Circular Motion Kinematics Centripetal Acceleration ll r r v1v1 v2v2 vv v2v2 -v 1 v = v 2 + (-v 1 ) An object moves around a circle at constant

Circular Motion KinematicsCircular Motion KinematicsCentripetal AccelerationCentripetal Acceleration

l

r r

v1

v2

vv2

-v1

v = v2 + (-v1)

An object moves around a circle at constant speed. By geometry the triangle formed by the radii and l is in the same proportion as the triangle formed by the velocity vectors and v

Triangles in the same proportion

v = l

v r

So… v = l v

r

v

Now… ac = v

t

= v l

r t

= v2

r

Note that the centripetal acceleration (v2/r) is directed towards the center of the circle

Circular Motion DynamicsCircular Motion DynamicsCentripetal ForceCentripetal Force

Fc = m v2

r

According to Newton’s second law (Fnet = ma) an object moving in a circle at constant speed must have a net force in the same direction as the acceleration….

This must be towards the center of the circle

The force that keeps a car moving around a bend in the road at constant speed is friction directed towards the center of the turn

The force that keeps the moon moving around the earth at a constant rate is the force of gravity directed towards the center of the Earth.


FFr = m vmax 2

r

A 1000 kg car rounds a curve on a flat road of radius 50m at a speed of 14m/s. Will the car make the turn of will it skid if a) the pavement is dry (s=0.6) and b) the pavement is icy (s=0.25)?

The car is only moving at 14m/s so it will be able to make the turn

m = 1000 kg r = 50m v = 14 m/s g = 9.8 m/s2 s,dry=0.6 s,icy=0.25

smg = m vmax2

r

So… g = vmax 2

r

and… gr = vmax

a) (0.6)(9.8m/s2)(50m) = vmax

17.14 m/s = vmax

The car is moving faster than this at 14m/s so it won’t be able to make the turn

b) (0.25)(9.8m/s2)(50m) = vmax

11.06 m/s = vmax


Fg = mm v 2

r

The Moon (m= 7.35 x 1022 kg) orbits the Earth at a distance of 384 x 106 m. a) What is the moon’s acceleration towards the Earth? b) How much does the Earth pull on the moon? c) Does the moon pull on the Earth more or less or the same as the Earth pulls on it? (The moon moves around the Earth one orbit every 27.4 days)

mm = 7.35 x 1022 kg r = 384 x 106 m T = 27.4 days v = ? Fg = ?

v = d / tFc v v = 2 (384 x 106m) / (27.4D x 24 H/D x 3600 s/H) = 1019 m/s

= (7.35 x 1022kg) (1019 m/s)2

(384 x 106 m)= 1.99 x 1020 N

= 2r / T

ag = v 2

r

= (1019 m/s)2

(384 x 106 m)

= 0.0027 m/s2

The moon pulls equally back on the Earth according to Newton’s third law of motion. The moon moves around the Earth because of its much smaller mass (inertia).

Circular Motion DynamicsCircular Motion DynamicsCentripetal Force - Banked TurnCentripetal Force - Banked Turn

Example: A car moving around a track which is on an incline

Fc = FNsin + FFrcos = FN sin + sFN cos

In this case, the weight of the car (mg) plus the vertical component of friction (FFr sin) is balanced by the vertical component of the normal force (FNcos ) because the vertical situation is static

FN

Fg = mg

FN sin

FFr

FFr cos

FN cos FN cos = mg + FFr sin

So……FN = mg / (cos - µssin )

The force towards the center of the turn (Fc) is provided by the horizontal component of the normal force and the horizontal component of the friction force

= FN (sin + scos)

m vmax 2 = (mg / (cos - µs sin)) (sin + scos)

r

FN cos = mg + µsFN sin

FN (cos - µssin) = mg

FFr sin

vmax = rg (sin + scos)

(cos - µs sin)

Circular Motion DynamicsCircular Motion DynamicsCentripetal Force - Banked TurnCentripetal Force - Banked Turn

A racetrack is banked up to 240. If the radius of the corner sections is 100m and the coefficient of static friction between the tires and the track is 1.5, what maximum speed can the racecar make the turn?

r = 100 m s = 1.5 = 240 g = 9.8 m/s2 vmax = ?

vmax = 76 m/s (170 mi/hr)

vmax = rg (sin + scos)

(cos - µssin)

vmax = (100m)(9.8 m/s2) (sin240 + (1.5)cos240)

(cos240 – (1.5)sin240)

Circular Motion DynamicsCircular Motion DynamicsUniversal GravitationUniversal Gravitation

The amount of gravitational force between two objects is due to the amount of mass each object has.

We can define a field as an area of force that surrounds an object. Gravitational fields exist around all masses.

