If  = 90o,  = rF
Torque
Torque is a vector quantity.
Counter clockwise rotation, F
torque is positive
F
Clockwise rotation, torque
is negative
A string is tied to a doorknob 0.80 m from
the hinge as shown in the figure. At the
instant shown, the force applied to the
string is 5.0 N. What is the torque on the
door?
hinge
Equilibrium
translational equilibrium – Fnet = 0.
 Fnet = 0, object at rest – static equilibrium.
 Fnet = 0, object moving with constant
velocity, - dynamic equilibrium.
 Recall,
equilibrium, net = 0.
 An object in equilibrium means it is in both
translational equilibrium and rotational
equilibrium.
 Conditions for equilibrium:
Fnet = 0 and net = 0
 Rotational
Equilibrium
Conditions for equilibrium:
Fnet = 0 and net = 0
 Torque
can be calculated about any desired axis.
 A judicious
 Choose
choice of axis helps.
axis at a point through which an
unknown force acts so that its torque does not
appear in the equation.
Center of Gravity

Is the point through which the
force of gravity acts.

Essentially the same as center
of mass.
M2 = 60 kg
M1 = 50 kg
1.5 m
2.5 m
pivot
The figure above is a snapshot of two
masses on a light beam placed on a pivot.
• What is the net torque about the pivot?
•
To be in rotational equilibrium, what
should have been the value of M2?
Example
The picture below shows two people lifting a
heavy log. Which of the two people is
supporting the greatest weight?
1. The person on the left is supporting the
greatest weight
2. The person on the right is supporting the
greatest weight
3. They are supporting the same weight
Example
The picture below shows two people (L &
R) lifting a 20-kg log. The length of the
log is 4.0 m and L is 1.0 m from the left
end while R is holding from the right
end. Find how much force each of them
have to exert.
L
R
Newton’s 2nd Law
Newton’s 2nd law for linear motion:
Fnet = ma
Eg. What force is required to change velocity of a
6 kg solid spherical object of radius 30 cm
object from 15 m/s to 25 m/s in 5 seconds?

Newton’s 2nd law for rotation:
net = I
Eg. What torque is required to change angular
velocity of a 6 kg solid spherical object of
radius 30 cm object from 15 rad/s to 25 rad/s in
5 seconds? [Assume spin about the axis]

Work Done by Torque
 Recall,
work done by constant force
W = F x cos 
Also, Wnet = K = ½ mvf2 – ½ mvi2
 For rotation, work done by constant torque:
W = Ftangential s
= Ftangential r 
=  
Also, Wnet = Krot = ½ If2 – ½ Ii2
Power, P = W/t =  /t
= 
Work Done by Torque
A 182-kg flywheel has an effective radius
of 0.62 m.
(a)How much torque will take it from rest
to a rotational speed of 120 rpm within
30.0 s?
(b) How much work will this take?
Rolling Object

A rolling object posses both Ktran (½ mv2CM)
and Krot (½ I2)
E1 = U1 + K1 = mgh + 0
h
E2 = U2 + K2
0 + ½ mv2 + ½ I2
• If it came down rolling: mgh = ½ mvA2 + ½ I2
• If it came down only skidding, mgh = ½ mvB2
Angular Momentum (L)
Recall: Linear Momentum: p = mv
Angular Momentum, L = I
Angular momentum of a rigid object rotating
about a fixed axis.
L is a vector quantity.
Units = kg-m2/s
Direction: If rotation is CCW, L is upward.
If rotation is CW, L is downward.
Conservation of Angular Momentum
If net external torque acting on a rotating object
is zero, total angular momentum will be
conserved.
If net = 0, Lbefore = Lafter
OR Iii = Iff
EG: Skater spinning with her arms extended away
from her body. Lbefore = Iii = (mri2)i
She then decides to pull her arms close to her body
so that her spinning radius becomes smaller
(rf < ri) and Lafter = Iff = (mrf2)f
Conservation of Angular Momentum
Since no external torque acts on her, principle of
conservation of angular momentum will hold:
Lbefore = Lafter or Iii = Iff or (mri2)i = (mrf2)f
Since ri > rf, i < f
ie, the skater will spin faster by just folding her
arms.
Example
• A 1000-kg merry-go-round of radius 8.0 m
is spinning at a uniform speed of 25 rpm.
If a 150-kg mass is gently placed at the
edge, what will be the final spin speed of
the merry-go-round?
Example
• Which of the forces in the figure below
produces the largest torque about the
rotation axis indicated? Assume all the four
forces have equal magnitudes.
(A) 2
(B) 3
(C) 4
(D) The torques are equal
4
3
2
Axis
1
A 250-kg merry-go-round of radius 2.2 m is
spinning at a uniform angular speed of 3.5 rad/s.
Calculate the magnitude of its angular momentum.
Use I = ½ MR2
93%
A.
B.
C.
D.
E.
605 kg-m2/s
1.93 x 103 kg-m2/s
7.41 x 103 kg-m2/s
2.12 x 103 kg-m2/s
963 kg-m2/s
3%
A.
0%
B.
5%
0%
C.
D.
E.
A 30.0-cm long wrench is used to generate a
torque of 14.6 N-m at a bolt. A force of 70.0 N is
applied at the end of the wrench at an angle of 
to the wrench. The angle 95%
is
A.
B.
C.
D.
44o
0.4o
9.2o
3.6o
2%
A.
B.
0%
C.
2%
D.
A 3.2-kg solid sphere of diameter 28 cm spins at
5.0 rad/s about an axis passing through its
center. How much work is needed to bring it to
80%
rest within 15 sec? [Use I = 2/5 x mr2]
A.
B.
C.
D.
E.
0.025 J
1.25 J
1.25x104 J
3.14 x 103 J
0.31 J
8%
5%
8%
0%
A.
B.
C.
D.
E.
Abel and Brian support a 68.0-kg uniform bar that
is 4.0 m long. Abel holds the bar 1.0 m from one
end, while Brian holds from the other end. How
much force does Abel have50%
to exert?
A.
B.
C.
D.
E.
444 N
45.3 N
1.33x103 N
333 N
222 N
30%
10%
A.
B.
5%
5%
C.
D.
E.