Motion in a Circle

Question 1

Define angular displacement

Answer 1

The angular displacement is the angle ‘swept out’ (in radians) of an object moving in circular motion

Question 2

Define angular velocity

Answer 2

The angular displacement per second or ω=θ/t

Question 3

Give the equation and conditions for the small angle apporximation and conditions

Answer 3

θ≈sinθ when θ is small and measured in radians

Question 4

Give the equation for the linear speed of a point moving in circular motion

Answer 4

v = rω

where r is the radius, v is speed and ω is the angular velocity

Question 5

Give the time period (T) for one revolution of an object moving in circular motion and thus from that, the frequency

Answer 5

T = 2π / ω
f = 1/T = 1 / (2π/ω) = ω / 2π
ω = 2πf

Question 6

Give the equation for the acceleration of an object in circular motion and in which direction is it accelerating?

Answer 6

a = (-) v²/ r = (-) rω²

The acceleration of an object in circular motion is always towards the centre

Question 7

Define centripetal force and give the direction it acts in

Answer 7

Centripetal force is the resultant force of an object moving round a circle at constant speed.

It acts towards the centre of the circle

Question 8

Give the equation for centripetal force

Answer 8

F = mv²/ r = mω²r

Question 9

For a car of mass m travelling over a hill will with curvature r, relate its weight and supporting force of the road, S, to the centripetal force acting upon it

Answer 9

mg - S = mv² / r

Question 10

For a car of mass m travelling at speed v round a round-a-bout with radius r, give;

i) the equation for the force of friction
ii) the equation for the limiting force of friction
iii) the equation for the limiting force of friction in terms of the friction coefficient

Answer 10

i) Friction F = mv² / r
ii) limiting Friction F₀ = mv₀² / r

iii) F₀ = μmg
therefore μmg = mv₀² / r

Question 11

For a car of mass m, on a banked track of angle θ with a support force on each tyre of N₁ and N₂ respectivley, prove that if there is no sideways friction of the speed v is such that: v² = grtanθ

Answer 11

Horizontally, (N₁ + N₂) sinθ acts as the centripetal force therefore (N₁ + N₂) sinθ = mv² / r
Vertically, (N₁ + N₂) cosθ balances the weight mg, therefore (N₁ + N₂) cosθ = mg
tanθ = (N₁ + N₂) sinθ / (N₁ + N₂) cosθ = mv² / mgr
Therefore, there is no sideways friction when;
tanθ = mv² / mgr
or when v² = grtanθ

Question 12

For a rollercoaster car going down a circular hill with radius of curvature r, on big dipper ride what force acts as the centripetal force

Answer 12

The difference between the support force of the tract and your weight
S - mg = mv² / r
Therefore S = mg + mv² / r