Circular Motion

Question 1

Which way does acceleration, velocity and force go in centripetal motion

Answer 1

Acceleration inwards to the circle
Force inwards to the circle
Velocity at right angle to the motion

Question 2

Angular velocity (ω) formula

Answer 2

ω=θ/t (units are rad s^-1)

Question 3

Period formula (T)

Answer 3

T=1/f

Question 4

Banked corners triangle

Answer 4

Hypotenuse = Fr (normal reaction force)
Opposite = Fr sin θ (horizontal component)
Adjacent = Fr cos θ (vertical component)
- vertical forces are balanced
- horizontal forces are unbalanced

Question 5

v formula for banked corners

Answer 5

v=√rg tanθ
minimum speed for a banked corner of angle θ (if there is no sideways friction)

Question 6

vertical circles (key differences from horizontal circles)

Answer 6

• the size of the normal reaction force acting on the object varies throughout the motion
• the speed of the object may not be constant as energy changes between Ek and Ep

Question 7

at the top of the loop the two forces acting on the car are:

Answer 7

the weight force (Fg)
the reaction force of the track pushing down on the car (Fr)

the total unbalanced force is Fg+Fr and is in the downwards direction. This unbalanced force provides the centripetal force for circular motion.

Question 8

the faster the car is travelling, the ___________ the centripetal force

Answer 8

larger (Fc is proportional to v^2). The weight force is constant so it is the reaction force that changes in size as the speed of the car changes.

Question 9

at the minimum speed, the reaction force at the top of the loop =

Answer 9

0

at the top of the loop, Fc=Fg
(mv^2)/r = mg
(v^2)/r = g
v = √rg

Question 10

why do you feel heavier at the bottom of the loop

Answer 10

Fc is directly upwards
Fc=Fr–Fg
Fr=Fc+Fg
Fr=(mv^2)/r+mg

as Fr > mg, you will feel heavier
- increase v, you will feel heavier
- decrease r, you will feel heavier
(Fc increases, so Fr increases)

Question 11

At the minimum speed for the loop how much heavier would you feel at the bottom?

Answer 11

Fr = (mv^2)/r + mg
as v = √rg,
Fr = (m(√rg)^2)/r +mg
= (mrg)/r + mg
= mg + mg
= 2mg
so you will feel twice as heavy

Question 12

mass on a string: tension (T) at top

Answer 12

at the top: T+mg = Fc
T+mg = (mv^2)/r – mg
T = (mv^2)/r – mg

Question 13

mass on a string: tension (T) at bottom

Answer 13

at the bottom: T–mg = Fc
T–mg = (mv^2)/r
T = (mv^2)/r + mg

Question 14

newton’s universal law of gravitation

Answer 14

“every single point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of their separation”

F = G (Mm/r^2)

Question 15

gravitational field definition

Answer 15

“a region of space where a mass experiences a force due to its mass”

Question 16

gravitational field strength (g)

Answer 16

“the force per unit mass experienced by a small point test mass placed in the field”
units are Nkg^–1

Question 17

what is a test mass

Answer 17

a test mass is one that has such a small mass that it does not change the gravitational field in which it is placed

Question 18

g=GM/r^2

Answer 18

g = F/m but F = GMm/r^2
so g = GMm/r^2/m
g = GM/r^2 (r is from the centre of the planet)

Question 19

orbital motion

Answer 19

for any orbiting object: centripetal force = gravitational force
Fc=Fg
(mv^2)/r = GMm/r^2
v^2=GM/r
v=√(GM/r)

this shows that the speed required to maintain an orbit depends on the radius of the orbit but NOT the mass of the satellite

Question 20

acceleration of a object orbiting

Answer 20

a = Fg/m (or Fc) (m = mass of object)
or
a = v^2/r

Question 21

why do astronauts feel weightless?

Answer 21

• weight depends on reaction force
• the ISS and astronaut are falling together so the reaction force = 0

Question 22

Kepler’s thrid law

Answer 22

derive:
r^3 = (GMT^2)/(4π^2)
G, M, 4 and π are all constants so r^3 and T^2
i.e. r^3/T^2 = constant (for anything orbiting a body of mass M)