Physics 1501: Lecture 19, Pg 1
Physics 1501: Lecture 19Physics 1501: Lecture 19
TodayToday’’s Agendas Agenda Announcements
HW#7: due Oct. 21
Midterm 1: average = 45 % …
TopicsRotational KinematicsRotational Energy Moments of Inertia
Physics 1501: Lecture 19, Pg 2
Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)
Angular Linear
And for a point at a distance R from the rotation axis:
x = Rv = Ra = R
Physics 1501: Lecture 19, Pg 3
Example: Wheel And RopeExample: Wheel And Rope A wheel with radius R = 0.4m rotates freely about a fixed
axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)
aa
R
Physics 1501: Lecture 19, Pg 4
Wheel And Rope...Wheel And Rope... Use a = R to find :
= a / R = 4m/s2 / 0.4m = 10 rad/s2
Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester.
= 0 + 0(10) + (10)(10)2 = 500 rad
aa
R
Physics 1501: Lecture 19, Pg 5
Rotation & Kinetic EnergyRotation & Kinetic Energy
Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).
The kinetic energy of this system will be the sum of the kinetic energy of each piece:
rr1
rr2rr3
rr4
m4
m1
m2
m3
Physics 1501: Lecture 19, Pg 6
Rotation & Kinetic Energy...Rotation & Kinetic Energy...
rr1
rr2rr3
rr4
m4
m1
m2
m3
vv4
vv1
vv3
vv2
which we write as:
Define the moment of inertiamoment of inertia
about the rotation axis I has units of kg m2.
So: but vi = ri
Physics 1501: Lecture 19, Pg 7
Lecture 19, Lecture 19, Act 1Act 1Rotational Kinetic EnergyRotational Kinetic Energy
I have two basketballs. BB#1 is attached to a 0.1m long rope. I spin around with it at a rate of 2 revolutions per second. BB#2 is on a 0.2m long rope. I then spin around with it at a rate of 2 revolutions per second. What is the ratio of the kinetic energy of BB#2 to that of BB#1?
A) 1/4 B) 1/2 C) 1 D) 2 E) 4
BB#1 BB#2
Physics 1501: Lecture 19, Pg 8
Rotation & Kinetic Energy...Rotation & Kinetic Energy...
The kinetic energy of a rotating system looks similar to that of a point particle:
Point ParticlePoint Particle Rotating System Rotating System
v is “linear” velocity
m is the mass.
is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 1501: Lecture 19, Pg 9
Moment of InertiaMoment of Inertia
Notice that the moment of inertia I depends on the distribution of mass in the system.The further the mass is from the rotation axis, the bigger the moment of inertia.
For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass).
We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics !
So where
Physics 1501: Lecture 19, Pg 10
Calculating Moment of InertiaCalculating Moment of Inertia
We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is:
where r is the distance from the mass
to the axis of rotation.
Example: Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
mm
mm
L
Physics 1501: Lecture 19, Pg 11
Calculating Moment of Inertia...Calculating Moment of Inertia...
The squared distance from each point mass to the axis is:
mm
mm
Lr
L/2
so
I = 2mL2
Using the Pythagorean Theorem
Physics 1501: Lecture 19, Pg 12
Calculating Moment of Inertia...Calculating Moment of Inertia...
Now calculate I for the same object about an axis through the center, parallel to the plane (as shown):
mm
mm
L
r
I = mL2
Physics 1501: Lecture 19, Pg 13
Calculating Moment of Inertia...Calculating Moment of Inertia...
Finally, calculate I for the same object about an axis along one side (as shown):
mm
mm
L
r
I = 2mL2
Physics 1501: Lecture 19, Pg 14
Calculating Moment of Inertia...Calculating Moment of Inertia...
For a single object, I clearly depends on the rotation axis !!
L
I = 2mL2I = mL2
mm
mm
I = 2mL2
Physics 1501: Lecture 19, Pg 15
Lecture 19, Lecture 19, Act 2Act 2Moment of InertiaMoment of Inertia
A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.Which of the following is correct:
(a)(a) Ia > Ib > Ic
(b)(b) Ia > Ic > Ib
(c)(c) Ib > Ia > Ic
a
b
c
Physics 1501: Lecture 19, Pg 16
Calculating Moment of Inertia...Calculating Moment of Inertia...
For a discrete collection of point masses we found:
For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.
We have to do anintegral to find I : r
dm
Physics 1501: Lecture 19, Pg 17
Moments of InertiaMoments of Inertia
Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center.
Some examples of I for solid objects:
RL
rdr
Physics 1501: Lecture 19, Pg 18
Moments of Inertia...Moments of Inertia... Some examples of I for solid objects:
Solid sphere of mass M and radius R,
about an axis through its center.R
Thin spherical shell of mass M and radius R, about an axis through its center.
R
Physics 1501: Lecture 19, Pg 19
Moments of InertiaMoments of Inertia
Some examples of I for solid objects:
Thin hoop (or cylinder) of mass M and
radius R, about an axis through its center,
perpendicular to the plane of the hoop.R
Thin hoop of mass M and radius R,
about an axis through a diameter.
R
Physics 1501: Lecture 19, Pg 20
Parallel Axis TheoremParallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass is known, = ICM
The moment of inertia about an axis parallel to this axis but a distance R away is given by:
IPARALLEL = ICM + MR2
So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.