It decreases in strength with distance in proportion to the inverse of the distance squared.

M1

M2

r

Fg

Fg Fg = G M1 M2

r122

G = Universal Gravitational Constant = 6.67 x 10-11 Nm2/kg2

M = Mass of objects (kg)

r = distance separating centers of masses (m)


What is the force of gravity acting on a 65kg man standing on the surface of planet Earth? (rearth = 6.38 x 106 m, MEarth = 5.97 x 1024 kg)

ME

MM

rE

Fg

Fg = G ME MM

rE2

Fg = 6.67 x 10-11Nm2/kg2 (5.97 x 1024kg)(65kg) (6.38 x 106m)2

Fg = 635.9 N = 636 N

Check: Fg = mM g = (65kg) (9.81N/kg) = 637.6 N = 638 N


Tides are created by the gravitational attraction of the sun and moon on Earth. Calculate the net force pulling on Earth during a) a new moon and b) a full moon and c) a first quarter moon. (Mmoon = 7.35 x 1022 kg, MEarth = 5.98 x 1024 kg, MSun = 1.99 x 1030 kg, rME = 3.84 x 108 m, rSE = 1.50 x 1011 m)

FSE = G MS ME

rSE2

= 6.67 x 10-11Nm2/kg2 (1.99 x 1030kg)(5.98 x 1024 kg) (1.50 x 1011m)2

= 3.53 x 1022 N

Earth

New Moon Full Moon

1st Quarter

Sun

FME = G MM ME

rME2

= 6.67 x 10-11Nm2/kg2 (7.35 x 1022kg)(5.98 x 1024 kg) (3.84 x 108m)2

= 1.99 x 1020 N


a) During a new moon

FSE = 3.53 x 1022 N

FE = FSE + FME

Earth

New Moon Full Moon

1st Quarter

Sun

FME = 1.99 x 1020 N

= 3.53 x 1022 N + 1.99 x 1020 N = 3.55 x 1022 N

b) During a full moon FE = FSE + FME = 3.53 x 1022 N - 1.99 x 1020 N = 3.51 x 1022 N

c) During 1st Quarter FE = FSE + FME

FE = (FSE2 + FME

2)

FE = 3.53 x 1022 N

FSE

FME

= ((3.53 x 1022 N)2 + (1.99 x 1020 N)2)

FE

= tan-1 (FME / FSE) = tan-1 (1.99 x 1022 / 3.53 x 1022 ) = 0.320

Circular Motion DynamicsCircular Motion DynamicsGravitational Field StrengthGravitational Field Strength

Field strength is defined as the amount of force per quantity.

M1

r12

g

= G M1 r12

2

gM = 6.67 x 10-11Nm2/kg2 7.35 x 1022 kg

(1.74 x 106m)2 = 1.62 N/kg

(about 1/6 that on the surface of Earth)

For a gravitational field around an object it is the number of Newton’s of force

acting on every kg of a second object placed in this field. g = Fg

M2

What is the gravitational field strength on the surface of the moon? (Mm = 7.35 x 1022 kg, rm = 1.74 x 106 m)

gM = G MM rM

2

Circular Motion DynamicsCircular Motion DynamicsGravitational Field StrengthGravitational Field Strength

MS

rM

g

gWD = 6.67 x 10-11Nm2/kg2 1.99 x 1030 kg

(1.74 x 106m)2 = 4.38 x 107 N/kg

(This is about 4.5 million times our surface gravity on Earth)

A typical white dwarf star, which once was an average star like our sun but is now in the last stages of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity (g) of this star? (Ms = 1.99 x 1030 kg)

gWD = G MS rM

2

Circular Motion DynamicsCircular Motion DynamicsGravitational PotentialGravitational Potential

Gravitational Potential Energy is the energy an object has due to its position in a gravitational field.

M1

r12

GP

now….g = G M1 r12

2

Like field strength, the amount of potential energy per kg of a second object placed at this position in the gravitational field is called gravitational potential

The gravitational potential energy of mass M2 at distance r12 can be defined as….

GPE = - G M1 M2 r12

GP = - G M1 r12

The GPE and GP get smaller the further you go from the mass M1. They both are reduced to zero at infinity. It should also be noted that the work done to move a mass between two points in a gravitational field is independent of the path taken. (proved by calculus)

The change in GPE (GPE) between two points at different distances from an object we have already expressed as - M2gdv according to the work-energy theorem

Circular Motion DynamicsCircular Motion DynamicsEscape SpeedEscape Speed

How much speed is needed to send a rocket (M2) soaring into space so that it escapes the pull of a planet (M1)?

