Chapter 8Chapter 8Rotational Motion of Rotational Motion of Solid ObjectsSolid Objects

What does a yo-yo have in common with a merry-go-

round?

What Is Rotational Motion?

How can we describe this type of motion?

Linear Motion vs. Rotational Motion

Linear motion involves an object moving from one point to another in a straight line.

Rotational motion involves an object rotating about an axis.Examples include a merry-go-round, the rotating

earth, a spinning skater, a top, and a turning wheel.

What causes rotational motion? Does Newton’s second law apply?

There is a useful analogy between linear motion and rotational motion. Rotational velocity is how fast the object is turning.

Units: revolutions per minute (rpm); degrees per second Analogous to linear velocity

Rotational displacement is how far the object rotates. Units: fractions of a complete revolution; degrees; radians 1 complete revolution = 360º = 2 radians Analogous to linear displacement: the straight-line distance traveled

by an object (including direction of travel)

Rotational acceleration is the rate of change of rotational velocity.

Units: revolutions per second per second (rev/s2); radians per second per second (rad/s2)

Analogous to linear acceleration

Constant acceleration equations for linear and rotational motion

v v0 at

d v0t 1

2at 2

0 t

0t 1

2t 2

Relationship between linear and rotational velocity

On a merry-go-round, a rider near the edge travels a greater distance in 1 revolution than one near the center. The outside rider is therefore traveling with a greater linear speed.

A merry-go-round is accelerated at a constant rate of 0.005 rev/s2, starting from rest.What is its rotational velocity at the end of 1 min?

= 0.005 rev/s2

0 = 0t = 60 s

= 0 + t = 0 + (0.005 rev/s2)(60 s) = 0.30 rev/s

a) 0.005 rev/sb) 0.03 rev/sc) 0.05 rev/sd) 0.30 rev/s

How many revolutions does the merry-go-round make in 1 minute?

= 0.005 rev/s2

0 = 0t = 60 s, = 0.30 rev/s

= 0t + 1/2 t2

= 0 + 1/2 (0.005 rev/s2)(60 s)2

= 9 rev

a) 1.5 revb) 3.0 revc) 9.0 revd) 18.0 rev

Torque and Balance

What causes the merry-go-round to rotate in the first place?

What determines whether an object will rotate?

If an unbalanced force causes linear motion, what causes rotational motion?

Torque and Balance

When is a balance balanced? Consider a thin but rigid beam supported by a

fulcrum or pivot point. If equal weights are placed at equal distances

from the fulcrum, the beam will not tend to rotate: it will be balanced.

To balance a weight twice as large as a smaller weight, the smaller weight must be placed twice as far from the fulcrum as the larger weight.

Both the weight and the distance from the fulcrum are important.

The product of the force and the distance from the fulcrum is called the torque.

Torque describes the tendency of a weight to produce a rotation.

The distance from the fulcrum to the point of application of the force must be measured in a direction perpendicular to the line of action of the force.

This distance is called the lever arm or moment arm.

A longer lever arm produces a greater torque.

For a force F and a lever arm l, the resulting torque is:

Which of the forces pictured as acting upon the rod will produce a torque about an axis perpendicular to the plane of the diagram

at the left end of the rod?

a) F1

b) F2 c) Both.d) Neither.

F2 will produce a torque about an axis at the left end of the rod. F1 has no lever arm with respect to the given axis.

The two forces in the diagram have the same magnitude. Which orientation will

produce the greater torque on the wheel?

F1 provides the larger torque. F2 has a smaller component perpendicular to the radius.

a) F1

b) F2 c) Both.d) Neither.

A 50-N force is applied at the end of a wrench handle that is 24 cm long. The force is applied in a direction perpendicular to the handle as shown. What is the torque applied to the nut by the wrench?

0.24 m 50 N = 12 N·m

a) 6 N·mb) 12 N·mc) 26 N·md) 120 N·m

What would the torque be if the force were applied half way up the handle instead of at the end?

0.12 m 50 N = 6 N·m

a) 6 N·mb) 12 N·mc) 26 N·md) 120 N·m

When the applied force is not perpendicular to the crowbar, for example, the lever arm is found by drawing the perpendicular line from the fulcrum to the line of action of the force.

We call torques that produce counterclockwise rotation positive, and torques that produce clockwise rotation negative.

Two forces are applied to a merry-go-round with a radius of 1.2 m as shown. What is the torque about the axle of the merry-go-round due to the 80-N force?

1.2 m 80 N = +96 N·m(counterclockwise)

a) +9.6 N·mb) -36 N·mc) +96 N·md) -36 N·m

What is the torque about the axle of the merry-go-round due to the 50-N force?

