University Physics: Mechanics

University Physics: Mechanics

Ch4. TWO- AND THREE-DIMENSIONAL MOTION

Lecture 6

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2013

6/2Erwin Sitompul University Physics: Mechanics

Uniform Circular Motion A particle is in uniform circular motion if

it travels around a circle or a circular arc at constant (uniform) speed.

Although the speed does not vary, the particle is accelerating because the velocity changes in direction.

The velocity is always directed tangent to the circle in the direction of motion.

The acceleration is always directed radially inward. Because of this, the acceleration associated with uniform

circular motion is called a centripetal (“center seeking”) acceleration.


Uniform Circular Motion The magnitude of this centripetal acceleration a is:

→

2va

r

where r is the radius of the circle and v is the speed of the particle.

In addition, during this acceleration at constant speed, the particle travels the circumference of the circle (a distance of 2πr) in time of:

2 rT

v

(centripetal acceleration)

(period)

with T is called the period of revolution, or simply the period, of the motion.


Centripetal Acceleration

r

r

v

v

r

tv

v

v

r

v

t

va

2


An object moves at constant speed along a circular path in a horizontal xy plane, with the center at the origin. When the object is at x = –2 m, its velocity is –(4 m/s) j. Give the object’s (a) velocity and (b) acceleration at y = 2 m.

Checkpoint

^

2 m

v1 = –4 m/s j ^→

v2 = –4 m/s i ^→

2va

r

2(4)

2 28m s

a = –8 m/s2 j ^→

a →

v1→

v2→


Fighter pilots have long worried about taking a turn too tightly. As a pilot’s body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to unconsciousness.

What is the magnitude of the acceleration, in g units, of a pilot whose aircraft enters a horizontal circular turn with a velocity of vi = 400i + 500j m/s and 24 s later leaves the turn with a velocity of vf = –400i – 500j m/s?

^^^ ^

2va

r

2 rT

v

2v

r T

2

vT

2 2(400) (500)v

640.312 m s

2640.312

48

283.818 m s

8.553g 29.8m sg

1

224 sT

Example: Fighter Pilot

→

→


An Aston Martin V8 Vantage has a “lateral acceleration” of 0.96g. This represents the maximum centripetal acceleration that the car can attain without skidding out of the circular path.

If the car is traveling at a constant speed of 144 km/h, what is the minimum radius of curve it can negotiate? (Assume that the curve is unbanked.)

2va

r

2vr

a

2

2

(40m s)

(0.96)(9.8m s )

170 m

• The required turning radius r is proportional to the square of the speed v

• Reducing v by small amount can make r substantially smaller

Example: Aston Martin


Relative Motion in One Dimension The velocity of a particle depends on the reference frame of

whoever is observing or measuring the velocity. For our purposes, a reference frame is the physical object to

which we attach our coordinate system. In every day life, that object is the ground.


Thom(p)son Encounters Relative Velocity


Relative Motion in One Dimension

Suppose that Alex (at the origin of frame A) is parked by the side of a highway, watching car P (the ”particle”) speed past. Barbara (at the origin of frame B) is driving along the highway at constant speed and is also watching car P.

Suppose that both Alex and Barbara measure the position of the car at a given moment. From the figure we see that

PA PB BAx x x

“The coordinate of P as measured by A is equal to the coordinate of P as measured by B plus

the coordinate of B as measured by A”



Taking the time derivative of the previous equation, we obtain

( ) ( ) ( )PA PB BA

d d dx x x

dt dt dt

“The velocity of P as measured by A is equal to the velocity of P as measured by B plus the

velocity of B as measured by A”

PA PB BAv v v



Here we consider only frames that move at constant velocity relative to each other.

In our example, this means that Barbara drives always at constant velocity vBA relative to Alex.

Car P (the moving particle), however, can accelerate.

( ) ( ) ( )PA PB BA

d d dv v v

dt dt dt

PA PBa a• Constant


Example: Relative Velocity

Suppose that Barbara’s velocity relative to Alex is a constant vBA = 52 km/h and car P is moving in the negative direction of the x axis.

