Physics Subject Area Test

MECHANICS:DYNAMICS

Uniform Circular Motion

Motion in a circular path at constant speed

Speed constant, velocity changing continually

Velocity changing direction, so there is acceleration

Called centripetal acceleration, since it is toward the center of the circle, along the radius

Value can be calculated by many formulas, first is

ac = v2/r

*Example

A bicycle racer rides with constant speed around a circular track 25 m in diameter. What is the acceleration of the bicycle toward the center of the track if its speed is 6.0 m/s?

ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2 r 12.5 m 12.5 m

Rotation and RevolutionRotation-Around an Internal Axis-Earth

rotates 24 hours for a complete turnLinear (tangential) versus rotational

speedLinear is greater on outside of disk

or merry-go-round, more distance per rotation

Linear is smaller in middle of disk, less distance per rotation.

Rotational speed is equal for bothRotations per minute (RPM)Linear speed is proportional to both

rotational speed and distance from the center

Rotation and RevolutionRevolution-Around an

External Axis-Earth revolves 365.25 days per trip around sun

Same relationship between linear and revolutional speeds as with rotational

Planets do not revolve at the same revolutional speeds around the sun

Period

Another important measure in UCM is period, the time for 1 rotation or revolution

Since x=v0t , this implies that vT = 2r and thus T= 2r/ v

Rearranging differently, v= 2r/ T and then inserting it into the acceleration equation

ac = v2/r = 42r/T2

*Example

Determine the centripetal acceleration of the moon as it circles the earth, and compare that acceleration with the acceleration of bodies falling on the earth. The period of the moon's orbit is 27.3 days.

According to Newton's first law, the moon would move with constant velocity in a straight line unless it were acted on by a force. We can infer the presence of a force from the fact that the moon moves with approximately uniform circular motion about the earth. The mean center-to-center earth-moon distance is 3.84 x 108 m.

*Exampleac= 42r = 42(3.84 x 108) T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 106 s

T2 (2.36 x 106)2

ac = 2.72 x 10-3 m/s2

The ratio of the moon's acceleration to that of an object falling near the

earth is ac = 2.72 x l0-3 m/s2 = 1 g 9.8 m/s2 3600

*Frequency

The number of revolutions per time unit

Value is the inverse of the period, 1/T

Units are sec-1 or Hertz (Hz)Inserting frequency into the ac equation

ac=42f 2r

*Example

An industrial grinding wheel with a 25.4-cm diameter spins at a rate of 1910 rotations per minute. What is the linear speed of a point on the rim?

The speed of a point on the rim is the distance traveled, 2r, divided by T, the time for one revolution. However, the period is the reciprocal of the frequency, so the speed of a point on the rim, a distance r from the axis of rotation, is

v = 2rf

v= (2)(25.4cm/2)(1910/1 min)(1min/60s)

v = 2540cm/s = 25.4 m/s.

*Angular Velocity

Velocity can be defined in terms of multiples of the radius, called radians

There are 2 radians in a circle, and so the angular velocity w = v/r

In terms of period w= 2/TIn terms of frequency w=2f

*ExampleAt the Six Flags amusement park near Atlanta.

The Wheelie carries passengers in a circular path with a radius of 7.7 m. The ride makes a complete rotation every 4.0 s. (a) What is a passenger's angular velocity due to the circular motion? (b) What acceleration does a passenger experience?

a) The ride has a period T = 4.0 s. We can use it to compute the angular velocity as

2= 2 rad = rad/s = 1.6 rad/sT 4.0 s 2.0

(b) Because the riders travel in a circle, they undergo a centripetal acceleration given by

ac= w2r = (/2 rad/s)2(7.7m) = 19m/s2. Notice that this is almost twice the acceleration of a body in free fall.

Angular Velocity and Acceleration

Any real object that has a definite shape can be made to rotate – solid, unchanging shape

Angular displacement -- q -- Radians around circular pathAngular velocity -- w --radians per second, angle between

fixed axis and point on wheel changes with timeAngular acceleration -- a -- increase of w , when angular

velocity of the rigid body changes, radians per seconds squared

Rotational Kinematics

Rotational velocity, displacement, and acceleration all follow the linear forms, just substituting the rotational values into the equations:

q= wot + 1/2 at2 wf2 = wo

2 + 2aqwf=w0 +at q=(w0 +wf) t/ 2q= x/r a= a/r w= v/r

*Example

The wheel on a moving car slows uniformly from 70 rads/s to 42 rads/s in 4.2 s. If its radius is 0.32 m:

a. Find a b. Find q c. How far does the car go?a. a= Dw = (42-70) rads/s = -6.7 rads/s2

Dt 4.2 sb. q= wot + 1/2 at2 = (70)(4.2) + 1/2(-6.7)(4.2)2 = 235 rads

c. q = x / r in rads so x = q r = (0.32)(235) = 75 m

*ExampleA bicycle wheel turning at 0.21 rads/s is brought to rest by the brakes in exactly 2 revolutions. What is its angular acceleration? q = 2 revs = 2(2) radians = 4 rads wf=0 rads/s wo= 0.21 rads/s

