rolling, torque, and angular momentum chapter 11 copyright © 2014 john wiley & sons, inc. all...

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Rolling, Torque, and Angular Momentum

Chapter 11

© 2014 John Wiley & Sons, Inc. All rights reserved.

11-1 Rolling as Translation and Rotation Combined

11.01 Identify that smooth rolling can be considered as a combination of pure translation and pure rotation.

11.02 Apply the relationship between the center-of-mass speed and the angular speed of a body in smooth rolling.

Learning Objectives

Figure 11-2



We consider only objects that roll smoothly (no slip) The center of mass (com) of the object moves in a

straight line parallel to the surface The object rotates around the com as it moves The rotational motion is defined by:

Figure 11-3



The figure shows how the velocities of translation and rotation combine at different points on the wheel



Figure 11-4



Figure 11-4

Answer: (a) the same (b) less than


11-2 Forces and Kinetic Energy of Rolling

Combine translational and rotational kinetic energy:



If a wheel accelerates, its angular speed changes

A force must act to prevent slip



If slip occurs, then the motion is not smooth rolling! For smooth rolling down a ramp:

1. The gravitational force is vertically down

2. The normal force is perpendicular to the ramp

3. The force of friction points up the slope



We can use this equation to find the acceleration of such a body

Note that the frictional force produces the rotation Without friction, the object will simply slide



Answer: The maximum height reached by B is less than that reached by A.

For A, all the kinetic energy becomes potential energy at h.

Since the ramp is frictionless for B, all of the rotational K stays rotational, and only the translational kinetic energy becomes potential energy at its maximum height.

11-3 The Yo-Yo

As yo-yo moves down string, it loses potential energy mgh but gains rotational & translational kinetic energy

Find linear acceleration of yo-yo accelerating down its string:

1.Rolls down a “ramp” of angle 90°

2.Rolls on an axle instead of its outer surface

3.Slowed by tension T rather than friction


11-3 The Yo-Yo

Replacing the values leads us to:

Example Calculate the acceleration of the yo-yo

o M = 150 grams, R0 = 3 mm, I

com = Mr2/2 = 3E-5 kg m2

o Therefore acom

= -9.8 m/s2 / (1 + 3E-5 / (0.15 * 0.0032)) = - .4 m/s2


11-4 Torque Revisited

Previously, torque was defined only for a rotating body and a fixed axis

Now we redefine it for an individual particle that moves along any path relative to a fixed point

The path need not be a circle; torque is now a vector Direction determined with right-hand-rule

Figure 11-10



The general equation for torque is:

We can also write the magnitude as:

Or, using the perpendicular component of force or the moment arm of F:



Answer: (a) along the z direction (b) along the +y direction (c) along the +x direction

Loosen a bolt

• Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt?

• Fa?

• Fb?

• Fc?

Loosen a bolt

• Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt?

• Fb!

Copyright © 2012 Pearson Education Inc.

Torque

• Let O be the point around which a solid body will be rotated.

O


Torque

• Let O be the point around which a solid body will be rotated.

• Apply an external force F on the body at some point.

• The line of action of a force is the line along which the force vector lies. F applied

Line of action

O


Torque

• Let O be the point around which a body will be rotated.

• The lever arm (or moment arm) for a force

is

The perpendicular distance from O to the line of action of the force

Line of action

O

L

F applied


Torque

Line of action

O

• The torque of a force with respect to O is the product of force and its lever arm.

t = F L

• Greek Letter “Tau” = “torque”

• Larger torque if

• Larger Force applied

• Greater Distance from O

L

F applied


Torque

t = F L

Larger torque if

• Larger Force applied

• Greater Distance from O


Torque

Line of action

O

• The torque of a force with respect to O is the product of force and its lever arm.

t = F L

• Units: Newton-Meters

• But wait, isn’t Nm = Joule??

L

F applied


Torque

• Torque Units: Newton-Meters

• Use N-m or m-N

• OR...Ft-Lbs or lb - feet

• Not…

• Lb/ft• Ft/lb• Newtons/meter• Meters/Newton


Torque

• Torque is a ROTATIONAL VECTOR!

• Directions:

• “Counterclockwise” (+)

• “Clockwise” (-)

• “z-axis”


Torque


• Directions:

• “Counterclockwise”

• “Clockwise”


Torque


• Directions:

• “Counterclockwise” (+)

• “Clockwise” (-)

• “z-axis”


Torque


• But if r = 0, no torque!


Torque as a vector

• Torque can be expressed as a vector using the vector cross product:

t= r x F

• Right Hand Rule for direction of torque.


Torque as a vector

• Torque can be expressed as a vector using the vector cross product:

t= r x F

• Where r = vector from axis of rotation to point of application of Force

q = angle between r and F

• Magnitude: t = r F sinq

F applied

Line of action

O r

Lever arm

q


Torque as a vector

• Torque can be evaluated two ways:

t= r x F

t= r * (Fsin q)

remember sin q = sin (180-q)

F applied

Line of action

O r

Lever arm

q

Fsinq


Torque as a vector

• Torque can be evaluated two ways:

t= r x F

t= (rsin q) * F

The lever arm L = r sin q F applied

Line of action

O r

Lever arm

r sin q

q


Torque as a vector

• Torque is zero three ways:

t= r x F

t= (rsin q) * F

t= r * (Fsin q)

If q is zero (F acts along r)

If r is zero (f acts at axis)

If net F is zero (other forces!)

