data booklet reference
TRANSCRIPT
Data booklet reference:
β’ π£ = ππ
β’ π =π£ 2
π=
4 2π
π 2
β’ πΉ =ππ£ 2
π= π2π
centripetal or radial acceleration
A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v.
ππ ππ ππ ππ π‘βπ ππππππ‘πππ ππ β π£, ππ πππππ‘π π‘ππ€πππ π‘βπ ππππ‘ππ ππ πππππ’πππ πππ‘πππ
2
c
va =
r
βΞΈ
βrβv βΞΈ
r1 r2
v2
P
O
Qv1
v2
v1
_
βv v1v2βv = β v2 = v1 + βv
velocity β tangent to the circle
centripetal acceleration β points toward the center
always perpendicular to each other
2
c c
mv F = ma =
r
useful relations: - speed(linear) and angular speed
2ΟRv =
T
2ΟΟ =
T
constant speed:
period T: time required for one complete revolution (s)
speed v = distance/time
angular speed Ο = angle swept/time
Translational variable Rotational variable Relationship
distance s angle swept (rad) s = r
tangential velocity
π£ =π
π‘
angular velocity (rad/s) π =
π
π‘
v = r Ο
tangentialacceleration a =
βπ£
π‘
angular acceleration (rad/s2) πΌ =
βπ
π‘a = r
v = s / t = ( r ) / t = r
Speed, v, depends on position r, but angular speed does not
A CAR ON A LEVEL (UNBANKED) SURFACE
Think of what would happen if there were no friction. Have you ever found yourself driving on ice? If you turn steering wheel nothing happens because there is little or no friction between the ice-covered road and the car's tires. You need frictional force to turn.
A 1200 kg car rounds the curve on a flat road of radius 45 m. If the coefficient of static friction between tires and the road is s = 0.82, what is the greatest speed the car can have in the curve without skidding?
So if you turn the steering wheel you are pushing the groundin the black direction and the ground is pushing back onyou in the orange direction.That is the friction force that the road exerts on thecar's tires pointing toward the center of the turn,responsible for turning the car.
Fn = mg Ffr = s Fn = s mg
2
fr cv
F = ma = ma = mr
2
sv
ΞΌ g = r
v = 19 m/s
A CAR ON A BANKED SURFACEβ¦
can a car negotiate a curve without friction (on ice)? YES, but the curve has to be banked!
If a roadway is banked at the proper angle, a car can round the corner without any assistance from friction between the tires and the road. The horizontal component of the normal (reaction) force is the net force. This force is centripetal force to turn the car.
Find the appropriate banking angle for a 900 kg car traveling at 20.5 m/s in a turn of radius 85.0 m so that NO friction is required.
πΉπsin π = mπ£2
πβ tanπ =
π£2
ππ
for given speed v and radius r of the turn we get angle of the bank independent of mass m
20v
ΞΈ = arctan = 26.7rg
πΉπcosπ = mπ β πΉπ =ππ
cosπ
for given ΞΈ and r:
if v < rg tan ΞΈ the car would slide down a frictionless banked curve
if v > rg tan ΞΈ the car would slide off the top
ROLLER COASTER
Top of a Hill 2 2
n nmv mv
mg - F = F = mg - R R
T v = Rg
minimum speed required to stay alive is mathematically given with Fn = 0 (the rider just barely looses contact with the seat
Upside-down at the Top of a loop
Fn = 0 danger when feeling weightless
min speed required to stay alive and not to fall down:
T v = Rg
Bottom of a Valley
2 2
n nmv mv
F - mg = F = mg + R R
The rider's hands and arms are hard to move. The rider's blood is even hard to move. Airplane pilots are in this situation as they pull out of a dive. Apparent weight of 6-7 times one's real weight -- can mean that not enough blood will be circulated to the brain and a pilot -- or other passenger -- may pass out.
πΉπ + ππ =ππ£2
π πΉπ =
ππ£2
π β ππ
AIRPLANE MAKING A TURN - BANKING
Lift force (L) generated by the airplane wings is not equal to mg anymore, but greater.
πΏ cos π = ππ πΏ sin π = ππ£2
π
π =π£2
π tanπ
As the plane banks (rolls), the lift vector begins to have a horizontal component.
kx = FC = mv 2/ r implies that as v increases, so does the centripetal force FC needed to move it in a circle. Thus, x increases.
F = kx (k = CONST).
kx = F k = F / x = 18 / 0.010 = 1800 Nm-1.FC = kx = 1800( 0.265 β 0.250 ) = 27 N.FC = v 2/ r v 2 = r FC = 0.265(27) = 7.155v = 2.7 ms-1.
Use v = r ( = CONST).
Use a = r 2 ( = CONST).
At Pr = R
v = Ra = R 2
At Qr = 2Rv = 2R = 2va = 2R 2 = 2a
EXAMPLE: The earth has a mass of M = 5.981024 kg and the moon has a mass of m = 7.361022 kg. The mean distance between the earth and the moon is 3.82108 m. What is the gravitational force between them?
SOLUTION: F = GMm / r 2.