Use angular equivalent of vf2 = vo

2 + 2ax which is

wf2 = wo

2 + 2aq

(0)2 = (0.21)2 + 2a(4)a = -(0.21)2 = -1.8 x10-3 rad/s2 2(4)

Forces in Circular Motion

Centripetal ForceForce toward the center from an object, holding it in circular motionAt right angle to the path of motion, not along its distance, therefore

does NO work on objectExamples

Gravitation between earth and moon Electromagnetic force between protons and electrons in an atomFriction on the tires of a car rounding a curve

Equation is Fc=mac = mv2/r

*ExampleApproximately how much force does the earth

exert on the moon? Moon’s period is 27.3 daysAssume the moon's orbit to be circular about a

stationary earth. The force can be found from F = ma. The mass of the moon is 7.35 x 1022 kg.

Fc= mac = m 42r

T2

Fc = (7.35 x 1022 kg)42(3.84 x 108m)

((27.3 days)(24 hr/day)(3600 s/hr))2

Fc=2.005 x 1020 N.

*Forces in Circular Motion

Centrifugal “Force”

Not a true force, but really the result of inertia

“Centrifugal force effect” makes a rotating object fly off in straight line if centripetal force fails

*Example

Imagine a giant donut-shaped space station located so far from all heavenly bodies that the force of gravity may be neglected. To enable the occupants to live a “normal” life, the donut rotates and the inhabitants live on the part of the donut farthest from the center. If the outside diameter of the space station is 1.5km, what must be its period of rotation so that the passengers at the periphery will perceive an artificial gravity equal to the normal gravity at the earth's surface?

The weight of a person of mass m on the earth is a force F = mg.The centripetal force required to carry the person around a circle of radius r is

F =mac = m 42r T2

We may equate these two force expressions and solve for the period T:mg =m42r

T2 T=2 =2 =55s = 0.92 min.rg

7509 81 2

mm s. /

*Banked Curves“Banking” road curves makes turns without skidding possibleFor angle q, there is a component of the normal force toward

the center of the curve, thus supplying the centripetal force. The other component balances the weight force.

FN sin q = mv2/r FN cos q = mg tan q= v2/gr thusly q = tan-1 (v2/gr) This equation can give the proper angle for banking a curve of

any radius at any linear speed

*Banked Curve ExampleA race track designed for average speeds of 240 km/h (66.7

m/s) is to have a turn with a radius of 975 m. To what angle must the track be banked so that cars traveling 240km/h have no tendency to slip sideways?

Determine q from q = tan-1 (v2/ g r)

= tan-1 (66.72/9.81(975))= 24.9o

*Law of Universal Gravitation

Newton’s first initiative for the Principia was investigating gravity

From his 3rd law, he proposed that each object would pull on any other object

He likewise noted differences due to distanceHis final relationship was that Force was proportional

to masses and inversely proportional to distance squared

Using a constant Fg = Gm1m2

r2

*Center of Gravity

Newton found that his law would only work when measuring from the center of both objects

This idea is called the center of gravitySometimes it is at the exact center of the objectSometimes it may not be in the object at all

All forces must be from the CG of one object to the CG of the other object

*Universal Gravitation Constant

G was elusive to find since gravity is a weak force if masses are small

Cavendish developed a device which made measurement of G possible

The value of G is 6.67 x 10-11 N m2

This puts Fg in Newtons kg2

G can be used then to find values of many astronomical properties

*Example

Consider a mass m falling near the earth's surface. Find its acceleration in terms of the universal gravitational constant G. The gravitational force on the body is F = GmME

r2

ME = mass of the earth r = the distance of the mass from the center of the earth, essentially the earth's radius.

The gravitational force on a body at the earth's surface is F = mg. mg= GmME or g =GME

r2 r2

Both G and ME are constant, and r does not change significantly for small variations in height near the surface of the earth. The right-hand side of this equation does not change appreciably with position on the earth’s surface, so replace r with the average radius of the earth RE

g = GME

RE2

*ExampleShow that Kepler’s third law follows from the law of universal gravitation. Kepler’s third

law states that for all planets the ratio (period)2/ (distance from sun)3 is the same.Make the approximation that the orbits of the planets are circles and that the orbital speed

is constant.The sun's gravitational force on any planet of mass m is

F= GmM r2

M =the mass of the sun. Because the mass of the sun is so much larger than the mass of the planet, we can assume, as Kepler did, that the sun lies at the center of the planetary orbit. The circular orbit implies a centripetal force. This net force for circular motion is provided by the gravitational force. Equating these two forces, we get

Fc=GmM = 42mr Rearranging gives T2 = 42

r2 T2 r3 GM

*Example

Use the law of universal gravitation and the measured value of the acceleration of gravity g to determine the average density of the earth. The density, r of an object is defined as its mass per unit volume: r = m/V where m is the mass of the object whose volume is V

