Download R - Uplift North Hills Prep

Data booklet reference:
• 𝑣 = 𝜔𝑟
•𝑎 =
•𝐹 =
𝑣2
𝑟
=
𝑚𝑣 2
𝑟
4  2𝑟
𝑇2
= 𝑚2𝑟
A particle is said to be in uniform circular motion if it travels in a
circle (or arc) with constant speed v.
centripetal or radial acceleration
𝑎𝑠 𝑎𝑐 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ∆𝑣,
𝑎𝑐 𝑝𝑜𝑖𝑛𝑡𝑠 𝑡𝑜𝑤𝑎𝑟𝑑 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
v1
P
Q
∆r
r1
v2
ac =
r
v2
v2
∆v
r2
∆θ
_v
∆θ
O
v2 = v1 + ∆v
∆v = v2 – v1
1
 velocity – tangent to the circle
 centripetal acceleration – points toward the center
 always perpendicular to each other
mv 2
Fc = mac =
r
useful relations: - speed(linear) and angular speed
 period T:
time required for one complete revolution (s)
speed v = distance/time
constant speed: v =
2πR
T
angular speed ω = angle swept/time
ω=
2π
T
Translational variable
Rotational variable
distance
angle swept (rad)
tangential
velocity
tangential
acceleration
s
𝑠
𝑣=
𝑡
a=
∆𝑣
𝑡
angular velocity
(rad/s)
angular acceleration
(rad/s2)
Relationship
s= r

𝜃
𝑡
v=rω
∆𝜔
𝑡
a=r
𝜔=
𝛼=
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.
So if you turn the steering wheel you are pushing the ground
in the black direction and the ground is pushing back on
you in the orange direction.
That is the friction force that the road exerts on the
car's tires pointing toward the center of the turn,
responsible for turning the car.
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?
Fn = mg
Ffr = s Fn = s mg
v2
Ffr = ma = mac = m
r
v2
μsg =
r
v = 19 m/s
A CAR ON A BANKED SURFACE…
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.
𝐹𝑛 cos𝜃 = m𝑔
⇒ 𝐹𝑛 =
𝑣2
𝐹𝑛 sin 𝜃 = m
𝑟
𝑚𝑔
cos 𝜃
𝑣2
⇒ tan 𝜃 =
𝑟𝑔
for given speed v and radius r
of the turn we get angle of the bank
independent of mass m
for given θ and r:
v2
θ = arctan
= 26.70
rg
if v < rg tan θ the car would slide down a frictionless banked curve
if v >
rg tan θ the car would slide off the top
ROLLER COASTER
Top of a Hill
mv2
mg - Fn =
R
mv2
 Fn = mg R
minimum speed required to stay alive is
mathematically given with Fn = 0 (the rider
just barely looses contact with the seat
v T = Rg
Bottom of a Valley
mv2
Fn - mg =
R
Upside-down at the Top of a loop
𝑚𝑣 2
𝑚𝑣 2
𝐹𝑛 + 𝑚𝑔 =
 𝐹𝑛 =
− 𝑚𝑔
𝑅
𝑅
Fn = 0 danger when feeling weightless
min speed required to stay alive
and not to fall down:
mv2
 Fn = mg +
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.
v T = Rg
AIRPLANE MAKING A TURN - BANKING
As the plane banks (rolls), the lift vector
begins to have a horizontal component.
Lift force (L) generated by the airplane wings is not
equal to mg anymore, but greater.
𝐿 cos 𝜃 = 𝑚𝑔
𝑣2
𝐿 sin 𝜃 = 𝑚
𝑅
𝑣2
𝑅=
𝑔 tan 𝜃
 F = kx (k = CONST).
 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.
kx = F  k = F / x = 18 / 0.010 = 1800 Nm-1.
FC = kx = 1800( 0.265 – 0.250 ) = 27 N.
FC = v 2/ r  v 2 = r FC = 0.265(27) = 7.155
v = 2.7 ms-1.
Use v = r  ( = CONST).
Use a = r  2 ( = CONST).
At Q
At P
r = R r = 2R
v = R v = 2R = 2v
a = R  2 a = 2R 2 = 2a
Objects moving in uniform circular motion feel a centripetal (centerseeking) force.
Data booklet reference:
𝑀𝑚
𝐹=𝐺 2
𝑟
𝐹
𝑔=
𝑚
𝑀
𝐹=𝐺 2
𝑟
 r is the distance between the centers of the masses.
m1
F12
r
F21
m2
FYI The radius of each
mass is immaterial.
EXAMPLE: The earth has a mass of M = 5.981024 kg and the moon has a
mass of m = 7.361022 kg. The mean distance between the earth and the
moon is 3.82108 m. What is the gravitational force between them?
SOLUTION: F = GMm / r 2.
F = (6.67×10−11)(5.981024 )(7.361022 ) / (3.82108)2
F = 2.011020 N.
What is the speed of the moon in its orbit about earth?
SOLUTION: FC = FG = mv 2 / r .
2.011020 = ( 7.361022 ) v 2 / 3.82108
v = 1.02103 ms-1.
What is the period of the moon (in days) in its orbit about earth?
SOLUTION: v = 2r / T.
T = 2r / v = 2( 3.82108 ) / 1.02103
= (2.35 106 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d.
PRACTICE: A 525-kg satellite is launched from the earth’s 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
M = 5.981024 kg
centers.
m = 7.361022 kg
SOLUTION:
gm
gM
Make a sketch.
d = 3.82108 m
Note that r = d / 2 = 3.82108 / 2 = 1.91 108 m.
gm = Gm / r 2
gm = (6.67×10-11)(7.361022)/(1.91108)2 = 1.3510-4 N.
gM = GM / r 2
gM = (6.67×10-11)(5.981024)/(1.91108)2 = 1.0910-2 N.
Finally, g = gM – gm = 1.0810 -2 N,→.
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 earth’s surface at all times, if desired.
Find the necessary orbital radius, and express it in terms of earth radii. RE =
6.37106 m.
SOLUTION:
𝑇 = (24 ℎ)(3600 𝑠 ℎ−1 ) = 86400 𝑠
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
2𝜋𝑟
𝑇
2
=𝐺

