general physics 1, lec 8 by/ t.a. eleyan 1 lecture 8 circular motion & relative velocity

General Physics 1, Lec 8 By/ T.A. Eleyan

1

Lecture 8

Circular Motion & Relative Velocity


2

Circular Motion

Consider an object moving at constant speed in a circle. The direction of motion is changing, so the velocity is changing (even though speed is constant).

Therefore, the object is accelerating.

The direction of the acceleration is toward the center of the circle and so we call it centripetal acceleration.

The magnitude of the acceleration is

r

vac

2


3

Centripetal Acceleration

radiansin measured if

2 travelledDistance

ˆ)sin(ˆ)cos(

ˆ)sin(ˆ)cos(

2

1

r

yvxvv

yvxvv

The best estimate of the acceleration at P is found by calculating the average acceleration for the symmetric interval 12.


4

2

2

Elapsed time t d/v 2

Components of Acceler

cos cos0

2sin sin sin

ation

2

x

y

y

θr/

v va

θr/vv v v

aθr/v r

va

r

v

0if

r

vac

2

Then,


5

Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?

Solution:

rac

2

t

r 2But

smt

rac /1093.5

4 32

2

Then

yeart

mearthr

1

10496.1)( 11


6

Tangential acceleration

The tangential acceleration component causes the change in the speed of the particle. This component is parallel to the instantaneous velocity, and is given by

Tangential and Radial acceleration

dt

dat

Note: If the speed is constant then the tangential acceleration is zero (uniform Circular Motion)


7

raa cr

2

The radial acceleration component arises from the change in direction of the velocity vector and is given by

Radial acceleration


8

Total acceleration

The total acceleration vector a can be written as the vector sum of the component vectors:

rt

rt

aaa

aaa

22

Since the component perpendicular to other


9

Example: A car exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this instant?


10

If the angle between

22

/072.0500

36sm

rar

222 /309.0 smaaa rt

5.13tan 1

t

r

a

a


11

Problem: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to round the bend. The radius of the curve is 150m. Compute the

acceleration at the train.


12

Example: A particle moves in a circular path 0.4m in radius with constant speed. If the particle makes five revolution

in each second of its motion, find:(a) The speed of the particle.

(b) Its acceleration.

(a) Since r =0.4m, the particle travels a distance 0f 2 r = 2.51m in each revolution. Therefore, it travels a distance of 12.57m in each second (since it makes 5 rev. in the second).v = 12.57m/1sec = 12.6 m/s

2

397ar

(b)


13

Centripetal Force

A string cannot push sideways or lengthwise.

A string in tension only pulls.

The string pulls the ball inward toward the center of the circle


14

What if we cut the sting?

The ball should move off with constant velocityThis means the ball will continue along the tangent to the circle.


15

Centripetal Force

If there is a centripetal acceleration, then the net force must also be a centripetal force:

r

vmmaF cc

2


16

The Conical Pendulum

As the ball revolves faster, the angle increases

What’s the speed for a given angle?

Example:


17

2

2

sin (1)

cos (2)

tan

tan

( sin )

sin tan

mvT

rT mg

then

v

rg

v rg

but r L

Lg


18

Problem: I rotate a ball at an angle of 30o. What is the centripetal acceleration? If the string is 1 meter long, how fast is it rotating?


19

ProblemDriving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the Figure. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump.

Nmg

a=v2/r


20

Relative Velocity Two observers moving relative to each other generally do not agree on

the outcome of an experiment For example, observers A and B below see different paths for the ball


21

Relative Velocity equations

The positions as seen from the two reference frames are related through the velocity

The derivative of the position equation will give the velocity equation

These are called the Galilean transformation equations

tvrr 0

0vvr


22

Central concept for problem solving: “x” and “y” components of motion treated independently.

Again: man on the cart tosses a ball straight up in the air. You can view the trajectory from two reference frames:

Reference frame

on the ground.

Reference frame

on the moving train.

y(t) motion governed by 1) a = -g y

2) vy = v0y – g t3) y = y0 + v0y – g t2/2

x motion: x = vxt

Net motion: R = x(t) i + y(t) j (vector)


23

Acceleration in Different Frames of Reference The derivative of the velocity equation will give the

acceleration equation v’ = v – vo

a’ = a

The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame.


24

Questions[1]You are on a train traveling 40 mph North. If you walk 5 mph

toward the front of the train, what is your speed relative to the ground?

A) 45 mph B) 40 mph C) 35 mph

[2]You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground?

A) 45 mph B) 40 mph C) 35 mph[3]You are on a train traveling 40 mph North. If you walk 5 mph

sideways across the car, what is your speed relative to the ground?

A) < 40 mph B) 40 mph C) >40 mph

general physics 1, lec 8 by/ t.a. eleyan 1 lecture 8 circular motion & relative velocity

Documents