From a previous example g = GME

RE2

Substitute for M an expression involving r, r = ME/V. If we take the earth to be a sphere of radius RE. Then

r = ME and ME= 4/3RE3 r

4/3RE3

The equation for g can then be rewritten in terms of the density as g = G(4/3RE

3 r) = 4/3GRE r

RE2

*Density Example

Upon rearranging, we find the density to ber = 3g 4REG

Inserting the numerical values, we getr = 3(9.81 m/s2) 4(6.38 x 106 m)(6.67 x 10-11 N m2/ kg2

r = 5.50 x 103 kg/m3

*Escape Velocity

Earth spacecraft must get entirely away from the earth to go on to other planetsThis requires giving a spacecraft enough energy to overcome the gravitational

potential energy of earthThis gives an equation such that

Where M and R varyaccording to the celestial object involved

v GMResc =

2

*Black Holes

If the escape velocity is equal to the speed of light, gravity will keep even light from escaping--the idea behind the black hole

Conjecture due to observations from spaceTheory is a supergiant star collapses in on itself creating super

strong gravity at a small point

*Black Holes

Gravity is great due to small distance with huge mass

Gravity only great near the object, at distance gravity is no different

*Keplers Laws

First Law: Each planet travels ina an elliptical path around the sun, and the sun is at one of the focal points

*Keplers Laws

Second Law: An imaginary line is drawn from the sun to any planet sweeps out equal areas in equal time intervals.

*Keplers Laws

Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun or

T2 ∝ r3

*Major Equations!!

ac=v2/r

ac=42r/T2

ac=42rf2

ac= w2r

w=v/r

w=2/T

w=2f

v=2r/T = 2rf

f=1/T

Fc=mac

Fc=mv2/r

Fg=GMm/r2

T=2

T2 = 42

r3 GM

rg

* General Definition of Torque

*Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by

*q = 0° or q = 180 °: torque are equal to zero*q = 90° or q = 270 °: magnitude of torque attain to the maximum

=rF sinq

*Torque Units and Direction*The SI units of torque are N.m

*Torque is a vector quantity*Torque magnitude is given by

*Torque will have direction*If the turning tendency of the force is counterclockwise,

the torque will be positive*If the turning tendency is clockwise, the torque will be

negative

The work done by the torque is given by

=rF sinq

W =q

*Moment of inertia*In rotation problems, its

not only the mass of an object that is important but also its location

*The spacial distribution of the mass of an object is called the

*Moment of inertia (I)

I =1/2 MR 2

* Newton’s Second Law for a Rotating Object

*When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration

*The angular acceleration is directly proportional to the net torque

*The angular acceleration is inversely proportional to the moment of inertia of the object

*The relationship is analogous to

=Ia

F =ma

*Extended Work-Energy Theorem

*The work-energy theorem tells us

*When Wnc = 0,

*The total mechanical energy is conserved and remains the same at all times

*Remember, this is for conservative forces, no dissipative forces such as friction can be present

ffii mgymvmgymv +=+ 22

21

21

Wnet =DKE +DPE

KE i + PE i =KE f + PE f

*Total Energy of a System*A ball is rolling down a ramp

*Described by three types of energy*Gravitational potential energy

*Translational kinetic energy

*Rotational kinetic energy

*Total energy of a system

2

21 wIKEr =

MghPE =

KE t =12

Mv 2

E =12

Mv 2 + Mgh +12

Iw 2

* Conservation of Mechanical Energy

*Conservation of Mechanical Energy

*Remember, this is for conservative forces, no dissipative forces such as friction can be present

( ) ( )t r g i t r g fKE KE PE KE KE PE+ + = + +

12

mv i2 + mgyi +

12

Iw i2 =

12

mv f2 + mgy f +

12

Iw f2

* Work-Energy in a Rotating System

*The work done on the body by the external torque equals the change in the rotational kinetic energy

*The work equals the negative of the change in potential energy

*Conservation of Energy in Rotational Motion

2

21 wIKW D=D=

W =K2 - K1 =12

Iw22 -

12

Iw12

-U2 +U1 =12

Iw22 -

12

Iw12

E =U +12

Iw 2 =constant

*Problem Solving Hints

*Choose two points of interest*One where all the necessary information is given

*The other where information is desired*Identify the conservative and non-conservative forces

*Write the general equation for the Work-Energy theorem if there are non-conservative forces*Use Conservation of Energy if there are no non-conservative forces

*Use v = w to combine terms*Solve for the unknown

*Buoyancy*The “upthrust” or buoyancy is equal to the weight

of the displaced fluid

*V = volume of liquid displaced*G = the acceleration of gravity

Fbouyant =rVg

Objects that are less dense than water will float

To find where an object sits in the water, find the ratio of its density to the density of water

roak

rwater

=percent submerged

*Bernoulli’s principleBernoulli's Principle: slower moving air below the wing creates greater pressure and pushes up.

*Moving fluids

* Q = volumetric flow rate

* A = cross sectional area

* v = fluid velocity

* = the angle between the direction of the fluid flow and a vector normal to A

* Mass flow rate (m)

* When a pipe is constricted, the mass flow rate is conserved

Q =Av cosq

q

VvAm rr ==

v1A1 =v2A2

* Bernoulli’s principle for an incompressible fluid

*constant=++ pghv rr 2

21

Where v = fluid velocityg = acceleration due to gravityh = heightp = pressure = fluid density

r