Equations for Projectile Motion

00 xvv

tvxx x00

tgvv yy 0

ygvv yy 220

2

200 2

1tgtvyy y

Horizontal Vertical

ax=0 ay = - g

vx= constant

Steps in Solving Problems

xx maF

yy maF

1. Draw free-body diagram for every object that is ”free”

2. Select coordinate system such that one of the axis is along the direction of acceleration

3. Write out the equations of motion for the x and y coordinate:

4. Step 2 should guarantee that the sum of the forces in at all but one direction equals zero.

5. Solve the equations simultaneously

Non-Conservative Forces

• If work is done by non-conservative forces:

2211NC EPEKPEEKW

• The sign of WNC is very important

•A motor adds energy so WNC is positive and E2 > E1

•Friction dissipates energy so WNC is negative and E2 < E1

21 EEWNC

Ch 6 5

Example 6-5 (43) The roller-coaster car shown is dragged up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4. Let point be the height where P.E. = 0

10 gym

)(2

121

2

2yygv

2222

1ygmmv

)(2212

yygv

21 EEWNC

2211 EPEKEPEKWNC

0

Ch 6 6

Example 6-5 (43) The roller-coaster car shown is dragged up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4. Let point be the height where P.E. = 0

)(2212

yygv

)035)(8.9)(2( 22 ms

mvs

m26

smmm

smv 12)2835)(8.9)(2( 23

smmm

smv 20)1535)(8.9)(2( 24

Continued

Conservation of Momentum

'22

'1211 vvvv 12

mmmm

In the collision of two isolated objects, the only forces are between the two objects—these are called internal forces. In this case:

initial momentum = final momentum

The total momentum of an isolated system of bodies remains constant.

F12 F21

Radial and Tangential Acceleration

tr

t

va

tan

A point on a rotating wheel always has centripetal acceleration and it will have tangential acceleration if the wheel has angular acceleration.

ra tan

rr

r

r

vaR

222

)(

raR2

Raaa tan

Kinematic Equations for Uniformly Accelerated Motion

The angular equations for constant angular acceleration are derived the same as for constant linear acceleration.

Rotational DynamicsTorque is necessary for angular acceleration

This is expressed as

I

where I is called the moment of inertia

Compare this with Newton’s Second Law:

amF

• τ is the equivalent of force for rotational motion

• I is the equivalent of mass for rotational motion

Example 8-5 A 1.5 kg mass is attached to a cord wrapped around a heavy pulley of mass 4.00 kg and radius 33.0 cm. The pulley is a solid cylinder. Calculate the acceleration of the mass.

m

I

R

aIRF

T

2R

IaF

TmaF

x

maFmgT

gm

TF

maR

Iamg

2

amR

Img

2

gmRI

ma

2