Lecture 12

Physics 207: Lecture 12, Pg 1

Lecture 12

Goals:Goals:

Assignment: Assignment: HW5 due Tuesday 10/19HW5 due Tuesday 10/19 For Monday: Read through Ch. 10.4For Monday: Read through Ch. 10.4

• Chapter 8: Chapter 8:

Solve 2D motion problems with frictionSolve 2D motion problems with friction

• Chapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time)


The swing….a test

at bottom of swing vT is max

Fr = m ac = m vT2 / r = T - mg

T = mg + m vT2 / r

T > mg

at top of swing vT = 0

Fr = m 02 / r = 0 = T – mg cos T = mg cos

T < mg

axis of rotation

vT

mg

T

mg

T

y

x


Example, Circular Motion Forces with Friction

(recall mar = m |vT | 2 / r Ff ≤ s N ) How fast can the race car go?

(How fast can it round a corner with this radius of curvature?)

mcar= 1600 kgS = 0.5 for tire/road r = 80 m g = 10 m/s2

r


Only one force is in the horizontal direction: static friction

x-dir: Fr = mar = -m |vT | 2 / r = Fs = -s N (at maximum)

y-dir: ma = 0 = N – mg N = mg

vT = (s m g r / m )1/2

vT = (s g r )1/2 = (x 10 x 80)1/2

vT = 20 m/s

Example

N

mg

Fs

mcar= 1600 kgS = 0.5 for tire/road r = 80 m g = 10 m/s2

y

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?


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

mg

Ff

What differs on a banked curve?


Banked Curves

Free Body Diagram for a banked curve.

Use rotated x-y coordinates

Resolve into components parallel and perpendicular to bank

N

mgFf

( Note: For very small banking angles, one can approximate that Ff is parallel to mar. This is equivalent to the small angle approximation sin = tan , but very effective at pushing the car toward the center of the curve!!)

mar

xx

y y


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.

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.

At what speed does the car lose contact?


Locomotion of a biped: Top speed


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


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


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)


Impulse and Momentum: A new perspective

Conservation Laws : Are there any relationships between mass and velocity that remain fixed in value?


Momentum Conservation

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

constant that implies 0 PP

dtd

dt

d

dt

md

dt

dmamEXT

P)vvF

( 0 if and EXTF

PPP


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 ?

vV

before after

x


Inelastic collision in 1-D: Example

What is the momentum of the bullet with speed v ?

Key question: Is x-momentum conserved ?

vV

before after

x

aaaa

vm

V)( 0 M v Mmm

P BeforeP Before P AfterP After

V)/1( v mM


Exercise Momentum is a Vector (!) quantity

A. Yes

B. No

C. Yes & No

D. Too little information given

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?



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 into the 10 kg cart

What is the final velocity of the cart?

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)

5.0 m30°

7.5 m

10 kg

2 kg



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

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 lands in the cart.

5.0 m

30°

7.5 m

N

mg

1) ai = g sin 30°

= 5 m/s2

2) d = 5 m / sin 30°

= ½ ai t2

10 m = 2.5 m/s2 t2

2s = t

v = ai t = 10 m/s

vx= v cos 30°

= 8.7 m/s

i

j

x

y

30°


Lecture 12

Assignment: Assignment: HW5 due Tuesday 10/19HW5 due Tuesday 10/19 Read through 10.4Read through 10.4

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Lecture 12

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