F applied

Line of action

O r

Lever arm

r sin q

q

Applying a torque

• Find t if F = 900 N and q = 19 degrees



Example Calculating net torque:

Figure 11-11


11-5 Angular Momentum

Angular counterpart to linear momentum

Note that particle need not rotate around O to have angular momentum around it

Unit of angular momentum is kg m2/s, or J s



To find direction of angular momentum, use right-hand rule to relate r & v to the result

To find magnitude, use equation for magnitude of a cross product:

This is equivalent to:



Angular momentum has meaning only with respect to a specified origin and axis.

L is always perpendicular to plane formed by position & linear momentum vectors



Angular momentum has meaning only with respect to a specified origin

It is always perpendicular to the plane formed by the position and linear momentum vectors

Answer: (a) 1 & 3, 2 & 4, 5

(b) 2 and 3 (assuming counterclockwise is positive)


11-6 Newton's Second Law in Angular Form

We rewrite Newton's second law as:

Torque and angular momentum must be defined with respect to the same point (usually the origin)



We rewrite Newton's second law as:

Note the similarity to the linear form:



Answer: (a) F3, F

1, F

2 & F

4 (b) F

3 (assuming counterclockwise is positive)


Sum to find angular momentum of a system of particles:

The rate of change of the net angular momentum is:

11-7 Angular Momentum of a Rigid Body



The rate of change of the net angular momentum is:

In other words, net torque is defined by this change:



We can find the angular momentum of a rigid body through summation:

The sum is the rotational inertia I of the body



Therefore this simplifies to:



Answer: (a) All angular momenta will be the same, because the torque is the same in each case (b) sphere, disk, hoop


11-8 Conservation of Angular Momentum

Since we have a new version of Newton's second law, we also have a new conservation law:

The law of conservation of angular momentum states that, for an isolated system,

(net initial angular momentum) = (net final angular momentum)


Since these are vector equations, they are equivalent to the three corresponding scalar equations

This means we can separate axes and write:



If the distribution of mass changes with no external torque, we have:



Examples A student spinning on a stool: rotation speeds up when arms

are brought in, slows down when arms are extended A springboard diver: rotational speed is controlled by tucking

her arms and legs in, which reduces rotational inertia and increases rotational speed

A long jumper: the angular momentum caused by the torque during the initial jump can be transferred to the rotation of the arms, by windmilling them, keeping the jumper upright



Answer: (a) decreases (b) remains the same (c) increases


The race of the rolling bodies

• Which one wins??

• WHY???

https://www.youtube.com/watch?v=tcs93mPn91E

https://www.youtube.com/watch?v=CHQOctEvtTY


Acceleration of a yo-yo

• We have translation and rotation, so we use Newton’s second law for the acceleration of the center of mass and the rotational analog of Newton’s second law for the angular acceleration about the center of mass.

• What is a and T for the yo-yo?


Acceleration of a rolling sphere

• Use Newton’s second law for the motion of the center of mass and the rotation about the center of mass.

• What are acceleration & magnitude of friction on ball?


Work and power in rotational motion

• Tangential Force over angle does work

• Total work done on a body by external torque is equal to the change in rotational kinetic energy of the body

• Power due to a torque is P = zz



• Calculating Power from Torque

• Electric Motor provide 10-Nm torque on grindstone, with I = 2.0 kg-m2 about its shaft.

• Starting from rest, find work W down by motor in 8 seconds and KE at that time.

• What is the average power?



• Calculating Power from Force

• Electric Motor provide 10-N force on block, with m = 2.0 kg.

• Starting from rest, find work W down by motor in 8 seconds and KE at that time.

• What is the average power?

• W = F x d = 10N x distance travelled

• Distance = v0t + ½ at2 = ½ at2

• F = ma => a = F/m = 5 m/sec/sec => distance = 160 m

• Work = 1600 J; Avg. Power = W/t = 200 W


Angular momentum • The angular momentum of a rigid body rotating about a

symmetry axis is parallel to the angular velocity and is given by L = I.

• Units = kg m2/sec (“radians” are implied!)

• Direction = along RHR vector of angular velocity.


Angular momentum • For any system of particles = dL/dt. • For a rigid body rotating about the z-axis z = Iz.

Turbine with I = 2.5 kgm2; w = (40 rad/s3)t2

What is L(t) & L @ t = 3.0 seconds; what is t(t)?


Conservation of angular momentum

• When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved).


A rotational “collision”


Angular momentum in a crime bust

• A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is w of the door?


Angular momentum in a crime bust

• A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is w of the door?

• Lintial = mbulletvbulletlbullet

• Lfinal = Iw = (I door + I bullet) w

• Idoor = Md2/3

• I bullet = ml2

• Linitial = Lfinal so mvl = Iw

• Solve for w and KE final


Gyroscopes and precession

• For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.


11-9 Precession of a Gyroscope

A non-spinning gyroscope falls



A spinning gyroscope instead rotates around a vertical axis

This rotation is called precession



• For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.


A rotating flywheel

• Magnitude of angular momentum constant, but its direction changes continuously.



Angular momentum of (rapidly spinning) gyroscope is:

Torque can only change direction of L, not magnitude

Only way direction can change along direction of torque without its magnitude changing is if it rotates around the central axis

Therefore it precesses instead of toppling over



The precession rate is given by:

True for a sufficiently rapid spin rate Independent of mass, (I is proportional to M) but does

depend on g Valid for a gyroscope at an angle to the horizontal as

well (a top for instance)

rolling, torque, and angular momentum chapter 11 copyright © 2014 john wiley & sons, inc. all...

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