M1

rp

ve

Determine the escape speed from the surface of earth

If the rocket just makes it to infinity and slows down all the way its final velocity (vf) will be zero and its final GPE will be zero so the total energy will be zero

1/2 M2 ve2 - G M1 M2 = 0

r

KEi + PEi = KEf + PEf = 0

ve = 2G M1

rp

rearth = 6.38 x 106 m MEarth = 5.97 x 1024 kg G = 6.67 x 10-11 Nm2/kg2

ve = 2G ME

rearth

= (2(6.67 x 10-11 Nm2/kg2) (5.97 x 1024 kg ))

6.38 x 106 m = 11.2 x 103 m/s (25000 mi/h)

Circular Motion DynamicsCircular Motion DynamicsEscape Speed - Black HoleEscape Speed - Black Hole

When a huge star many times our own sun’s mass runs out of nuclear fuel gravity causes it to collapse in on itself eventually resulting in an incredibly dense piece of matter.

MBH

R

Light has a speed (c) of 3 x 108 m/s. If the escape speed is greater than this not even light will escape. Karl Schwarzchild calculated the radius around a black hole at which light wouldn’t be able to escape. This boundary is called the event horizon (R)

c = 2G MBH

R

c = 3 x 108 m/s MBH = 5.97 x 1030 kg G = 6.67 x 10-11 Nm2/kg2

R = 2G MBH

c2

= 2(6.67 x 10-11 Nm2/kg2) (5.97 x 1030 kg )

(3 x 108)2 m = 8850m (8.9 km)

If a black hole has a mass of 3 times our own sun’s mass and a radius of 2km determine its event horizon (Ms = 1.99 x 1030 kg)

So ….R = 2G MBH

c2

Circular Motion DynamicsCircular Motion DynamicsSatellites and WeightlessnessSatellites and Weightlessness

Satellites are freefalling towards the earth just as a dropped ball falls towards earth but they have such high tangential velocities that as they fall they follow the curvature of the planet.

Motion of projectile without gravity

If the tangential speed is high enough the projectile will fall “around the earth”

The earth’s surface drops about 5m for every 8km horizontally. How fast should a projectile be fired to go into a low orbit around the earth?

A projectile falls about 5m in 1s so it should be fired at 8km/s (18000 mi/h)


How can someone in the space shuttle be experiencing weightlessness when the spacecraft is experiencing nearly as much gravitational acceleration (g) as someone on the surface?

An airborne athlete and an astronaut both experience weightlessness


Our experience of weight is really the normal force of the surface we are in contact with pushing back on our body

You are made aware of this idea when traveling on a high speed elevator or riding on an amusement park ride.

The normal force has a small magnitude at the top of the loop (where the rider often feels weightless) and a large magnitude at the bottom of the loop (where the rider often feels heavy).

Circular Motion DynamicsCircular Motion DynamicsApparent WeightlessnessApparent Weightlessness

Consider a roller coaster track that has a series of hills and dips as shown below. The black arrows show that the centripetal acceleration is directed towards the center of the circular shaped arcs as the car moves along the track.

The forces acting on the car at positions A, B and D are shown below


The radius of the hill at B is 20m. A) What speed would the 1000 kg coaster car have to go over the top of the hill for the passengers to feel weightless? B) If the car went faster than this, what would the passenger feel if they were wearing a harness?

When the passengers feel weightless the force applied by the seat/harness (Fapp) would be zero.

FC = Fg + Fapp = m v2 / r

Fapp = FN r = 20 m m = 1000 kg g = 9.8 m/s2

Fapp = m v2 / r - Fg

Fapp = m v2 / r - mg = m (v2 / r - g)

When g = v2 / r Fapp = 0N

v = g r = (9.8m/s2 )(20m) = 14 m/s (31 mi/h)

If the car went faster than this, the needed FC would increase so a force greater than Fg would be needed to act downwards (FC = Fg + Fapp). The harness would have to push down on the riders shoulders to provided the needed extra force


c) If the radius of the track at the bottom of the dip (between d and e) is 50 m and the car enters this section at 35 m/s, determine i) the apparent weight and ii) the number of “g”’s of centripetal acceleration experienced by a 60 kg passenger

Fapp = FN r = 50 m m = 60kg v = 20 m/s g = 9.8 m/s2

The amount of centripetal acceleration experienced is given by aC = v2 / r

FC = FN - Fg = m v2 / r

FN = m v2 / r + Fg

FN = m v2 / r + mg = m (v2 / r + g)

= 2058 N FN = (60kg) ((35 m/s)2 / 50 m + 9.8 m/s2 )

aC = v2 / r = (35 m/s)2 / 50 m = 24.5 m/s2

This is 24.5 m/s2 / 9.8 m/s2 = 2.5 “g”’s

Circular Motion DynamicsCircular Motion DynamicsKepler’s Third LawKepler’s Third Law

Kepler’s third law relates the force of gravity using Newton’s Universal Law of Gravitation with centripetal force.

