Transcript Physics 207: Lecture 2 Notes

Lecture 12
Goals:
• Chapter 8:
•
Solve 2D motion problems with friction
Chapter 9: Momentum & Impulse
 Solve problems with 1D and 2D Collisions
 Solve problems having an impulse (Force vs. time)
Assignment:
 HW5 due Tuesday 10/19
 For Monday: Read through Ch. 10.4
Physics 207: Lecture 12, Pg 1
The swing….a test
axis of rotation
T
T
y
vT
q
mg
x
mg
at top of swing vT = 0
at bottom of swing vT is max
Fr = m 02 / r = 0 = T – mg cos q
T = mg cos q
Fr = m ac = m vT2 / r = T - mg
T = mg + m vT2 / r
T < mg
T > mg
Physics 207: Lecture 12, Pg 2
Example, Circular Motion Forces with
Friction
(recall mar = m |vT | 2 / r Ff ≤ ms N )

How fast can the race car go?
(How fast can it round a corner with this radius of curvature?)
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
r
Physics 207: Lecture 12, Pg 3
Example

Only one force is in the horizontal direction: static friction
x-dir: Fr = mar = -m |vT
y-dir: ma = 0 = N – mg
|2/
y
r = Fs = -ms N (at maximum)
N = mg
vT = (ms m g r / m )1/2
vT = (ms g r )1/2 = (0.5 x 10 x 80)1/2
vT = 20 m/s
N
Fs
mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
Physics 207: Lecture 12, Pg 4
x
Zero Gravity Ride
A rider in a “0 gravity ride” finds herself
stuck with her back to the wall.
Which diagram correctly shows the
forces acting on her?
Physics 207: Lecture 12, Pg 6
Banked Curves
In the previous car scenario, we drew the following free
body diagram for a race car going around a curve on a
flat track.
n
Ff
mg
What differs on a banked curve?
Physics 207: Lecture 12, Pg 7
Banked Curves
Free Body Diagram for a banked curve.
Use rotated x-y coordinates
Resolve into components parallel and
perpendicular to bank
N
mar
Ff
q
y
x
mg
( Note: For very small banking angles, one can approximate
that Ff is parallel to mar. This is equivalent to the small angle
approximation sin q = tan q, but very effective at pushing the
car toward the center of the curve!!)
Physics 207: Lecture 12, Pg 8
Navigating a hill
Knight concept exercise: A car is rolling over the top of a hill
at speed v. At this instant,
A. n > w.
B. n = w.
C. n < w.
D. We can’t tell about n without
knowing v.
At what speed does the car lose
contact?
This occurs when the normal force goes to zero or, equivalently, when
all the weight is used to achieve circular motion.
Fc = mg = m v2 /r  v = (gr)1/2 ½ (just like an object in orbit)
Note this approach can also be used to estimate the maximum walking
speed.
Physics 207: Lecture 12, Pg 13
Locomotion of a biped: Top speed
Physics 207: Lecture 12, Pg 14
How fast can a biped walk?
What about weight?
(a) A heavier person of equal
height and proportions
can walk faster than a
lighter person
(b) A lighter person of equal
height and proportions
can walk faster than a
heavier person
(c) To first order, size doesn’t
matter
Physics 207: Lecture 12, Pg 15
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
Physics 207: Lecture 12, Pg 18
How fast can a biped walk?
Given a model then what does the physics
say?
Choose a position with the simplest
constraints.
If his radial acceleration is greater than g
then he is on the wrong planet!
Fr = m ar = m v2 / r < mg
Otherwise you will lose contact!
ar = v2 / r  vmax = (gr)½
vmax ~ 3 m/s !
(So it pays to be tall and live on Jupiter)
Olympic record pace over 20 km 4.2 m/s
(Lateral motion about hips gives 2.3 m/s more)
Physics 207: Lecture 12, Pg 21
Impulse and Momentum: A new perspective
Conservation Laws : Are there any relationships
between mass and velocity that remain fixed in
value?
Physics 207: Lecture 12, Pg 22
Momentum Conservation

FEXT



dv d (mv) dP

 ma  m 

dt
dt
dt

and if FEXT  0


dP
 0 implies that P  constant
dt

P
Momentum conservation (recasts
Newton’s 2nd Law when
net external F = 0) is an important principle (usually when
forces act over a short time)

It is a vector expression so must consider Px, Py and Pz
 if Fx (external) = 0 then Px is constant
 if Fy (external) = 0 then Py is constant
 if Fz (external) = 0 then Pz is constant
Physics 207: Lecture 12, Pg 23
Inelastic collision in 1-D: Example

A block of mass M is initially at rest on a frictionless horizontal
surface. A bullet of mass m is fired at the block with a muzzle
velocity (speed) v. The bullet lodges in the block, and the
block ends up with a final speed V.

In terms of m, M, and V :
What is the momentum of the bullet with speed v ?
x
v
V
before
after
Physics 207: Lecture 12, Pg 24
Inelastic collision in 1-D: Example
What is the momentum of the bullet with speed v ?

mv
 Key question: Is x-momentum conserved ?
P After
P Before
mv  M 0  (m  M )V
aaaa
v
V
after
x
before
v  (1  M / m)V
Physics 207: Lecture 12, Pg 25
Exercise
Momentum is a Vector (!) quantity


A.
B.
C.
D.
A block slides down a frictionless ramp and then falls and
lands in a cart which then rolls horizontally without friction
In regards to the block landing in the cart is momentum
conserved?
Yes
No
Yes & No
Too little information given
Physics 207: Lecture 12, Pg 26
Exercise
Momentum is a Vector (!) quantity
x-direction: No net force so Px is conserved.
 y-direction: Net force, interaction with the ground so
depending on the system (i.e., do you include the Earth?)
Py is not conserved (system is block and cart only)

2 kg
5.0 m
30°
Let a 2 kg block start at rest on a
30° incline and slide vertically a
distance 5.0 m and fall a distance 7.5 m
7.5 m into the 10 kg cart
10 kg
What is the final velocity of the cart?
Physics 207: Lecture 12, Pg 27
Exercise
Momentum is a Vector (!) quantity


1) ai = g sin 30°
= 5 m/s2
x-direction: No net force so Px is conserved
y-direction: vy of the cart + block will be zero
and we can ignore vy of the block when it
2) d = 5 m / sin 30°
lands in the cart.
j
= ½ ai Dt2
N
i 5.0 m
30° mg
30°
Initial
Final
Px: MVx + mvx = (M+m) V’x
M 0 + mvx = (M+m) V’x
V’x = m vx / (M + m)
= 2 (8.7)/ 12 m/s
V’x = 1.4 m/s
10 m = 2.5 m/s2 Dt2
2s = Dt
v = ai Dt = 10 m/s
vx= v cos 30°
= 8.7 m/s
7.5 m
y
x
Physics 207: Lecture 12, Pg 28
Lecture 12

Assignment:
 HW5 due Tuesday 10/19
 Read through 10.4
Physics 207: Lecture 12, Pg 31