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Chapter 7 Rotational Motion and the Law of

Gravity

71 Measuring Rotational Quantities

Rotational Motion

- Motion of a body that spins about an axis.

Axis of rotation

- Is the line about which the rotation occurs.

Angles can be measured in radians (rads).

- Radians an angle whose arc length is equal to

its radius, which approximately equal to 57.3

degrees.

1 radian 57.3o

- s
- ----
- r

1 radian 57.3o

- s
- ----
- r

s arc length r length of radius q angle of

rotation

- The angle of 360o is one revolution. (1 rev

360o) - One revolution is equal to the circumference of

the circle of rotation.

Circumference is 2prso,1 rev 360o 2p rads

- Radius is half the diameter (d).

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Angular Displacement (Dq)

- The angle through which a point, line, or body is

rotated in a specified direction and about a

specified axis.

Describes how far an object has rotated.

- Change
- Angular in arc length
- Displacement --------------
- radius

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- Counterclockwise (CCW) rotation is considered ()
- Clockwise (CW) rotation is considered (-)

- Ex 1 John Glenn, in 1962, circled the earth 3

times in less than 5 hours. If his distance from

the center of the earth was 6560 km, what arc

length did he travel through?

- G Dq 3 rev 6p rads, r 6560km
- U Ds ?
- E Ds Dq r
- S Ds (6p)(6560)
- S Ds 123653 km

Angular Speed (w)

- The rate at which a body rotates about an axis,

usually expressed in radians per second.

(rads/sec)

- Angular
- Average displacement
- Angular -------------------
- Speed time interval

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- Ex 2 In 1975, an ultra-fast centrifuge attained

an average angular speed of 2.65 x 104 rads/sec.

What was the angular displacement after 1.5 sec?

- G wavg 2.65 x 104 rads/sec, Dt 1.5 sec
- U Dq ?
- E Dq wavg Dt
- S Dq (2.65 x 104)(1.5)
- S Dq 39750 radians

How fast is the earth rotating?

- The Earth rotates at a moderate angular velocity

of 7.2921159 10 -5 rads/s

- Ex 3 A water supply tunnel in NYC is 1.70 x 105

m long and a radius of 2 m. If an electronic eye

sensor moves around the perimeter of the tunnel

at an average angular speed of 5.9 rads/sec. How

long will it take the beetle to travel the

equivalent of the length of the tunnel?

- G wavg 5.9 rads/sec, r 2 m, Ds 1.70 x 105

m - U Dt ?
- E Dt Dq / wavg
- We need to find Dq.

- Dq Ds / r
- Dq 1.70 x 105 m / 2 m
- Dq 85000 rads

- S Dt 85000 rads
- 5.9 rads/sec

- S Dt 85000 rads
- 5.9 rads/sec
- S Dt 14406.8 sec

Angular Acceleration (a)

- The time rate of change of angular speed,

expressed in rads/sec2 .

wf - wia -------- Dt

- wf final angular speed
- wi initial angular speed
- a angular acceleration
- Dt elapsed time

Equation is rearranged just like the linear

acceleration equation

- ALWAYS multiply by Dt first! Then isolate what

you are looking for.

Ex 4 A brake is applied to a flywheel for 5

seconds, reducing the angular velocity of the

wheel uniformly from 2.2 rads/sec to 1.8

rads/sec. What is the angular acceleration?

G Dt 5 sec, w2 1.8 rads/sec, and w1 2.2

rads/sec

- U a ?
- E w2 - w1
- a --------
- Dt

- 1.8 rads/sec 2.2 rads/sec
- a ------------------------
- 5 sec
- a - 0.08 rads/sec2

In comparing angular and linear quantities, they

are similar.

Looking back to Ch 2 equations

- Dx viDt ½ a Dt2
- Dx ½ (vi vf)Dt
- vf vi aDt
- vf2 vi2 2aDx

We apply angular notation to these equations and

get

- Dq wiDt ½ a Dt2
- Dq ½ (wi wf)Dt
- wf wi aDt
- wf2 wi2 2aDq

- Ex 5 A wheel on a bike moves through 11.0 rads

in 2 seconds. What is the angular acceleration of

the initial angular speed is 2.0 rads/sec?

- G Dq 11 rads, Dt 2 sec, wi 2 rads/sec
- U a ?
- E Dq wiDt ½ aDt2
- a 2(Dq - wiDt) / Dt2
- S a 211 (2)(2) /22
- S a 3.5 rads/sec2

7-2 Tangential and Centripetal Acceleration

Objects in motion have a tangential speed.

- Tangent a line that lies in a plane of circle

that intersects at one point.

Tangent line

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Tangential Speed (vt)

- The instantaneous linear speed of an object

directed along the tangent to the objects path.

- r distance from axis
- angular speed
- vt tangential speed

- Ex 6 The radius of a CD is 0.06 m. If a microbe

riding on the discs rim has a tangential speed

of 1.88 m/s, what is the microbes angular speed?

