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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 1

Physics 141.Lecture 17.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 2

Physics 141.Lecture 17.

• Course information.

• Quiz

• Topics to be discussed today (Chapter 11):• Rotational Variables

• Rotational Kinetic Energy

• Torque

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 3

Physics 141.Course information.

• Homework set # 7 is due on Friday 11/4 at noon.

• Homework set # 8 is due on Friday 11/11 at noon.

• Lab report # 4 is due on Wednesday 11/9 at noon.

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 4

The Personal Response System (PRS).Quiz.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 5

Rotational variables.

• The variables that are used todescribe rotational motion are:

• Angular position q

• Angular velocity ω = dq /dt

• Angular acceleration α = dω/dt

• The rotational variables arerelated to the linear variables:

• Linear position l = Rq

• Linear velocity v = Rω

• Linear acceleration a = Rα

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 6

Rotational variables.

• Things to consider when lookingat the rotation of rigid objectsaround a fixed axis:

• Each part of the rigid object hasthe same angular velocity.

• Only those parts that are locatedat the same distance from therotation axis have the same linearvelocity.

• The linear velocity of parts of therigid object increases withincreasing distance from therotation axis.

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 7

Rotational variables.

• Note: the acceleration at = rα isonly one of the two component ofthe acceleration of point P. Thetwo components of theacceleration of point P are:

• The radial component: thiscomponent is always present sincepoint P carried out circular motionaround the axis of rotation.

• The tangential component: thiscomponent is present only whenthe angular acceleration is notequal to 0 rad/s2.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 8

Rotational variables.

Angular velocity and acceleration are vectors! They have a magnitude and a direction. The direction of ω is found using the right-hand rule.The angular acceleration is parallel or anti-parallel to the angular velocity:

If ω increases: parallelIf ω decreases: anti-parallel

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 9

Rotational kinetic energy.

• Since the components of a rotating object have a non-zero(linear) velocity we can associate a kinetic energy with therotational motion:

• The kinetic energy is proportional to the rotational velocityω. Note: the equation is similar to the translational kineticenergy (1/2 mv2) except that instead of being proportional tothe the mass m of the object, the rotational kinetic energy isproportional to the moment of inertia I of the object:

Note: units of I: kg m2

K =

12

mivi2

i∑ =

12

mi ωri( )2

i∑ =

12

miri2

i∑⎛

⎝⎜⎞

⎠⎟ω 2 =

12

Iω 2

I = miri

2

i∑

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 10

The moment of inertia I.Calculating I.

• The moment of inertia of anobjects depends on the massdistribution of object and on thelocation of the rotation axis.

• For discrete mass distribution itcan be calculated as follows:

• For continuous mass distributionswe need to integrate over themass distribution:

I = miri

2

i∑

I = r 2dm∫

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 11

Calculating the moment of inertia.Sample problem.

• Consider a rod of length L andmass m. What is the moment ofinertia with respect to an axisthrough its center of mass?

• Consider a slice of the rod, withwidth dx, located a distance xfrom the rotation axis. The massdm of this slice is equal to

axis

L/2L/2

x

dx

dm =

mL

dx

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 12

Calculating the moment of inertia.Sample problem.

• The moment of inertia dI of thisslice is equal to

• The moment of inertia of the rodcan be found by adding thecontributions of all of the slicesthat make up the rod:

axis

L/2L/2

x

dx

dI = x2dm =

mL

x2dx

I =mL

x2dx−L / 2

L / 2

∫ =m3L

L2

⎛⎝⎜

⎞⎠⎟

3

− −L2

⎛⎝⎜

⎞⎠⎟

3⎡

⎣⎢⎢

⎤

⎦⎥⎥=

112

mL2

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 13

Calculating the moment of inertia.Parallel-axis theorem.

• Calculating the moment ofinertial with respect to asymmetry axis of the object is ingeneral easy.

• It is much harder to calculate themoment of inertia with respect toan axis that is not a symmetryaxis.

