2

Rotational Kinematics

3

Linear Motion

Rotational Motion

position x angular position

velocity v = dx/dt angular velocity

acceleration a = dv/dt angular acceleration

mass m moment of inertia

linear momentum

p=mv angular momentum

force F = ma torque

work work

power P = Fv power

kinetic energy kinetic energy

The analogies between translational and rotational motion

4

Angular VariablesConsider a planar object rotating about an axis perpendicular to its plane. The position is described as a point on the object by the coordinates r and Á, where Á is the angle measured with respect to the axis. When the object moves through an angle Á, the point moves a distance s along the arc. We define the angle Á in radians as

s = r Á

5

The linear velocity in meters per second of a point as moves around a circle is called the tangential velocity

We define the angular velocity ! in radians per second as ! = d Á/dt. Thus v = r !If a point is accelarating along its path with tangential acceleration ®, then

We define the angular acceleration ® in radians per second,Thus

It is common to describe rotating objects by their frequency of revolution in revolutions per second. Since 2 rev is 2¼ rad, then

6

Problems 1.) An electric drill rotates at 1600 rev/min. Through what angle does it turn in 4 ms? If it reaches this speed from rest in 0.32 s, what is its average angular acceleration?

2.) The moon goes around the earth in about 27.3 days. What is its angular velocity? a.) Convert 27.3 days to seconds

b.) Plug in the numbers

7

Moment of InertiaThe moment of inertia represents the effort you need to get something to turn. Consider a continuous object to be composed of many small pieces of mass dm. We can write:

If the mass is spread throughout the volume with density ½, the mass in a volume dV is dm = ½ dV. For a surface mass density ¾, dm = ¾ dA. For a line mass density ¸, dm = ¸ dx. In these cases the moment of inertia can be written:

or or

8

Derivation of the Moment of Inertia for a uniform round disk of thickness b and radius R.

So the moment of inertia is:

9

l – h

h

x

axis

0

Derivation of the Moment of Inertia for a uniform rod of dimensions of length l and radius r

10

We have a rod of uniform composition, same density everywhere, and it has a mass m and length l. We place the axis of rotation at O, a distance h from one of the ends. Simple enough, now we pick an element of volume of a short segment of length dx and cross-sectional area A, a distance x from O. That means:

Since we need to find the total rotational inertia of the entire rod, we need to integrate from x = -h to x = l - h:

11

If the axis of rotation is at the end of the rod, h = 0, which simplifies the equation greatly. Thin Rod About Axis Through One End Perpendicular to Length

If the axis of rotation is at the center, h = l/2, which also simplifies the equation greatly.

12

Thin Rod About Axis Through Center Perpendicular to Length

13

Moment of Inertia I

hoop or cylindrical shell about its axis

solid cylinder or disk

rod about perpendicular axis through end

rectangular plate about perpendicular axis through center

solid sphere

spherical shell

14

Problems

1.) Calculate the moment of inertia of a solid cylinder of mass M and radius R about its axis.

Solution

and

15

The angular momentum with respect to the origin of a particle with position r and momentum p = mv is defined as

because ! is perpendicular to r , . If the angle between r and p is µ, then the magnitude of L is

The time rate of change of the of the angular momentum is

The cross-product of velocity and momentum is zero, because these vectors are parallel. The magnitude of the torque is

Angular Momentum and Torque

16

Problems

1.) You push a merry-go-round at its edge, perpendicular to the radius. If the merry-go-round has a diameter of 4.0 m and you push with a force of 160 N, what torque are you applying? Solution

2.) You apply force on a wrench to loosen a pipe. If the wrench is 22 cm longand you apply 120 N at an angle of with respect to the wrench, what torque are you applying?Solution

3.) Because the force of gravity is directed along the line joining the centers of the sun and the earth, the force is negligible. The earth revolves in a slightlyelliptical orbit around the sun. It is 1.47 £ 108 km away from and travelling at a speed of 30.3 km/s when it is nearest to the sun (perihelion). The earth’s farthestdistance from the sun (aphelion) is 1.52 £ 108 km. How fast is the earth moving at its aphelion?

17

Solution