𝑣2
𝑀
=𝐺
𝑟
𝑀
𝑟
−115.981024
𝐺𝑀
6.6710
𝑟3 = 2 𝑇2 ⇒ 𝑟3 =
(86400)2
2
4𝜋
4𝜋
𝑟 = 42250474 𝑚 = 6.63 𝑅𝐸
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
2𝜋𝑟
𝑇
2
=𝐺
𝑀
𝑟
 𝑣2 = 𝐺
⇒ 𝑟3 =
𝑀
𝑟
𝐺𝑀 2
𝑇
4𝜋 2
Kepler’s third law originally said that the square of the period was proportional
to the cube of the radius – and nothing at all about what the constant of
proportionality was. Newton’s law of gravitation was needed for that!
𝑚𝑣 2
𝑀𝑚
▪ 𝐹𝑐 =
=𝐺 2
𝑟
𝑟
▪
2𝜋𝑟
𝑇
2
=𝐺
 𝑣2 = 𝐺
𝑀
𝑟
𝑀
𝑟
2
2
4𝜋
4𝜋
▪ 𝑇2 =
𝑟3 ⇒ 𝑇 =
𝐺𝑀
𝐺𝑀
3/2
𝑟 3/2
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
 𝑟=𝐺
𝑀
𝐺 2
𝑟𝑋
𝑣𝑥 𝑣𝑌2
=
=
=
𝑟𝑦 𝐺 𝑀 𝑣𝑥2
𝑣𝑌2
𝑟𝑋
𝑟𝑦
3
= 64 ⇒
𝑀
𝑣2
2𝜋𝑟𝑌
𝑇𝑌
2𝜋𝑟𝑋
𝑇𝑋
𝑟𝑋
=4
𝑟𝑦
2
𝑇𝑋 𝑟𝑌
=
𝑇𝑌 𝑟𝑋
2
8𝑟𝑌
=
𝑟𝑋
2
Since the satellite is in circular orbit FC = mv 2/ r.
Since the satellite’s weight is holding it in orbit, FC = mg.
Thus mv 2/ r = mg.
Finally g = v 2/ r.
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.
(a)
State the name of the force that provides the centripetal force
for the motion of the stars.
 It is the gravitational force.
(b) By considering the force acting on one of the stars, deduce that
the orbital period T is given by the expression
 M1 experiences FC = M1v12/ R1.
 Since v1 = 2R1/ T, then v12 = 42R12/ T 2.
 Thus FC = FG  M1v12/ R1 = GM1M2 / (R1+R2) 2.
M1(42R12/ T 2) / R1 = GM1M2 / (R1+R2) 2
42R1(R1+R2) 2 = GM2T 2

T2
42
=
R (R +R ) 2
GM2 1 1 2
Note that FG = GM1M2 / (R1+R2) 2.
R
M1 1
R2
P
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.
 From (b)
T 2 = (42 / GM2)R1(R1+R2) 2.
 From symmetry
T 2 = (42/ GM1)R2(R1+R2) 2.
(42/ GM2)R1(R1+R2) 2 = (42/ GM1)R2(R1+R2) 2
(1 / M2)R1 = (1 / M1)R2
M1 / M2 = R2 / R1
Since R2 > R1, we see that M1 > M2.
R1
M1
R2
P
M2