AP Physics CMontwood High School

R. Casao

•When a wheel moves along a straight track, the center of the wheel moves forward in pure translation. •A point on the rim of the wheel traces out a complex path called a cycloid.

For a wheel passing at constant speed while rolling smoothly (no sliding):◦ the center of mass O of the wheel moves forward

at constant speed vcom.◦ the point P where the wheel makes contact with

the surface also moves forward at speed vcom so that it remains directly below the center of mass O.

During time interval t, bothO and P move forward bya distance s. Observerssee the wheel rotate thru anangle about the center of the wheel.

t1 = 0 t2 = t

The point on the wheel P that was touching the surface at t = 0 s moves through an arc length s.

Equations: s = R·; vcom = ds/dt; = d/dt Differentiating the arc length equation wrt

time: ds/dt = R·d/dt to give us vcom = R·.

t1 = 0 t2 = t

The rolling motion of a wheel is a combination of purely translational and purely rotational motions.

Every point on the wheel rotates about the center with angular speed .

Every point on the outside edge of the wheel has linear speed vcom.

Purely translational motion: every point on the wheel moves forward with speed vcom.

Combination of rotational motion and translational motion produces the rolling motion of the wheel.

The portion of the wheel at the bottom (point P) is stationary and the top of the wheel is moving at speed 2·vcom. The wheel is moving fastest near the top than near the bottom because the spokes are more blurred at the top than at the bottom.

Consider the rolling motion about an axis thru point P: K = 0.5·Ip·²◦ Ip is the rotational inertia of wheel about axis thru

point P and is the angular speed of the wheel.◦ Apply the parallel-axis theorem to determine Ip:

Ip = Icom + M·h²; for the wheel, h = R Substituting:

K = 0.5·Icom·² + 0.5·M·R²·² Because vcom = R·,

K = 0.5·Icom·² + 0.5·M·vcom²

22com )RM(I0.5K

A rolling object has two types of kinetic energy: a rotational kinetic energy (0.5·Icom·²) due to its rotation about its center of mass and a translational kinetic energy (0.5·M·vcom²) due to translation of its center of mass.

Forces of Rolling If a wheel rolls at constant speed it has no tendency to

slide at the point of contact P and no frictional force acts there; acom = 0 at point P.

If a net force acts on the rolling wheel to speed it up or slow it down, the net force causes an acceleration of the center of mass along the direction of travel.

The wheel also rotates faster or slower, so an angular acceleration occurs about the center of mass.

These accelerations tend to make the wheel slide at point P and a frictional force must act on the wheel at P to oppose the tendency to slide.

If the wheel does NOT slide, the force is a static frictional force fs and the motion is smooth rolling motion.

Smooth rolling motion: acom = R· If the wheel does slide when the net force acts

on it, the frictional force that acts at P is a kinetic frictional force and the motion is not

smooth rolling motion.

For a uniform body of mass M and radius R rolling smoothly down a ramp at angle along an x axis.

The normal force N acts atpoint P but has been shiftedup to the center of mass. Static frictional force fs acts

at point P and is directedup the incline. If the bodywere to slide down the incline, the frictional force would oppose that motion.

acom

Motion along the x-axis (up as positive):fs – M·g·sin = M·acom

Angular motion: Torque = I·; ◦ N and Fg·sin pass thru the pivot at the center of

mass and do not produce a torque.◦ fs applied perpendicularly a distance R at point P and

produces a torque.◦ Torque = F·r = fs·R; fs·R = Icom·◦ -acom = R·; acom is negative because it is directed in

the negative direction on the x-axis. Unlike the linear motion problems we have done in the past where we took the direction of motion as positive, with the rotation involved, we take the positive direction as the positive direction of the rotation (ccw = positive; cw = negative).

Solve for and substitute: =-acom/R

fs·R = Icom· -acom/R

2com

com

2com

com

2com

com

2comcom

com

com2comcom

coms

2comcom

s

RMI

1

sinθga;

RI

M

sinθgMa

sinθgMRI

Ma

sinθgMR

aIaM

aMsinθgMR

aI

aMsinθgMfR

aIf

As yo-yo rolls down its string from height h, potential energy m·g·h is converted to both rotational (0.5·Icom·²) and translational kinetic energy (0.5·M·vcom²). As it climbs back up, kinetic energy is converted to potential energy.

The string is looped around the axle and when the yo-yo hits the bottom of the string, an upward force on the axle from the string stops the descent.

The yo-yo spins with the axle inside the loop with only rotational kinetic energy.

y

acom

The yo-yo spins until you jerk on the string to cause the string to catch on the axle and allow the yo-yo to climb back up.

