lecture 10

Physics 207: Lecture 10, Pg 1

Lecture 10

Goals:Goals:

Employ Newton’s Laws in 2D problems with circular motion

Assignment: HW5, (Chapters 8 & 9, due 3/4, Wednesday)

For Tuesday: Finish reading Chapter 8, start Chapter 9.


Uniform Circular Motion

For an object moving along a curved trajectory, with non-uniform speeda = ar (radial only)

ar

v|ar |=

vT2

r

Perspective is important


Non-uniform Circular Motion

For an object moving along a curved trajectory, with non-uniform speeda = ar + at (radial and tangential)

ar

at

dt

d| |v|aT |=

|ar |= vT

2

r


Circular motion Circular motion implies one thing

|aradial | =vT2 / r


Key steps

Identify forces (i.e., a FBD)

Identify axis of rotation

Apply conditions (position, velocity, acceleration)


Example The pendulum

Consider a person on a swing:

When is the tension on the rope largest? And at that point is it :

(A) greater than(B) the same as(C) less than

the force due to gravity acting on the person?

axis of rotation


Example Gravity, Normal Forces etc.

vT

mg

T

at bottom of swing vT is max

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

T = mg + m vT2 / r

T > mg

mg

T

at top of swing vT = 0

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

T < mg

axis of rotation

y

x


Conical Pendulum (very different) Swinging a ball on a string of length L around your head

(r = L sin ) axis of rotation

Fr = mar = T sin

Fz = 0 = T cos – mg

so

T = mg / cos (> mg)

mar = mg sin / cos

ar = g tan vT2/r vT = (gr tan )½

Period:t = 2 r / vT =2 (r cot

/g)½


A match box car is going to do a loop-the-loop of radius r.

What must be its minimum speed vt at the top so that it can manage the loop successfully ?

Loop-the-loop 1


To navigate the top of the circle its tangential velocity vT must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching).

Loop-the-loop 1

Fr = mar = mg = mvT2/r

vT = (gr)1/2

mg

vT


The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point?

Hint: The car is constrained to the track.

Loop-the-loop 2

mg

v

N

Fr = mar = mvB2/r = N - mg

N = mvB2/r + mg


Once again the car is going to execute a loop-the-loop. What must be its minimum speed at the bottom so that it can make the loop successfully?

This is a difficult problem to solve using just forces. We will skip it now and revisit it using energy considerations later on…

Loop-the-loop 3


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


A horizontal disk is initially at rest and very slowly undergoes constant angular acceleration. A 2 kg puck is located a point 0.5 m away from the axis. At what angular velocity does it slip (assuming aT << ar at that time) if s=0.8 ?

Only one force is in the horizontal direction: static friction

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

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 0.5)1/2

vT = 2 m/s = vT / r = 4 rad/s

Another Example

mpuck= 2 kgS = 0.8 r = 0.5 m g = 10 m/s2

y

xN

mgFs


UCM: Acceleration, Force, Velocity

F av


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

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


Banked Curves, Testing your understanding

Free Body Diagram for a banked curve.

Use rotated x-y coordinates

Resolve into components parallel and perpendicular to bank

N

mgFf

At this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you accelerate in that direction.

How does the radial acceleration change?

mar

yy

x x


Locomotion: how fast can a biped walk?


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



What about height?

(a) A taller person of equal weight and proportions can walk faster than a shorter person

(b) A shorter person of equal weight and proportions can walk faster than a taller person

(c) To first order, height doesn’t matter



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









(most likely from back foot, no tension along the red line)

3. Need a normal force as well









(Notice the new direction)



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 “in orbit”

Fr = m ar = m v2 / r < mg

Otherwise you will lose contact!

ar = v2 / r vmax = (gr)½

vmax ~ 3 m/s !

(And it pays to be tall and live on Jupiter)


Orbiting satellites vT = (gr)½


Geostationary orbit


Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth.

Fgravity is proportional to r-2 and so little g is now ~10 m/s2 / 50

vT = (0.20 * 42000000)½ m/s = 3000 m/s At 3000 m/s, period T = 2 r / vT = 2 42000000 / 3000 sec =

= 90000 sec = 90000 s/ 3600 s/hr = 24 hrs

Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary)

Great for communication satellites

(1st pointed out by Arthur C. Clarke)

lecture 10

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