Data booklet reference:

• 𝑣 = 𝜔𝑟

• 𝑎 =𝑣 2

𝑟=

4 2𝑟

𝑇 2

• 𝐹 =𝑚𝑣 2

𝑟= 𝑚2𝑟

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.

𝑎𝑠 𝑎𝑐 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ∆ 𝑣, 𝑎𝑐 𝑝𝑜𝑖𝑛𝑡𝑠 𝑡𝑜𝑤𝑎𝑟𝑑 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛

2

c

va =

r

∆θ

∆r∆v ∆θ

r1 r2

v2

P

O

Qv1

v2

v1

_

∆v v1v2∆v = – v2 = v1 + ∆v

velocity – tangent to the circle

centripetal acceleration – points toward the center

always perpendicular to each other

2

c c

mv F = ma =

r

useful relations: - speed(linear) and angular speed

2πRv =

T

2πω =

T

constant speed:

period T: time required for one complete revolution (s)

speed v = distance/time

angular speed ω = angle swept/time

Translational variable Rotational variable Relationship

distance s angle swept (rad) s = r

tangential velocity

𝑣 =𝑠

𝑡

angular velocity (rad/s) 𝜔 =

𝜃

𝑡

v = r ω

tangentialacceleration a =

∆𝑣

𝑡

angular acceleration (rad/s2) 𝛼 =

∆𝜔

𝑡a = 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

μ g = 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.

𝐹𝑛sin 𝜃 = m𝑣2

𝑟⇒ tan𝜃 =

𝑣2

𝑟𝑔

for given speed v and radius r of the turn we get angle of the bank independent of mass m

20v

θ = arctan = 26.7rg

𝐹𝑛cos𝜃 = m𝑔 ⇒ 𝐹𝑛 =𝑚𝑔

cos𝜃

for given θ and r:

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 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.

𝐹𝑛 + 𝑚𝑔 =𝑚𝑣2

𝑅 𝐹𝑛 =

𝑚𝑣2

𝑅− 𝑚𝑔

AIRPLANE MAKING A TURN - BANKING

Lift force (L) generated by the airplane wings is not equal to mg anymore, but greater.

𝐿 cos 𝜃 = 𝑚𝑔 𝐿 sin 𝜃 = 𝑚𝑣2

𝑅

𝑅 =𝑣2

𝑔 tan𝜃

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 – 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

Objects moving in uniform circular motion feel a centripetal (center-seeking) force.

Data booklet reference:

𝐹 = 𝐺𝑀𝑚

𝑟2

𝑔 =𝐹

𝑚

𝐹 = 𝐺𝑀

𝑟2

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 = (6.67×10−11)(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 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 centers.

SOLUTION:

Make a sketch.

Note that r = d / 2 = 3.82108 / 2 = 1.91 108 m.

gm = Gm / r 2

gm = (6.67×10-11)(7.361022)/(1.91108)2 = 1.3510-4 N.

gM = GM / r 2

gM = (6.67×10-11)(5.981024)/(1.91108)2 = 1.0910-2 N.

Finally, g = gM – gm = 1.0810 -2 N,→.

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 earth’s surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. RE = 6.37106 m.

SOLUTION:

𝑇 = (24 ℎ)(3600 𝑠 ℎ−1 ) = 86400 𝑠

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺𝑀

𝑟

2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟

𝑟3 =𝐺𝑀

4𝜋2𝑇2 ⇒ 𝑟3 =

6.6710−115.981024

4𝜋2(86400)2

𝑟 = 42250474 𝑚 = 6.63 𝑅𝐸

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!

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺𝑀

𝑟

2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟⇒ 𝑟3 =

𝐺𝑀

4𝜋2 𝑇2

▪ 𝐹𝑐 =𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺𝑀

𝑟

▪2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟

▪ 𝑇2 =4𝜋2

𝐺𝑀𝑟3 ⇒ 𝑇 =

4𝜋2

𝐺𝑀

3/2

𝑟3/2

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑟 = 𝐺𝑀

𝑣2

𝑟𝑋𝑟𝑦

=

𝐺𝑀𝑣𝑥

2

𝐺𝑀𝑣𝑌

2

=𝑣𝑌

2

𝑣𝑥2 =

2𝜋𝑟𝑌𝑇𝑌

2𝜋𝑟𝑋𝑇𝑋

2

=𝑇𝑋𝑟𝑌𝑇𝑌𝑟𝑋

2

=8𝑟𝑌𝑟𝑋

2

𝑟𝑋𝑟𝑦

3

= 64 ⇒𝑟𝑋𝑟𝑦

= 4

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.

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