physics 1501: lecture 19 today ’ s agenda

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


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


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


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


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


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


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


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.


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


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


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



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



Finally, calculate I for the same object about an axis along one side (as shown):

mm

mm

L

r

I = 2mL2



For a single object, I clearly depends on the rotation axis !!

L

I = 2mL2I = mL2

mm

mm

I = 2mL2


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



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


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


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


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


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.

physics 1501: lecture 19 today ’ s agenda

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