F = (6.67Γ10β11)(5.981024 )(7.361022 ) / (3.82108)2
F = 2.011020 N.
r is the distance between the centers of the masses.
m1 m2
rF12 F21
FYI The radius of each
mass is immaterial.
What is the speed of the moon in its orbit about earth?
SOLUTION: FC = FG = mv 2 / r .
2.011020 = ( 7.361022 ) v 2 / 3.82108 v = 1.02103 ms-1.
What is the period of the moon (in days) in its orbit about earth?
SOLUTION: v = 2r / T.
T = 2r / v = 2( 3.82108 ) / 1.02103
= (2.35 106 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d.
PRACTICE: A 525-kg satellite is launched from the earthβs surface to a height of one earth radius above the surface. What is its weight (a) at the surface, and (b) at altitude?
SOLUTION:
(a) AT SURFACE: gsurface = 9.8 m s-2.
F = mg = (525)(9.8) = 5145 N. W=5100 N
(b) AT ALTITUDE: gsurface+R = 2.45 m s-2.
F = mg = (525)(2.46) = 1286.25 N W=1300N
EXAMPLE: Find the gravitational field strength at a point between the earth and the moon that is right between their centers.
SOLUTION:
Make a sketch.
Note that r = d / 2 = 3.82108 / 2 = 1.91 108 m.
gm = Gm / r 2
gm = (6.67Γ10-11)(7.361022)/(1.91108)2 = 1.3510-4 N.
gM = GM / r 2
gM = (6.67Γ10-11)(5.981024)/(1.91108)2 = 1.0910-2 N.
Finally, g = gM β gm = 1.0810 -2 N,β.
M = 5.981024 kgm = 7.361022 kg
gMgm
d = 3.82108 m
PRACTICE: Two spheres of equal mass
and different radii are held a distance d apart. The gravitational field strength is measured on the line joining the two masses at position x
which varies.
Which graph shows the variation of g with x correctly?
There is a point between M and m where g = 0.
Since g = Gm / R 2 and Rleft < Rright, then gleft > gright at the surfaces of the masses.
EXAMPLE: A satellite in geosynchronous orbit takes 24 hours to orbit the earth. Thus, it can be above the same point of the earthβs surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. RE = 6.37106 m.
SOLUTION:
π = (24 β)(3600 π ββ1 ) = 86400 π
ππ£2
π= πΊ
ππ
π2 π£2 = πΊπ
π
2ππ
π
2
= πΊπ
π
π3 =πΊπ
4π2π2 β π3 =
6.6710β115.981024
4π2(86400)2
π = 42250474 π = 6.63 π πΈ
Keplerβs third law originally said that the square of the period was proportional to the cube of the radius β and nothing at all about what the constant of proportionality was. Newtonβs law of gravitation was needed for that!
ππ£2
π= πΊ
ππ
π2 π£2 = πΊπ
π
2ππ
π
2
= πΊπ
πβ π3 =
πΊπ
4π2 π2
βͺ πΉπ =ππ£2
π= πΊ
ππ
π2 π£2 = πΊπ
π
βͺ2ππ
π
2
= πΊπ
π
βͺ π2 =4π2
πΊππ3 β π =
4π2
πΊπ
3/2
π3/2
ππ£2
π= πΊ
ππ
π2 π = πΊπ
π£2
ππππ¦
=
πΊππ£π₯
2
πΊππ£π
2
=π£π
2
π£π₯2 =
2πππππ
2πππππ
2
=ππππππππ
2
=8ππππ
2
ππππ¦
3
= 64 βππππ¦
= 4
Since the satellite is in circular orbit FC = mv 2/ r.
Since the satelliteβs weight is holding it in orbit, FC = mg.
Thus mv 2/ r = mg.
Finally g = v 2/ r.
It is the gravitational force.
This question is about gravitation. A binary star consists of two stars that each follow circular orbits about a fixed point P as shown. The stars have the same orbital period T. Each star may be considered to act as a point mass with its mass concentrated at its centre. The stars, of masses M1 and M2, orbit at distances R1 and R2 respectively from point P.
(b) By considering the force acting on one of the stars, deduce that
the orbital period T is given by the expression
(a) State the name of the force that provides the centripetal force
for the motion of the stars.
M1 experiences FC = M1v12/ R1.
Since v1 = 2R1/ T, then v12 = 42R1
2/ T 2.
Thus FC = FG M1v12/ R1 = GM1M2 / (R1+R2)
2.
42R1(R1+R2)2 = GM2T
2
T 2 = R1(R1+R2)2
42
GM2
M1(42R1
2/ T 2) / R1 = GM1M2 / (R1+R2)2
Note that FG = GM1M2 / (R1+R2)2.
M2M1
R1
R2
P
M2M1
R1
R2
P From (b)
T 2 = (42 / GM2)R1(R1+R2)2.
From symmetry
T 2 = (42/ GM1)R2(R1+R2)2.
(1 / M2)R1 = (1 / M1)R2
M1 / M2 = R2 / R1
Since R2 > R1, we see that M1 > M2.
(c) The star of mass M1 is closer to the point P than the star of mass M2.
Using the answer in (b), state and explain which star has the larger mass.
(42/ GM2)R1(R1+R2)2 = (42/ GM1)R2(R1+R2)
2