-(1.2 m 50 N) = -60 N·m

(clockwise)

a) +60 N·mb) -60 N·mc) +120 N·md) -120 N·m

What is the net torque acting on the merry-go-round?

96 N·m (counterclockwise) - 60 N·m (clockwise) = +36 N·m (counterclockwise)

a) +36 N·mb) -36 N·mc) +96 N·md) -60 N·me) +126 N·m

We want to balance a 3-N weight against a 5-N weight on a beam. The 5-N weight is placed 20 cm to the right of a fulcrum. What is the torque produced by the 5-N weight?

a) +1 N·mb) -1 N·mc) +4 N·md) +4 N·m

F = 5 N = - Fll = 20 cm = 0.2 m = - (5 N)(0.2 m)

= -1 N·m

How far do we have to place the 3-N weight from the fulcrum to balance the system?

a) 2 cmb) 27 cmc) 33 cmd) 53 cm

F = 3 N l = / F = +1 N·m = (+1 N·m) / (3 N)

= 0.33 m = 33 cm

The center of gravity of an object is the point about which the weight of the object itself exerts no torque.

We can locate the center of gravity by finding the point where it balances on a fulcrum.

For a more complex object, we locate the center of gravity by suspending the object from two different points, drawing a line straight down from the point of suspension in each case, and locating the point of intersection of the two lines.

If the center of gravity lies below the pivot point, the object will automatically regain its balance when disturbed.

The center of gravity returns to the position directly below the pivot point, where the weight of the object produces no torque.

How far can the child walk

without tipping the

plank?

For a uniform plank, its center of gravity is at its geometric center.

The pivot point will be the edge of the supporting platform. The plank will not tip as long as the counterclockwise torque

from the weight of the plank is larger than the clockwise torque from the weight of the child.

The plank will verge on tipping when the magnitude of the torque of the child equals that of the plank.

An 80-N plank is placed on a dock as shown. The plank is uniform in density so the center of

gravity of the plank is located at the center of the plank. A 150-N boy standing on the plank walks

out slowly from the edge of the dock. What is the torque exerted by the weight of the plank about

the pivot point at the edge of the dock?a) +80 N·mb) -80 N·mc) +160 N·md) -160 N·me) +240 N·mf) -240 N·m

1 m 80 N = +80 N·m(counterclockwise)

How far from the edge of the dock can the 150-N boy walk until the plank is just on the verge of

tipping?

a) 0.12 mb) 0.23 mc) 0.53md) 1.20 m

80 N·m / 150 N = 0.53 m

Rotational Inertia and Newton’s Second Law

In linear motion, net force and mass determine the acceleration of an object.

For rotational motion, torque determines the rotational acceleration.

The rotational counterpart to mass is rotational inertia or moment of inertia. Just as mass represents the resistance to a change in linear

motion, rotational inertia is the resistance of an object to change in its rotational motion.

Rotational inertia is related to the mass of the object. It also depends on how the mass is distributed about the

axis of rotation.

Simplest example: a mass at the end of a light rod

A force is applied to the mass in a direction perpendicular to the rod.

The rod and mass will begin to rotate about the fixed axis at the other end of the rod.

The farther the mass is from the axis, the faster it moves for a given rotational velocity.

Simplest example: a mass at the end of a light rod

To produce the same rotational acceleration, a mass at the end of the rod must receive a larger linear acceleration than one nearer the axis.

It is harder to get the system rotating when the mass is at the end of the rod than when it is nearer to the axis.

Rotational Inertia and Newton’s Second Law

The resistance to a change in rotational motion depends on: the mass of the object; the square of the distance of the mass from the axis of

rotation. For an object with its mass concentrated at a point:

Rotational inertia = mass x square of distance from axisI = mr2

The total rotational inertia of an object like a merry-go-round can be found by adding the contributions of all the different parts of the object.

Rotational Inertia and Newton’s Second Law

Newton’s second law for linear motion:

Fnet = ma Newton’s second law for rotational motion:

The net torque acting on an object about a given axis is equal to the rotational inertia of the object about that axis times the rotational acceleration of the object.

net = I The rotational acceleration produced is equal to the torque

divided by the rotational inertia.

Example: a baton with a mass at both ends

Most of the rotational inertia comes from the masses at the ends.

A torque can be applied at the center of the rod, producing a rotational acceleration and starting the baton to rotate.

If the masses were moved toward the center, the rotational inertia would decrease and the baton would be easier to rotate.