- G r 0.06 m, vt 1.88 m/s
- U w ?
- E w vt / r
- S w 1.88 m/s / 0.06 m
- S w 31.3 rads/sec

Tangential Acceleration (at)

- The instantaneous linear acceleration of an

object directed along the tangent to the objects

circular path

- at tangential acceleration
- r distance from axis
- a angular acceleration

- Ex 7 To crack a whip requires its tip to move

at supersonic speed. This was achieved with a

whip 56.24 m. If the tip moved in a circle and it

accelerated from 6.0 rads/sec to 6.3 rads/sec in

0.6 sec. What would the tangential acceleration

be?

- G r 56.24 m, Dt 0.6 sec
- wi 6 rads/sec,
- wf 6.3 rads/sec,
- U at ?
- E at ra

We need to find angular acceleration (a)

- E wf - wi
- a --------
- t

- S 6.3 - 6.0
- a --------
- 0.6
- S a 0.5 rads/sec2

- S at (56.24)(0.5)
- S at 28.12 m/s2

Centripetal Acceleration (ac)

- Acceleration directed toward the center of a

circular path.

- r distance from axis (m)
- ac centripetal accel. (m/s2)
- vt tangential speed (m/s)

- r distance from axis (m)
- ac centripetal accel. (m/s2)
- w angular speed (rads/sec)

- Ex 8Calculate the orbital radius of the Earth,

if its tangential speed is 29.7 km/s and the

centripetal acceleration acting on the Earth is 5

x 10-3 m/s2?

- G vt29.7 km/s 29700 m/s
- ac 5 x 10-3 m/s2
- U r ?
- E vt2
- r ----
- ac

- S 297002
- r ------------
- 5 x 10-3
- S r 1.764 x 10 11m

Tangential and centripetal accelerations are

perpendicular.

ac

at

atotal

The total acceleration can be found by using the

Pythagorean Theorem.

The total acceleration can be found by using the

Pythagorean Theorem.

If you need to find the angle (q) use the inverse

tangent function (tan-1)

7-3 Causes of Circular Motion

If a mass undergoes an acceleration, there must

be a force causing it. (Newtons 2nd Law)

A force directed toward the center is necessary

for circular motion.

This force causing circular motion is called

Centripetal Force (Fc)

Fc

r

m

v

With a centripetal force, an object in motion

will be accelerated and change its direction.

Work Done by the Centripetal Force

- Since the centripetal force on an object is

always perpendicular to the objects velocity,

the centripetal force never does work on the

object - no energy is transformed.

Without a centripetal force, an object in motion

continues along a straight-line path.

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Centrifugal Force

- centrifugal force is a fictitious force - it is

not an interaction between 2 objects, and

therefore not a real force.Nothing pulls an

object away from the center of the circle.

Centrifugal Force

- What is erroneously attributed to centrifugal

force is actually the action of the objects

inertia - whatever velocity it has (speed

direction) it wants to keep.

From Newtons 2nd Law F ma

- Fc mac

From Newtons 2nd Law F ma

- Fc mac

From Newtons 2nd Law F maFc mac

- Ex 9 Joe attaches a 0.5 kg mass to a 1m rope.

Joe swings the rope in a horizontal circle above

his head with a speed of 4 m/s. What force is

exerted to keep the mass moving in a circle?

- G m 0.5 kg
- vt 4 m/s
- r 1 m
- U Fc ?
- E Fc m(vt2 / r)

- S
- Fc .5 kg x(4 m/s)2/(1 m)
- S Fc 8 N

Gravitational Force

- The mutual force of attraction between particles

of matter.

This is a field force that always exists between

any two objects, regardless of their sizes.

The gravitational force (Fg) depends upon the

distance between the two masses.

Newtons Law of Universal Gravitation

- m1 mass 1 (kg)
- m2 mass 2 (kg)
- r distance between center of masses. (m)
- G Constant of Universal Gravitation.

Units will cancel out with others leaving just

force units of Newtons (N).

The universal gravitational constant (G) can be

used to find the gravitational force between any

two particles.

This is what is known as the inverse-square law.

- The force between the two masses decreases as the

masses move farther apart.

The acceleration due to gravity, g, decreases

with distance from the earths center as 1/d2.

As an object moves away from the center of the

Earth, by a distance equivalent to the earths

radius.

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- Ex 10 What is the force of attraction between

Adam and Jason, gravitational of course. If they

are 12 m apart and Jason mass is 82 kg and Adam

is 74 kg.

- G m1 82 kg, m2 74kg, r 12 m,
- G 6.673x 10-11 n-m2/kg2
- U Fg ?

- G m1 82 kg, m2 74kg, r 12 m,
- G 6.673x 10-11 n-m2/kg2
- U Fg ?
- E

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- S Fg 2.81 x 10 -9 N

- Ex 11 The sun has a mass 2 x 10 30 kg and a

radius of 7 x 10 5 km. What mass must be located

on the suns surface for gravitational force of

470 N to exist between the mass and the sun?

- G m1 2 x 10 30 kg,
- r 7 x 10 5 km7 x 108 m
- G 6.673x 10-11 n-m2/kg2
- Fg 470 N
- U m2 ?

E

E

S m2 1.7 kg

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