• However, we can make a hardproblem easier by using theparallel-axis theorem:

Easy

HardIcmI

I = Icm + Mh2

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 14

Calculating the moment of inertia.Sample problem.

• Consider a rod of length L andmass m. What is the moment ofinertia with respect to an axisthrough its left corner?

• We have determined the momentof inertia of this rod with respectto an axis through its center ofmass. We use the parallel-axistheorem to determine the momentof inertia with respect to thecurrent axis:

x

dx

axis

L

I = Icm + m L

2⎛⎝⎜

⎞⎠⎟

2

=1

12mL2 +

14

mL2 =13

mL2

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 15

3 Minute 2 Second Intermission.

• Since paying attention for 1 hourand 15 minutes is hard when thetopic is physics, let’s take a 3minute 2 second intermission.

• You can:• Stretch out.• Talk to your neighbors.• Ask me a quick question.• Enjoy the fantastic music.• Solve a WeBWorK problem.

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 16

Torque.

• Consider a force F applied to anobject that can only rotate.

• The force F can be decomposedinto two two components:• A radial component directed

along the direction of the positionvector r. The magnitude of thiscomponent is Fcosq. Thiscomponent will not produce anymotion.

• A tangential component,perpendicular to the direction ofthe position vector r. Themagnitude of this component isFsinq. This component will resultin rotational motion.

Ar

φ

F

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 17

Torque.• If a mass m is located at the position

on which the force is acting (and weassume any other masses can beneglected), it will experience a linearacceleration equal to Fsinφ/m.

• The corresponding angularacceleration α is equal to

• Since in rotation motion the momentof inertia plays an important role, wewill rewrite the angular accelerationin terms of the moment of inertia:

Ar

φ

F

α = rF sinφ

mr 2 = rF sinφI

α = F sinφ

mr

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 18

Torque.

• Consider rewriting the previousequation in the following way:

rFsinf = Iα• The left-hand-side of this equation is

called the torque τ of the force F:τ = Iα

• This equation looks similar toNewton’s second law for linearmotion:

F = ma• Note:

linear rotationalmass m moment Iforce F torque τ

Ar

φ

F

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 19

Torque.

• In general the torque associated witha force F is equal to

• The arm of the force (also called themoment arm) is defined as rsinq.The arm of the force is theperpendicular distance of the axis ofrotation from the line of action of theforce.

• If the arm of the force is 0, the torqueis 0, and there will be no rotation.

• The maximum torque is achievedwhen the angle q is 90°.

!τ = rF sinθ = !r ×!F

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 20

Rotational motion.Sample problem.

• Consider a uniform disk withmass M and radius R. The disk ismounted on a fixed axle. A blockwith mass m hangs from a lightcord that is wrapped around therim of the disk. Find theacceleration of the falling block,the angular acceleration of thedisk, and the tension of the cord.

• Expectations:• Linear acceleration should

approach g when M approaches 0kg.

m

M, R

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 21

Rotational motion.Sample problem.

• Start with considering the forcesand torques involved.

• Define the sign convention to beused.

• The block will move down andwe choose the positive and wechoose the positive y axis in thedirection of the linearacceleration.

• The net force on mass m is equalto

T

mg

a

T

R

ma = mg − T

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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 22

Rotational motion.Sample problem.

• The net torque on the pulley is equalto

• The resulting angular acceleration isequal to

• Assuming the cord is not slipping wecan determine the linear acceleration:

T

mg

a

T

R

τ = RT

α =τI=

RT12

MR2=

2TMR

a = αR = 2

TM

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 23

Rotational motion.Sample problem.

• We now have two expressions fora:

• Solving these equations we find:

T

mg

a

T

R

Note: a = g when M = 0 kg!!!

a = 2

TM

a =

mg − Tm

= g −Tm

T =

MM + 2m

mg

a =

2mM + 2m

g

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 17, Page 24

Done for today!

Landing at Amsterdam Airport.