Download Collisions and rotational kinematics

Collisions and rotational kinematics
•  Spring break starts tomorrow!
•  Finish up collisions today
•  Start Chapter 10 (Rotational Motion).
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Collision summary
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In all collisions,
mA vAi + mBvBi = mA vAf + mBvBf
momentum is conserved:
This is a vector equation.
Components are conserved.
mA vAxi + mBvBxi = mA vAxf + mBvBxf
mA vAyi + mBvByi = mA vAyf + mBvByf
For elastic collisions, kinetic
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1
energy is conserved.
mA vAi
+
2
In perfectly inelastic collisions,
!
!final
velocities are the same vAf = vBf
1
2
2
2
mBvBi2 = 12 mA vAf
+ 12 mBvBf
vAyf = vByf
Components:
vAxf = vBxf
Remember momentum and velocities have signs
However kinetic energy is always positive
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Elastic collision example
Two air carts have equal mass and springs attached. Cart A is
moving toward cart B with speed vAi. Cart B is initially at rest.
What is the speed of the carts after the collision?
The carts are on a track and have springs (conservative force).
This is an elastic collision in 1D which gives us two equations:
mAvAi + mBvBi = mAvAf + mBvBf
1
1
1
1
2
2
2
2
m
v
+
m
v
=
m
v
+
m
v
2 A Ai
2 B Bi
2 A Af
2 B Bf
Since masses are equal, can cancel them out. Also, vBi=0
2
2
2
vAi = vAf + vBf
vAi
= vAf
+ vBf
2
2
2
2
+ 2vAf vBf + vBf
= vAf
+ vBf
or 2vAf vBf = 0
Substituting gives vAf
Either vAf or vBf is 0 and the other must equal vAi. Since the
carts can’t pass through each other, cart A must stop and
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cart B travels with velocity vAi.
Clicker question 1
Set frequency to BA
m
A big ball, mass M=10m, speed v,
v
10m
strikes a small ball, mass m, at rest.
Could the following occur: The big
ball comes to a complete stop and the
10m
small ball takes off with speed 10v?
A.  Yes, this could occur
B.  No, it would violate conservation of momentum
C.  No, it would violate conservation of energy
D.  Need more information.
Momentum Pi = 10mv
is conserved: Pf = 10mv
m
10v
2
2
1
K
=
10mv
=
5mv
Kinetic energy is i 2
not conserved: K f = 1 m(10v)2 = 50mv 2
2
We can lose kinetic energy during a collision (that is the definition
of an inelastic collision) but we cannot gain kinetic energy.
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Other examples of elastic collisions
Newton’s cradle is a good
example of elastic collisions
Need to conserve both kinetic
energy and momentum
The basketball and tennis ball experiment is a little
complicated to work out so I will just set it up and then jump
to the answer. The full details will be in the posted slides.
The text book works out the results
of elastic collisions in more depth.
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An elastic example
A tennis ball is put on top of a basketball and both
are dropped from a height of 1 m. How high will the
tennis ball go? Assume all collisions are elastic.
Can use conservation of energy to get velocity
after falling 1 m. K1 +U1 = K 2 +U 2 becomes
mgh = 12 mv 2 so v = 2gh = 2 ⋅10 m/s2 ⋅1 m = 4.5 m/s
Basketball hits first. What happens?
If no friction losses then by conservation of energy the upward
speed after bounce equals the downward speed before bounce
Or, elastic collisions conserve kinetic energy. Same answer.
Is momentum conserved? Only if you consider Earth
as part of the system
An elastic example
So we have a 1D elastic collision between basketball moving
at vB = 4.5 m/s and tennis ball moving at vT = –4.5 m/s
The collision conserves momentum: mBvB1 + mT vT1 = mBvB2 + mT vT2
Elastic collision conserves
kinetic energy:
1
2
2
2
2
2
mBvB1
+ 12 mT vT1
= 12 mBvB2
+ 12 mT vT2
Solve this problem in the reference
frame where the basketball is at rest
This gives mT vT1 = mBvB2 + mT vT2 and
vB1 = 0 m/s
vT1 = −9.0 m/s
2
2
2
mT vT1
= mBvB2
+ mT vT2
An elastic example
2
2
2
m
v
=
m
v
+
m
v
m
v
=
m
v
+
m
v
We have
T T1
B B2
T T2 and
T T1
B B2
T T2
2
2
2
2
2
mBvB2
= mT vT1
− mT vT2
= mT ( vT1
− vT2
) = mT (vT1 − vT2 ) (vT1 + vT2 )
Factoring:
mBvB2 = mT vT1 − mT vT2 = mT ( vT1 − vT2 )
Divide the equations to get vB2 = vT1 + vT2 Substitute in to get
mBvT1 + mBvT2 = mT vT1 − mT vT2
Rearrange:
mT vT2 + mBvT2 = mT vT1 − mBvT1 becomes ( mT + mB ) vT2 = ( mT − mB ) vT1
(
mT − mB )
vT1
Solve: vT2 =
(mT + mB )
Substitute back in to get
(
mT − mB )
2mT
vB2 = vT1 + vT2 = vT1 +
(mT + mB ) vT1 = mT + mB vT1
An elastic example
mT − mB )
So in the basketball rest frame we have
(
vT2 =
vT1
a post collision tennis ball velocity of
( mT + mB )
Masses and initial velocities:
mB = .4 kg
vB1 = 0 m/s
mT = .07 kg vT1 = −9.0 m/s
.07 kg −.4 kg)
(
vT2 =
(−9.0 m/s) = 6.3 m/s
(.07 kg + .4 kg)
This is with respect to the basketball which is moving
up at 4.5 m/s. Therefore the tennis ball speed with
respect to the ground is 6.3 m/s + 4.5 m/s = 10.8 m/s.
