Transcript Physics 106P: Lecture 1 Notes

Physics 101: Lecture 18
Rotational Dynamics

Today’s lecture will cover Textbook Sections 9.4 - 9.6:
Quick review of last lecture:
torque, center of gravity, equilibrium
Moment of inertia
Newton’s second law for rotational motion
Rotational work and kinetic energy
Angular momentum
Physics 101: Lecture 18, Pg 1
Torque and Equilibrium
Find Mpaint so that
plank is in equilibrium.
d1
Find total torque
about this axis
d2
d3
FA
Mpaintg
mg
Mg
t(mg) = mgd1
t(FA) = 0
t(Mg) = -Mgd2
t(paint) = -Mpaintgd3
Total torque = 0 = mgd1 –Mgd2 - Mpaintgd3
Thus, when Mpaint=(md1-Md2)/d3
then the plank is in equilibrium.
Physics 101: Lecture 18, Pg 2
Moment of Inertia

When torque is the analogue of force what is the
analogue of mass or what is the measure of inertia of a
rigid body in rotational motion about a fixed axis ?
Consider a tangential force FT acting on a particle with
mass M rotating on a circular path with radius R. The
torque is given by
t = FT R = M aT R = M a R R = (M R2) a
I=M R2 is called the moment of inertia of the particle.
For any rigid body : I= S (m r2)
SI unit: [kg m2]
Any rigid body has an unique total mass, but the moment
of inertia depends on how the mass is distributed with
respect to the axis of rotation.
Physics 101: Lecture 18, Pg 3
Moments of Inertia of Common Objects
Hollow cylinder or hoop about central axis
I = MR2
Solid cylinder or disk about central axis
I = MR2/2
Solid sphere about center
I = 2MR2/5
Uniform rod about center
I = ML2/12
Uniform rod about end
I = ML2/3
Physics 101: Lecture 18, Pg 4
Concept Question
The picture below shows two different dumbbell shaped objects.
Object A has two balls of mass m separated by a distance 2L, and
object B has two balls of mass 2m separated by a distance L.
Which of the objects has the largest moment of inertia for rotations
around the x-axis?
m
1. A
CORRECT
2. B
2m
3. Same
2L
L
x
2m
m
A
I = mL2 + mL2
= 2mL2
B
I = 2m(L/2)2 + 2m(L/2)2
= mL2
Physics 101: Lecture 18, Pg 5
Newton’s 2nd Law

If a net torque is applied to a rigid body rotating about
a fixed axis, it will experience an angular acceleration:
S text = I a
a in rad/s2
For the same net torque, the angular acceleration is the
larger the smaller the moment of inertia.
Physics 101: Lecture 18, Pg 6
Rotational Work and Kinetic Energy
Work done by a constant torque in turning an object through an
angle q :
WR = t q
q in rad SI Unit: [J]
Translational kinetic energy:
KEtrans = 1/2 m v2
Rotational kinetic energy:
KErot = 1/2 I 2
 in rad/s
SI Unit: [J]
Rotation plus translation:
KEtotal = KEtrans + KErot = ½ m v2 + 1/2 I 2
Physics 101: Lecture 18, Pg 7
Conservation of Mechanical Energy

The total mechanical energy for a rigid body with mass
m and moment of inertia I is given by
the sum of translational and rotational kinetic energy
and gravitational potential energy:
E = KEtrans + KErot + E pot =
= ½ m v2 + ½ I 2 + m g h
If the work done by non-conserving forces and torques
is zero, the total mechanical energy is conserved (final
equals initial total energy) :
Ef = E0 if Wnc=0
Physics 101: Lecture 18, Pg 8
Angular Momentum

The rotational analogue to momentum (1-dim: p = m v) in linear
motion is angular momentum:
L=I
 in rad/s
SI unit: [kg m2/s]
Conservation of Angular Momentum:
If the net average external torque is zero, the angular
momentum is conserved, i.e. the final and initial angular momenta
are the same :
Lf = L0 if S tave,ext = 0
Physics 101: Lecture 18, Pg 9
See text: chapters 8-9
Rotation Summary
(with comparison to 1-d linear motion)
Angular
a = constant, t0=0s
 = 0 + a t
q = q0 + 0 t + ½ a t2
2 = 02 + 2 a (q-q0)
Linear
a = constant, t0 =0 s
v = v0 + a t
x = x0 + v0 t + ½ a t2
v2 = v02 + 2 a (x-x0)
St=Ia
W=tq
KErot 
1
I 2  L2 / 2 I
2
L=I
See Table 8.1, 9.2
SF=ma
W=Fs
KEtrans 
1
mv 2  p 2 /( 2m)
2
p=mv
Physics 101: Lecture 18, Pg 10