Equations for Projectile Motion Horizontal Vertical a x =0 a y = - g v x = constant.
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Transcript of Equations for Projectile Motion Horizontal Vertical a x =0 a y = - g v x = constant.
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