(a) If Alex measures a constant vPA = –78 km/h for car P, what velocity vPB will Barbara measure?

52 km hBAv

PA PB BAv v v

P moving in the negative direction

78 km hPAv

PB PA BAv v v ( 78) (52) 130km h


Example: Relative Velocity

(b) If car P brakes to a stop relative to Alex (and thus relative to the ground) in time t = 10 s at constant acceleration, what is its acceleration aPA relative to Alex?

Suppose that Barbara’s velocity relative to Alex is a constant vBA = 52 km/h and car P is moving in the negative direction of the x axis.

0, 78 km h ,PAv 10 st

0,PA PAPA

v va

t

0 ( 78)

7.8 km h s10

22.167 m s

(c) What is the acceleration aPB of car P relative to Barbara during the braking?

0,PB PBPB

v va

t

52 ( 130)7.8 km h s

10

22.167 m s

0,AB PBv v

t


Relative Motion in Two Dimensions In this case, our two observers are again watching a moving

particle P from the origins of reference frames A and B, while B moves at a constant velocity vBA relative to A.

The corresponding axes of these two frames remain parallel, as shown, for a certain instant during the motion, in the next figure.

PA PB BAr r r

PA PB BAv v v

PA PBa a

The following equations describe the position, velocity, and acceleration vectors:

→


20 m

Example: Sail Through the River

A boat with the maximum velocity of 5 m/s aims to cross the river from F to H. H is located directly on the other side of the river, 20 m to the east of F.The speed of current is 1.5 m/s due south. F H

1.5 m/s

(a) Determine how the boat driver should direct the boat so that it can sail due east directly from F to H;

WGvBWv

BGv

1.5 m sWGv

5 m sBWv

1sin

WG

BW

v

v 1 1.5

sin5

17.46

• The sailor should direct the boat 17.46° north of due east


Example: Sail Through the River

A boat with the maximum velocity of 5 m/s aims to cross the river from F to H. H is located directly on the other side of the river, 20 m to the east of F.The speed of current is 1.5 m/s due south.

(b) Calculate the time of trip from F to H.

WGvBWv

BGv

2 2( ) ( ) BG BW WGv v v

FH BGx v t

• The time of trip in presence of current is somehow greater than the one when the boat sails in still water, (20 m) / (5 m/s) = 4 s.

4.77 m s

20 mF H

1.5 m/s

204.193 s

4.77 FH

BG

xtv


Trivia: Frog Crossing the River

A certain species of frog has a unique characteristic. Every time the frog jump forward for 3.0 m, it will jump backward for 1.0 m. The frog never jumps forward twice in a row.

If the frog must cross a river 17.0 m wide, how many jumps does it need?

Solution:Only one jump. Afterwards the frog will swim until it reaches

the other side of the river!


Example: Plane Moves West

A plane moves due east while the pilot points the plane somewhat south of east, toward a steady wind that blows to the northeast. The plane has velocity vPW relative to the wind, with an airspeed (speed relative to wind) of 215.0 km/h, directed at angle θ south of east.The wind has velocity vWG relative to the ground with speed 65.0 km/h, directed 20.0° east of north.What is the magnitude of the velocity vPG of the plane relative to the ground, and what is θ.

→

→

→


Example: Plane Moves West

PG PW WGv v v

215km hPWv

65km h 70WGv

km h 0PG PGv v

sin sin 70PW WGv v cos cos 70PG PW WGv v v

215 sin 65 sin 70

16.50

215 cos( 16.50 ) 65 cos70

228.38 km h


Exercise Problems

1. A cat rides a mini merry-go-round turning with uniform circular motion. At time t1 = 2 s, the cat’s velocity is v1 = 3i + 4j m/s, measured on a horizontal xy coordinate system. At t2 = 5 s, its velocity is v2 = –3i – 4j m/s.

What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cats average acceleration during the time interval t2 – t1?

^ ^

^ ^

2. A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking 10.0 s. What is the ratio of the man’s running speed to the sidewalk’s speed?

Answer: (a) 5.236 m/s2; (b) –2i – 2.667 j m/s2.

Answer: 1.67

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