Topic 3 Dynamics ALLsno.phy.queensu.ca/~potato/APSC111/Lecture_15_10_2010.pdfAPSC 111 Dynamics Page...

APSC 111 Dynamics Page 3.125

ExampleA crate with a mass of 150 kg is pulled by a force F which is exerted at 15° above the horizontal.

If = 0.50, how large must F be to move the crate?

Free body diagram

F15

s

gF Mg

FN

fF


If = 0.50, how large must F be to move the crate?

Components/coordinates

F15s

gF Mg

FN

fFx

y

and

sinyF F cosxF F

APSC 111 Dynamics .127

gF Mg

FN

fF

x

y

and

cos 0 fxF F F

sin 0yF N F Mg

f sF N

substitute for Ncos sF N

cos sin 0s

F F Mg

671 Ncos sin

s

s

MgF


Example (cont)

At what angle should F be applied if the crate is to move with the smallest possible force F?

F

cos sins

s

MgF

2( 1) sin cos 0(cos sin )

ss

s

MgdFd

A crate with a mass of 150 kg is pulled by a force F which is exerted at some angle above the horizontal.

sin coss tan s 1tan ( ) 26.6s

657 Ncos 26.6 sin 26.6

s

s

MgF


ExampleConsider the following “train” what are the tensions in the strings?

31

2 F

1M g

12TfF1N

1

12 1 1a kxF T N m

1 1 0 yF N m g

12 1 1a kT m g m


31

2 F

2M g

23TfF

2N

23 2 12 2a kxF T N T m

2 2 0 yF N m g

12 1 1a kT m g m

212T 23 2 12 2a kT N T m

23 2 1 1 2a a k kT m g m g m m

23 1 2( )( a) kT m m g


31

2 F

3M g

FfF

3N

12 1( a) kT m g

323T

1 2 3( )( a) kF m m m g

Identical formulation gives:

23 1 2( )( a) kT m m g


31

2 F

12 ( a) kT g

a6( ) kF g

23 a3( ) kT g

And with masses of 1, 2 and 3 kg

1: 3 : 6 ratio of tensions

If constant velocity: (a=0) still get 1: 3 : 6 ratio of tensions

If frictionless as well: there is no tension at all.


Circular Motion with Friction

What is the maximum velocity the steel cylinder can have before it slips off the turntable? The mass of the cylinder is 0.2 kg, the coefficient of static friction between the cylinder and the turntable is 0.080, and the cylinder is located 0.15 m from the center of the turntable.

Mg

NWhat forces act on the cylinder?


What about centripetal force? The cylinder has an acceleration

Towards the centre, so there is a force

Towards the centre. What causes this force and what about friction? Where do we draw that? Mg

NWhat forces act on the cylinder?

2va R

2

centvF Ma MR

centF


The centripetal force is the frictional term. Without friction, the cylinder would never travel in a circle.

but,

and,

so,

Mg

N

cent fF F

fF

f s sF N Mg

2

cent RvF Ma MR

2

svM MgR

2sv Rg

sv Rg

0.34 m/sv


The Centripetal Force

It is very important to realize that this is not some new force. It is simply a way of describing a net force which points towards the centre during circular motion. The force that produces the centripetal force could be:• Friction, keeping a loose object on a rotating object.• Tension in a rope, for an object being swung• Gravitational pull to keep an object in orbit• … and many more

In each case where there is circular motion, there will be some cause for the centripetal force that you may need to determine.


In the game of tetherball, the In the game of tetherball, the struck ball whirls around a pole. In struck ball whirls around a pole. In what direction does the what direction does the net forcenet force on on the ball point?the ball point?

14%2%72%3%8% 1. toward the top of the pole

2. toward the ground3. along the horizontal component of the tension force4. along the vertical component of the tension force5. tangential to the circle

W

T

Tetherball


In the game of tetherball, the In the game of tetherball, the struck ball whirls around a pole. In struck ball whirls around a pole. In what direction does the what direction does the net forcenet force on on the ball point?the ball point?

W

T

The vertical component of the tension balances the weight.

The horizontal component of tension provides the

centripetal force that points toward the center of the circle.

1. toward the top of the pole2. toward the ground3. along the horizontal component

of the tension force4. along the vertical component of

the tension force5. tangential to the circle


Dynamics of Uniform Circular Motion

The Conical pendulum.

A small ball of mass m, suspended by a cord of length l, revolves in a circle of radius r = l sin θ, where θ is the angle the string makes with the vertical.

(a) In what direction is the acceleration of the ball, and what causes the acceleration?

(b) Calculate the speed of the ball in terms of l, θ, g, and m.


Dynamics of Uniform Circular Motion

The Conical pendulum.

0yF cosy TF F Mg

cosTF Mg

2

xvF Ma Mr

sinx TF F

2

sinTvF Mr

2sincos

vMg Mr

2 tan sin tanv rg g


Highway Curves: Banked and Unbanked

When a car goes around a curve, there must be a net force toward the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.



If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.


As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than the static.

2. The static frictional force can point toward the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.



Example: Skidding on a curve.

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume:

(a)the pavement is dry and the coefficient of static friction is μs = 0.60;

(b) the pavement is icy and μs = 0.25.


From the vertical we learn:

0y NF F mg From the horizontal we learn:

And with:

f s NF F

One obtains: 2

srv mgm or sv r g

2

x fvF ma F mr


When coefficient is 0.60,

sv r g

50 0.6 9.8 17.1 m/s 62 km/hv

When coefficient is 0.25,

sv r g

50 0.25 9.8 11.1 m/s 40 km/hv Therefore car in example would slip

Therefore car in example would not slip


Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed at which the entire centripetal force is supplied by the horizontal component of the normalforce, and no friction is required. This occurs when:


And with:

0 cosy NF F mg so

2sincos

vmg mr

or 21tan v

rg

Note: independent of mass, friction, so valid for all vehicles in

all conditions


Velocity Dependent Forces: Terminal Velocity

• The drag force that acts on an object moving through a fluid depends on the velocity of the object.

• The velocity dependence is rather complicated.– For objects moving at low speeds:

– For objects moving at high speeds:bvfD

2cvfD


Where c is a constant. In fact c is better given by:

12 Dc c A

A constant dependent on shape. Usually in range 0.5 to 1

Density. Hence much more drag in water compared to air

Cross-sectional area. Less drag for streamlined objects


Terminal Speed• Consider an object that is dropped from

a height and falling through the air• Initially v = 0 so a = g• As its speed increases FD also increases

• This causes a to decrease.

• Eventually mg and FD will be equal in magnitude.• When this happens: a = 0

• The speed at which this happens is the Terminal Speed.

Mg

DF

DF Ma Mg F


Example Problem• If a bullet is shot straight up in the air, will it reach

a terminal velocity while returning to earth, and therefore be “safe”.

• Take, cd = 0.3, m = 0.01 kg and A=5.8Χ10-5 m2 (for an AK 47 bullet) and ρ = 1.3 kg/m3. Muzzle velocity is 710 m/s

2

kg5

0.01 9.8193 335

1.13 10

ms m km

T s hm

kgmgvc

51 12 2 0.3 1.3 5.8 10Dc c A

5= 1.13 10 kg/m

Topic 3 Dynamics ALLsno.phy.queensu.ca/~potato/APSC111/Lecture_15_10_2010.pdfAPSC 111 Dynamics Page...

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