The rotational kinetic energy of the yo-yo at the bottom of the string can be increased by throwing the yo-yo downward so that it starts down the string with initial speeds vcom and instead of rolling down from rest.

The yo-yo rolls on the axle of radius Ro.

The yo-yo is slowed by the tension force T from the string on the axle.

y

acom

Acceleration equation: the yo-yo has the same downward

acceleration when it is climbing back up the string because the forces acting on it are still those shown in the figure.

Torque Revisited We previously defined torque for a rigid body

that rotated about a fixed axis in a circle. To define the torque for an individual particle

that moves along any path relative to a fixed point (rather than a fixed axis), the path doesn’t have to be a circle and the torque has to be written as a vector.

2com

com

RMI

1

ga

The figure shows a particle at point A in the xy plane with a single force F acting on the particle.

The particle’s position relative to the origin O is given by the position vector r.

Torque equation: Fxr

To find the direction of the torque, slide the force vector without changing its direction until its tail is at the origin O so that it is tail to tail with the position vector.

Use the right hand rule to rotate the position vector r into the force vector F. The thumb points in the direction of the torque.

The magnitude of the torque is given by:

is the angle between the position vector r and the force vector F.

Angular Momentum The figure shows a particle of mass m with linear momentum p = m·v as it passes thru point a in the xy plane. The angular momentum l of the particle with respect to theorigin O is:

sinθFr

vxrmpxrl

r is the position vector of the particle wrt O. As the particle moves wrt O in the direction of

its momentum p, the position vector r rotates around O.

To have angular momentum, the particle itself does not have to rotate around O. Unit for angular momentum:

kg·m²/s = J·s The direction of the angular momentum vector is found by sliding the vector p until its tail is at the origin O.

Use the right-hand rule, rotating the vector r into vector p. The thumb points in the direction of the angular momentum vector.

The magnitude of the angular momentum vector is: l = r·m·v·sin , where is the angle between r and p when the two vectors are tail to tail.

Angular momentum only has meaning wrt a specific origin.

The direction of the angular momentum vector is always perpendicular to the plane formed by the position vector and the linear momentum vector.

Fnet = dp/dt expresses the close relation between force and linear momentum for a single particle.

There is also a close relationship between torque and angular momentum: Tnet = dl/dt

The vector sums of all the torques acting on a particle is equal to the time rate of change in the angular momentum of that particle.

The torques and angular momentum are both defined wrt the same origin.

For a system of particles with respect to an origin, the total angular momentum L of the system is the vector sum of the angular momenta l of the individual particles:

Over time, the angular momenta of individual particles may change due to interactions within the system between individual particles or because of outside influences on the system.

n

iillllL

1321 ...

n

i

i

dtld

dtLd

1

Since Tnet = dl/dt, then

The rate of change of the system’s angular momentum L is equal to the vector sum of the torques on the individual particles.

These torques include internal torques due to forces between particles and external torques due to forces on the particles from bodies outside of the system.

The forces between particles always occur in 3rd law pairs so their torques cancel each other, so the only torques that can change the total angular momentum L of a system are external

torques.

n

iinetT

dtLd

1,

dtLd

Tnet

2nd law for rotation of a system of particles: the net external torque Tnet acting on a system of particles is equal to the time rate of change of the system’s total angular momentum L.

The torque and the system’s angular momentum must be measured from the same origin.

Angular Momentum of a Rigid Body Rotating About a Fixed Axis

For a system of particles that form a rigid body that rotates about a fixed axis rotating with constant angular speed : L = I·

If the net external torque acting on a system is zero, the angular momentum L of the system remains constant, no matter what changes take place within the system: Li = Lf; Ii·wi = If·wf

Because momentum is a vector quantity , the conservation of momentum has to be considered in all three dimensions (x, y, z). Depending on the torques acting on a system, the angular momentum of the system might be conserved in only 1 or 2 directions but not in all directions.

i

The figure shows a student seated

on a stool that can rotate freely about a

vertical axis. The student who has been set into

rotation at an initial angular speed , holds

two dumbb

Example:

ells

in h

is outstretched hands. His

angular momentum vector lies along the

rotation axis, pointing upward.

L

The student then pulls in his hands as shown in fig.b. This action reduces the

rotational inertia from an initial value to a smaller final value .

No net external torque acts on the student-st

i fI I

ool system. Thus the

angular momentum of the system remains unchanged.

Angular momentum at : Angular momentum at :

Since 1

i i i i f f f f

i ii f i i f f f i f

ffi

f

t L I t L I

I IL L I I I I

I I

The rotation rate of the student in fig.b is faster

i

(11-16)