Rotational Motion

Assume that you are replacing the bearings in the wheel of your bicycle.  To test if you have done a good job, you hold the axle vertically and give it a spin.  The axle at the center of the wheel is stationary.  The wheel as a whole has no translational motion.  But the wheel is turning around the axle.  Every point on the wheel undergoes circular motion about an axis of rotation.  Such motion around an axis of rotation is called rotational motion.

How do we describe rotational motion?

Assume you make a chalk mark on the rim of the wheel and you originally orient the wheel so that the chalk mark is facing you.  This is your reference orientation.  The angular position of the wheel describes its orientation relative to this reference orientation.  As the wheel rotates, this angular position is changing.  The angular displacement measures how far it has rotated from its reference orientation.

It is often convenient to orient a coordinate system such that the z-axis coincides with the axis of rotation and the x-axis defines the reference orientation.  Then the angular displacement q of a point P on the wheel is the angle q a line from the axis of rotation to the point P makes with the x-axis.

Angles can be measured in units of degrees or radians.  360 degrees = 2p radians.  When describing rotational motion it is most convenient to measure angles in units of radians.

To find out how fast the wheel is rotating, we measure its angular speed w.  The average angular speed is given by

.

The instantaneous angular speed is given by

.

Every point on the wheel has the same angular speed.  The units of angular speed are rad/s or s-1, because radians are not dimensional.

Assume that you turn the axle of the spinning wheel from vertical to horizontal.  The wheel is still spinning with the same angular speed, but its angular velocity w has changed.  Angular velocity is a vector quantity.  Its magnitude is the angular speed, and its direction is the direction of the axis of rotation.  There is, however, a subtlety we have to take care of.  Assume that the axis of rotation is vertical.  What is the sense of rotation?  Is the wheel spinning clockwise or counterclockwise as viewed from above?  Just saying the axis is vertical does not tell us the sense of rotation.

To specify the sense of rotation we use a convention called the right-hand rule.  If the fingers of your right hand are curling to indicate which way the wheel is turning, then the thumb of you right hand is pointing in the direction of the axis of rotation.  This is the direction of the angular velocity.

Problems:

If a car's wheels are replaced with wheels of a larger diameter, will the reading of the speedometer change?  Explain!
Solution:
The sensor for the speedometer senses the angular speed of the wheel.  Using vnominal = rnominalw, the speedometer displays the correct speed if the tires have the nominal radius.  If you put larger tires on your car, then your actual speed vactual = ractualw is greater that the displayed speed vnominal = rnominalw.
What is the magnitude of the angular velocity, w, of the second hand of a clock?  What is the direction of w as you view a clock hanging vertically?  What is the magnitude of the angular acceleration of the second hand?
Solution:
The second hand of a clock goes through an angular displacement of 2p in one minute.  Its angular speed is w = 2p/60s = 0.105/s.  The direction of w is perpendicular to the face of the clock, pointing into the face of the clock.  The average angular acceleration of the second hand is zero.
What is the angular speed, in radians per second, of
(a) the Earth in its orbit about the Sun, and
(b) the Moon in its orbit about the Earth?
Solution:
(a) The earth orbits the sun once every year.  The angular speed of the earth in its orbit about the sun is
w = 2p/year = 2p/(365´24´60´60s) = 2´10-7/s.
(b) The moon orbits the earth once 27.3 days.  The angular speed of the moon in its orbit about the earth is
w=2p/(27.3 days)=2p/(27.3´24´60´60s) = 2.7´10-6/s.

The angular acceleration a is define as the rate of change of the angular velocity.  The average angular acceleration is given by

,

and the instantaneous angular acceleration is given by

.

The angular velocity changes whenever the angular speed or the orientation of the axis of rotation changes.  The angular acceleration is a vector.

If we change the number of revolutions the wheel makes per second, then each point on the wheel has angular acceleration.  Assume the angular velocity points in the z-direction, and over a one second time interval we change the angular speed of the wheel from p/s to 2p/s. The angular acceleration is

,

i.e. it has magnitude p/s and points in the z-direction.

If we turn the axis of rotation from vertical to horizontal, then each point on the wheel has angular acceleration.  Assume the wheel is rotating with angular velocity (2p/s)k about the z-axis.  If over a time interval of 1s we reorient the axis of rotation from vertical to horizontal, so that the wheel is now rotating with angular velocity (2p/s)i about the x-axis, then the angular acceleration is given by

.

The magnitude is , and its direction makes an angle of 315o with the x-axis.

 

Kinematic Equations

For motion with constant angular acceleration we have

Dw = aDt, wf = wi + aDt.

This is a vector equation.  It yields an equation for each Cartesian component.  For the z-component we have

wzf = wzi + azDt.

If an object is rotating about the z-axis, and there are no other components of w and a, we often drop the index z and just write

wf = wi + aDt.

The angular displacement q about the z-axis is then given by

.

For motion with constant angular acceleration we therefore have

,

wf = wi + a(tf-ti).

These equations are of the same form as the equations for linear motion with constant acceleration. For motion along the x-axis we have

,

vf = vi + a(tf-ti).

If we replace x by q and a by a, then the kinematic equations for linear motion along the x-axis transform into the kinematic equations for rotational motion about the z-axis.

When a wheel rotates about the z-axis, each point on the wheel has the same angular speed.  The linear speed v of a point P, however, depends on its distance from the axis of rotation.
When a point P goes through an angular displacement of 2p, then its distance traveled is 2pr.
When a point P goes through an angular displacement of p, then its distance traveled is pr.
When a point P goes through an angular displacement of q, then its distance traveled is qr.

In terms of the angular speed w, the speed v of the point P therefore is

,

if r is constant; v is the tangential velocity of the point P.