This force of gravity that the earth exerts on the moon is the centripetal force that the moon requires to orbit the earth. That means that…

With this equation, we can find the velocity of the moon around the earth given the mass of the earth and the radius of the moon’s revolution. This relationship is the same for any planet revolving around the sun.

Fg = Fc

G ME m / r2 = m vt2 / r or…..vt

2 = GME / r

Since the period of a planet, T, can be related to velocity by v = 2r / T, we’re all set. Now we just solve for T.

This is Kepler’s third law. Using this equation, we can find the period of a planet if we know the radius of its revolution and the mass of the planet it revolves around.

(2r / T)2 = GM / r So…..T2 = (42 r3) / (GM).

Find the time for the earth to revolve around the sun, given that the sun’s mass is 1.991 • 10 30 kg and that the earth is 1.496 • 1011 m from the sun.

or…..T2 r3

T2 = [4 • 2 • (1.496x1011 m )3] / [(6.673x10-11 N•m2/kg2) • (1.991x1030 kg)] = 9.953 x 1014 s2.

T = 3.155 x 107 s = 365.14 days. (as expected!!)

Circular Motion DynamicsCircular Motion DynamicsTorqueTorque

The ability of a force to rotate an object about some axis is measured by quantity called torque, .

Torque like force is a vector quantity. Its magnitude is:

= Frperp

where rperp is the lever arm and is the perpendicular distance from the axis of rotation to a line drawn along the direction of force.

If there are several forces acting on the object then the net torque is obtained by summing the torques produced by each of these forces., thus

net = = 1 + 2 +…..= F1r1perp + F2r2perp +…..

F

rrperp

A student pushes on a door furthest from the hinges (1.5m) with a force of 15N at an angle of 350 to the plane of the door. Determine the torque on the door.

= Frperp = F r sin = (15N) (1.5m) sin350

= 13 Nm Note: Positive torque is assigned counterclockwise and negative torque clockwise

Circular Motion DynamicsCircular Motion DynamicsRotational EquilibriumRotational Equilibrium

The first condition for equilibrium that we have discussed is namely, the sum of all the external forces is zero. With no net forces acting on an object it obeys Newton’s first law, I.e. no accelerations and thus no changes in motion. (F= 0) The object could still rotate and change its rate of rotation if the torques don’t add to zero.

The second condition for equilibrium, therefore, is that there are no changes in rotation. This happens when the sum of the external torques adds to zero, so we have

= 0

Position of the Axis of Rotation

If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque; the location of the axis is completely arbitrary.

The Center of Gravity

The center of gravity is that point in or near an object where all the torques due to the weight of the object add to zero no matter how the object is orientated. It is what we usually call the balance point. The x component is found from the equation

xcg = m1x1 + m2x2 +… = i mi+xi

m1 + m2 +… M

The y and z components of the center of gravity are found in a completely analogous manner.

The center of gravity of a symmetrical body that is homogenous must lie on the axes of symmetry.

Circular Motion DynamicsCircular Motion DynamicsRotational EquilibriumRotational Equilibrium

A hungry 700 N bear walks out on a beam in an attempt to retrieve some “goodies” hanging at the end. The beam is uniform, weighs 200 N, and is 6.00 m long; the goodies weigh 80.0 N. a) Draw a FBD for the beam. b) When the bear is at x=1.00 m, find the tension in the wire and the components of the reaction force at the hinge. c) If the wire can withstand a maximum tension of 900 N, what is the maximum distance the bear can walk before the wire breaks?

700 N700 N

HH

VV

XX

3m3m 3 m3 m

80 N80 N200 N200 N

TTxx = T cos60 = T cos6000 = 0.5 T = 0.5 T

TTyy = T sin60 = T sin6000 = 0.866T= 0.866T

b) If x = 1 m then

left end = (-700 N)(1m) - (200 N)(3 m) - (80 N)(6 m) + (0.866T)(6 m)

Equating this to zero gives: T = 342 N

From Fx = 0, H = 0.5 T = 171 N

From Fy = 0, V = 980 N - 0.866T = 683 N

a)

c) If T = 900 N

left end = (-700 N)(X) - (200 N)(3 m) - (80 N)(6 m) + (779.4 N)(6 m)

Equating this to zero and solving for x gives: x = 5.13 m

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Circular Motion Kinematics Centripetal Acceleration ll r r v1v1 v2v2 vv v2v2 -v 1 v = v 2 + (-v 1 ) An object moves around a circle at constant