2ndLE Lecture 27 - R10 Rotational Dynamics (2)

7/25/2019 2ndLE Lecture 27 - R10 Rotational Dynamics (2)

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Lecture 27: Rotational Dynamics II

Objectives1. Apply Newtons 2nd law of rotation and conservation of

energy to obtain qualitative and quantitative conclusions

on the motion of a system that involves rotating about a

moving axis.2. Analyze work and power delivered to a rotating system.


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Again: v = vcmalso v = rand a = r

Sliding (slipping only) K= mv2

Rolling (w/o slipping) K= mv2 + I2

where I depends on the shape of the object

Recall: surface with friction (rolling without slipping)


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Newtons second law

3

=

Rotational analog of Newtons second law

= Relationship of translational and angular velocitiesand acceleration, rotating without slipping:

= = Energy relationships: K = K

translation+ K

rotation

= +


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Sample Problem 1: Unwinding Cable

A cable is wrapped several times around a uniform solid cylinder

that can rotate about its axis. The cylinder has diameter 0.120m

and mass 50kg. The cable is pulled with a force of 9.0N. Assumingthat the cable unwinds without stretching or slipping, what is its

acceleration? Icm of solidcyl= MR2/2


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First calculate the angular acceleration of the cylinder using

Newtons 2ndlaw: = .Torque (

z= FR) is positive as it tends to cause a

counterclockwise rotation:

Therefore the acceleration is:


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Sample Problem 2: (Tipler)

A cue stick hits a cue ball horizontally a distance x above

the center of the ball. Find the value of x for which thecue ball will roll without slipping from the beginning.

Express your answer in terms of the radius R of the ball.

for a solid sphere: = 2

5

Cue ball possess both linear and angular acceleration (use this to

relate x with R; = )The net torque about the center of the ball with lever arm, x is:

=


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For varying torque:

=

For a constant torque and finite

change in angle,

= = Unit of work:Joules (J)

Work and Power in rotational motion

Power is the rate of doing work: = Ave. Power: =

Or for Instantaneous Power: =

a tangential forceF

tanis involved


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Sample Problem 3 (Young and Freedman):

The power output of an automobile engine is advertised to

be 200hp (1.49x105W) at 6000rpm. What is the

corresponding torque?

Using formula for power:

Convert angular velocity to rad/s

=


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Given:

= 10.0Nm

I = 2.0kgm2

t = 8.0s0= 0; 0=0

Required: P

Sample Problem 4 (Young and Freedman):

An electric motor exerts a constant torque of 10Nm on a

grindstone mounted on its shaft. The moment of inertia of the

grindstone about the shaft of 2.0kgm2.(a) If the system starts from rest, find the work done by the

motor in 8.0s and the kinetic energy at the end of this time.

(b) What was the average power delivered by the motor?

= =10

Since: I = 2.0kgm2

, therefore:

= 5.0rad/s2

Note: motion here is only rotational.

Using the rotational version of Newtons 2ndLaw:

l d h l l


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constant acceleration, 0= 0 and 0=0 the total angular

displacement is:

Therefore, the total work done by torque is:

Again, 0= 0 , the angular velocity and KE after 8.0s are:

The power becomes: P = W/t


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Summary: Work, Power and Rotational Dynamics

Work with constant torque: = =

Average Power: = Instantaneous Power: =

Rotational analog of Newtons 2ndlaw:

=


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2ndLong exam

26 October (Mon, 12:00-02:00PM)

Coverage: Ch 6 to 10 (+ dot and cross product)

Problem set for chapters 6,7 and 8 will

be emailed on FridayDeadline: Oct. 27 (Tue)


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Seatwork :

2 points per letter

(1pt magnitude, 1pt direction)

(a) 40.0Nm

(+z)

(b) 34.6Nm

(+z)(c) 20.0Nm

(+z)

(d) 17.3Nm

(-z)(e) Zero

(f) Zero


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15

Young and Freedman, 10.21What fraction the total kinetic energy is rotational for

the following objects rolling without slipping on a

horizontal surface?

SW13: Uniform solid cylinder ( = )

=

=

=12

2 = 1212

2 90

= 14

2 =

12 2

=+= + 2 = 2


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16

Young and Freedman, 10.21What fraction the total kinetic energy is rotational for

the following objects rolling without slipping on a

horizontal surface?

SW14: Uniform sphere ( = )

=

=

7

=12

2 = 1225

2 90

= 15

2 =

12 2

=+= + 2 = 7 2


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Seatwork- solve problems in your

notebooks

- write the answers only inyour bluebook

- indicate the date

October 15, 20141. Blah?

2. Blah blah!

3. Blah blah blah!

4. Blah blah blah blah!

17


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Conversion: 1rev = 2rad Uniform disk: Icm= MR2

(1) A flywheel used for storing energy consists of a uniformdisk of mass 1.50x105kg and diameter 4.40m that rotates

at 3000rpm (314rad/s) about its center of mass. What its

kinetic energy?

(a) 1.11x109J (b) 1.79x1010J (c) 4.82x1010J

(2) A car delivers 175Nm of torque at 5000rpm. What is the

power output of the car at that engine speed?(a) 91.5KW (b) 103KW (c) 2.00kW


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4-5) The flywheel of an engine has moment of inertia

2.50kgm2about its rotation axis. The flywheel rotates at

41.9rad/s in 8.00s starting from rest.(4) What is the angular acceleration of the flywheel?

(5) What is the torque applied on the flywheel?

(3) For a hoop lying in the -plane, which of the followingrequires more workbe done by an external agent to accelerate

the hoop from rest to an angular speed

?

a. Rotation about the -axis through the center of the hoopb. Rotation about the axis parallel to the -axis passing

through a point on the hoopsrimc. Rotation about an axis parallel to the

-axis a distance

3

from the center of the loop


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Seatwork

- seatwork will be checked at the

end of the class

- if wrong, write the correct

answer- in checking: place the score

above the checkers name

- the checker must sign under

his/her name & student number

October 15, 20141. Blah?

2. Blah blah!

3. Blah blah blah!

4. Blah blah blah blah!

Score: 3/4Checked by:

(signed)

Albert Einstein Jr.

(2013-24601)

X Bleh!!!

20


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Seatwork 1

(b)


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Seatwork 2:

(a)

2ndLE Lecture 27 - R10 Rotational Dynamics (2)

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