Two 0.2-kg masses are located at either end of a 1-m long, very light and rigid rod as shown. What is the rotational inertia of

this system about an axis through the center of the rod?

a) 0.02 kg·m2 b) 0.05 kg·m2

c) 0.10 kg·m2

d) 0.40 kg·m2

I = mr2

= (0.2 kg)(0.5m)2 x 2 = 0.10 kg·m2

Rotational inertias for more complex shapes:

Conservation of Angular Momentum

How do spinning

skaters or divers

change their

rotational velocities

?

Angular Momentum

Linear momentum is mass (inertia) times linear velocity: p = mv

Angular momentum is rotational inertia times rotational velocity:

L = I Angular momentum may also be called rotational

momentum. A bowling ball spinning slowly might have the same angular

momentum as a baseball spinning much more rapidly, because of the larger rotational inertia I of the bowling ball.

Conservation of Angular Momentum

Linear momentum is conserved if the net external force acting on the system is zero.

Angular momentum is conserved if the net external torque acting on the system is zero.

Inertia I : net I L IIf net 0, L constant

KE 1

2I 2

Inertia m : Fnet ma p mvIf Fnet 0, p constant

KE 1

2mv 2

Angular momentum is conserved by changing the angular velocity

When the masses are brought in closer to the student’s body, his rotational velocity increases to compensate for the decrease in rotational inertia.

He spins faster when the masses are held close to his body, and he spins more slowly when his arms are outstretched.

Angular momentum is conserved by changing the angular velocity

The diver increases her rotational velocity by pulling into a tuck position, thus reducing her rotational inertia about her center of gravity.

Kepler’s Second Law

Kepler’s second law says that the radius line from the sun to the planet sweeps out equal areas in equal times.

The planet moves faster in its elliptical orbit when it is nearer to the sun than when it is farther from the sun.

Kepler’s Second Law

This is due to conservation of angular momentum.

The gravitational force acting on the planet produces no torque about an axis through the sun because the lever arm is zero: the force’s line of action passes through the sun.

Kepler’s Second Law

When the planet moves nearer to the sun, its rotational inertia about the sun decreases.

To conserve angular momentum, the rotational velocity of the planet about the sun must increase.

Riding a bicycle and other amazing feats

Why does a bicycle remain upright when it

is moving but promptly falls

over when not moving?

Angular momentum is a vector

The direction of the rotational-velocity vector is given by the right-hand rule.

The direction of the angular-momentum vector is the same as the rotational velocity.

Inertia I, rotational velocity Angular momentum: L I

Angular momentum and bicycles

The wheels have angular momentum when the bicycle is moving.

For straight line motion, the direction of the angular-momentum vector is the same for both wheels and is horizontal.

To tip the bike over, the direction of the vector must change, requiring a torque.

Angular momentum and bicycles

If the bike is not perfectly upright, a gravitational torque acts about the line of contact of the tires with the road.

As the bike begins to fall, it acquires a rotational velocity and angular momentum about this axis.

If the bike tilts to the left, the change in angular momentum points straight back.

Angular momentum and bicycles

If the bike is standing still, the gravitational torque causes the bike to fall.

Angular momentum and bicycles

When the bike is moving, the change in angular momentum caused by the gravitational torque adds to the angular momentum already present from the rotating tires.

This causes a change in the direction of the total-angular momentum vector which can be accommodated by turning the wheel.

A student holds a spinning bicycle wheel while sitting on a stool that is free to rotate. What happens if the

wheel is turned upside down?

To conserve angular momentum, the original direction of the angular-momentum vector must be maintained.

A student holds a spinning bicycle wheel while sitting on a stool that is free to rotate. What happens if the

wheel is turned upside down?

The angular momentum of the student and stool, +Ls, adds to that of the (flipped) wheel, -Lw, to yield the direction and magnitude of the original angular momentum +Lw.

A student sits on a stool holding a bicycle wheel with a rotational velocity of 5 rev/s about a vertical axis. The

rotational inertia of the wheel is 2 kg·m2 about its center and the rotational inertia of the student and wheel and platform about the rotational axis of the platform is 6 kg·m2. What is the initial angular momentum of the

system?

a) 10 kg·m2/s upwardb) 25 kg·m2/s

downwardc) 25 kg·m2/s upwardd) 50 kg·m2/s

downwardL = I = (2 kg·m2)(5 rev/s)

= 10 kg·m2/s upward from plane of wheel

If the student flips the axis of the wheel, reversing the direction of its angular-momentum vector, what is the rotational velocity of the student and stool about their

axis after the wheel is flipped?

a) 1.67 rad/sb) 3.33 rad/sc) 60 rad/sd) 120 rad/s

= L / I = (20 kg·m2/s) / (6 kg·m2)= 3.33 rad/s

in direction of original angular velocity

Where does the torque come from that accelerates the student and the stool?

The student exerts forces on the handles when he flips the wheel, producing the torque.