Basketball and tennis ball
Before collision, the tennis ball is moving down at 4.5 m/s
After collision the tennis ball is moving up at 10.8 m/s
What height will the tennis ball reach?
1 mv2 = mgh so h = v
2
2g
2
2
(
10.8 m/s)
=
= 5.8 m
2 ⋅10 m/s2
19 foot height from just a 3 foot drop?!
Inelastic vs elastic collisions
We have two balls with different material properties:
The sad ball just goes squash and lies there like a lump
The happy ball bounces right back like a champ
We will use each ball connected to the end of a
pendulum to try to knock over a block of wood.
Clicker question 2
Set frequency to BA
Two balls of equal mass m are used as the bob of a
pendulum and are released from the same point. Which
ball is more likely to knock over the block of wood?
A.  The ball that splats on the wood (sad ball)
B.  The ball that bounces off the wood (happy ball)
C.  It doesn’t make a difference which ball is used
Both balls have the same mass and velocity just before
hitting the block and therefore the same momentum: p = mv
Conservation of momentum:
mballvball,i + 0 = mballvball,f + mblock vblock,f
Sad ball ends with p ≈ 0 so by conservation of momentum, the
block gains momentum Δp ≈ mv (it gets an impulse of mv).
Happy ball ends with p ≈ −mv (because it bounces off). By
conservation of momentum, the block gains momentum of Δp ≈ 2mv.
Angular kinematics
Early in the semester, we dealt with kinematics which
involved displacement, velocity, and acceleration.
Angular kinematics is the same thing but for objects
which are rotating (rather than translating).
For something to rotate, it must have an axis
about which it rotates like the axle for a wheel.
Only sensible place for the origin is along the axis.
Use polar coordinates (r,θ) instead
of Cartesian coordinates (x,y).
Axis is in the z-direction.
r
θ
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Angular kinematics
θ is a measure of angular position
Δθ is a measure of angular displacement
Δθ
r
θ
In angular motion (and many other areas of physics) it is
much more convenient to use radians (abbreviated rad)
rather than degrees or revolutions to measure angles.
To convert you just need to remember
1 revolution (1 rev) =
2π radians (2π rad) =
360 degrees (360°)
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Angular velocity
Angular velocity tells us how fast (and in
what direction) something is spinning.
Δθ
ω z, avg =
Δt
ωz =
Δθ
r
θ
dθ
dt
The z subscript indicates the axis is in the
z-direction (and the rotation is therefore in the xy plane)
Counterclockwise is positive
Also have angular acceleration which describes how the
spinning rate changes
Δω
dω d 2θ
α z, avg =
Δt
αz =
dt
=
dt 2
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Angular kinematics
The same equations which were derived for constant
linear (also known as translational) acceleration apply for
constant angular (also known as rotational) acceleration
Constant linear
acceleration only!
vx = v0 x + at
x = x0 + v0 x t + 12 ax t 2
2
x
2
0x
v = v + 2ax ( x − x0 )
Constant angular
acceleration only!
ω z = ω0 z + α zt
θ = θ 0 + ω 0 z t + 12 α z t 2
ω z2 = ω 02z + 2α z (θ − θ 0 )
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Relationships to linear velocity
If we want the linear displacement or velocity of a point on a
rotating object, we need r and either θ or ω.
What is the speed of the tire rim if the radius
r s
is 0.35 m and the tire rotates at 20 rad/s?
θ
Radians measure distance around a unit circle
Therefore s = rθ (only for θ in radians!)
ds
dθ
=r
Time derivative of both sides (r is constant):
dt
dt
v = rω . Linear speed is radius times angular speed
Rim speed is v = rω = 0.35 m ⋅ 20 rad/s = 7 m/s
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