The tangential acceleration a of a point P moving along a circular path is given in terms of its angular acceleration by

.

Its radial acceleration is

.

The total acceleration is given by

.

Problems:

An airliner arrives at the terminal and the engines are shut off.  The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s.  The engines rotation slows with with an angular acceleration of magnitude 80 rad/s2.
(a) Determine the angular speed after 10s.
(b) How long does it take the rotor to come to rest?
Solution:
(a) In this problem the initial angular velocity wi and the angular acceleration a are given.  If we choose the direction of the initial angular acceleration to be the z-direction, then wf = wi-a(tf-ti), since a is in the negative z-direction.
At t = 0, wi = 2000/s.
At t = 10s we have w = 2000/s-(80/s2)10s = 1200/s.
(b) Setting wf = wi-a(tf-ti) = 0 we find the time it takes the rotor to come to rest.
2000/s - (80/s2)t = 0,  t = (2000/80)s = 25s is the time it takes the rotor to come to rest.
A rotating wheel requires 3s to rotate 37 revolutions.  Its angular velocity at the end of the 3s interval is 98 rad/s.  What is the constant angular acceleration of the wheel?
Solution:
Using

,

with qi = 0, we have 37´2p = wi3s + (1/2)a(3s)2.

Using wf = wi + a(tf-ti) we have 98/s = wi + a(3s).  We solve this equation for wi, wi = 98s - a(3s), and insert it into the first equation.

37´2p = (98/s)(3s) - a(3s)2 + (1/2)a(3s)2, 74p = 294 - a9s2 + a4.5s2,
4.5s2a = 294 - 74p, a = 13.67/s2 is the constant acceleration of the wheel.

A car accelerates uniformly from rest and reaches a speed of 22m/s in 9s.  If the diameter of a tire is 58cm, find
(a) the number of revolutions the tire makes during this motion, assuming no slipping, and
(b) the final rotational speed of the tire in revolutions per second.
Solution:
The acceleration of the car is a = v/t = (22m/s)/(9s) = 2.44m/s2.
The distance traveled in 9s is d = (1/2)at2 = ((1/2) ´ 2.44´81)m = 99m.
The circumference of the tire is p0.58m = 1.82m.
The number of revolution made by the wheel is 99/1.82 = 54.3.
The final linear speed of the tire is v = 22m/s.  Using v = wr, w = v/r the final angular speed is w = 75.9/s.  The number of revolutions per second is w/2p = 12/s.

A rotating object has kinetic energy, even when the object as a whole has no translational motion.  If we consider the object made up of a collection of particles, then each particle i has kinetic energy Ki = (1/2)mivi2.

The total kinetic energy of the rotating object is therefore given by

.

We write

.

The quantity in parenthesis is called the moment of inertia  of the object about the axis of rotation.  When an object is rotating about an axis, its rotational kinetic energy is K = (1/2)Iw2.

Rotational kinetic energy = (1/2) moment of inertia ´ (angular speed)2.

When the angular velocity of a spinning wheel doubles, its kinetic energy increases by a factor of four.

When an object has translational as well as rotational motion, its total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy.

The moment of inertia of an object depends on the mass of the object, and on how this mass is distributed with respect to the axis of rotation.  The farther the bulk of the mass is from the axis of rotation, the greater is the rotational inertia (moment of inertia) of the object.  The units of the moment of inertia are units of mass times distance squared, for example kgm2.

Imagine two wheels with the same mass.  One is a solid wheel, the mass is evenly distributed throughout the structure, while the other has most of the mass concentrated near the rim.

moment.gif (2343 bytes)

The wheel with the mass near the rim has the greater moment of inertia.

The moment of inertia is defined with respect to an axis of rotation.  For example, the moment of inertia of a circular disk spinning about an axis through its center perpendicular to the plane of the disk differs from the moment of inertia of a disk spinning about an axis through its center in the plane of the disk.

Problems:

Three particles are connected by rigid rods of negligible mass lying along the y-axis as shown.

If the system rotates about the x-axis with angular speed of 2 rad/s, find
(a)  the moment of inertia about the x-axis and the total rotational kinetic energy evaluated from (1/2)Iw2, and
(b) the linear speed of each particle and the total kinetic energy evaluated from å(1/2)mivi2.
Solution:
(a) The moment of inertia is .  Here ri is the perpendicular distance of particle i from the x-axis.
I = 4kg9m2 + 2kg 4m2 + 3kg16m2 = 92kgm2.
The rotational kinetic energy is K = (1/2)Iw2 = 46´4/s2 = 184J.
(b) The linear speed of particle i is vi = wri.
The linear speed of the 4kg mass is v = 6m/s, and its kinetic energy is (1/2)mv2 = 72J.
The linear speed of the 2kg mass is v = 4m/s, and its kinetic energy is (1/2)mv2 = 16J.
The linear speed of the 3kg mass is v = 8m/s, and its kinetic energy is (1/2)mv2 = 96J.
The sum of the kinetic energies of the three particles is 184J.

The four particles in the figure below are connected by rigid rods.

The origin is at the center of the rectangle.  If the system rotates in the x-y plane about the z-axis with an angular speed of 6 rad/s, calculate
(a) the moment of inertia of the system about the z-axis and
(b) the rotational energy of the system.
Solution:
(a) The moment of inertia is .  Here ri is the perpendicular distance of particle i from the z-axis.
Each particle is a distance from the axis of rotation.
I = (3kg + 2kg + 4kg + 2kg)13m2 = 143kgm2.
(b) The rotational kinetic energy is K = (1/2)Iw2 = 71.5´36/s2 = 2574J.

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