# CIRCULAR MOTION - revision notes and calculations

Free essay example:

Circular Motion

Let a body moves in a circle as shown in the figure below:

First of all, we define several terms: -

(1) The angular position θ: - This is the angle moved from the initial position.

Unit: rad

(2) The angular velocity ϖ: - This is the angle rotated per unit time.

Unit: rad s-1. i.e. ϖ = Δθ/Δt

(3) Period T: - is the time required for the body to move one revolution. i.e. T = 2π/ϖ.

(4) Speed V: - is the distance moved per unit time. i.e. V = s/Δt, but in circle relationship S = r Δθ ,

Therefore, V = r Δθ/Δt = rω, or V = rω .

(5) Angular acceleration of a circle :-In circular motion, velocity always changes even the speed is keep constant, therefore, there should be an acceleration, a. This value of a can be deduced from the change of velocity.

Consider a particle moves in a circular path with constant speed V from A to B in a short time interval Δt as shown below: -

The change of velocity from A to B is denoted in the vector diagram as ΔV such that: ΔV = VB - VA

Since, the magnitude of VA = magnitude of VB = VΔθ (if Δθ is very small)

Therefore, a = change of velocity/time taken = ΔV/Δt = VΔθ/Δt = Vϖ

i.e. a = Vϖ = V2/r = rω2 (in magnitude)

Note: (I) For direction, if Δt is made so small that A and B almost coincide, then, the vector ΔV is perpendicular to VA (or VB) i.e. the direction is pointing to the centre O.

(II) The acceleration is pointing to the centre, which is therefore known as centripetal acceleration.

(III) According to Newton's 2nd law, acceleration is due to force, namely the resultant force

Therefore, F = ma = mV2/r = mrω2

This is known as the centripetal force

(IV) We must be clear of the cause -result relationship. i.e. we must have centripetal force first, this produces a centripetal acceleration which couples with a tangential velocity to start the circular motion.

i.e. Cause Result

Centripetal acceleration + imply Circular motion

Tangential Velocity

(V) In our usual calculation, we must consider the net force acting on the "free body diagram' of the rotating object being the centripetal force.

(6) Examples of Circular Motion

(i) Motion of Bicycle Rider Round Circular Track

The centripetal force is provided by frictional force F

Therefore, F = mV2/r (1)

This frictional force produce a clockwise moment about G, which must be balanced by the anticlockwise moment produced by R.

i.e. Fh = Ra = mga (2)

Therefore, a/h = F/mg = tanθ

Where θ is the angle of inclination to the vertical?

tanθ = mV2/rmg = V2/rg

i.e. tanθ ∝ V2 and tanθ ∝ 1/r

This means that the quicker the velocity to turn a sharp angle the larger is θ (bend dipper) if the frictional force is not enough to provide centripetal force, skidding occurs.

(ii) Motion of car (or Train) Round Circular Track

Suppose a car (or train) is moving with velocity V round a horizontal circular track of radius r, and let R1 and R2 be the respective normal reactions at wheels And B, while F1 and F2 are the corresponding frictional forces.

The centripetal force is provides by friction

∴ F1 + F2 = mV2/r (1)

Sum of all vertical forces equal zero

∴ R1 + R2 = mg (2)

For the car not to be overturn, sum of all moments about G = 0

∴(F1 + F2)h + R1a - R2a = 0 (3)

From these 3 equations:

∴ R1 = m/2(g - V2h/ra)

R2 = m/2(g + V2/ra)

Note: (1) R2 never vanishes since it always has a positive value, but if V2 = arg/h (i.e. g = V2h/ra), R1 = 0. That means the car is about to overturn outwards. In order to keep the car not overturn, V2 < arg/h

(2) For racing car, the value of V is inevitably large, and the corner is inevitably sharp, the design is such as to maintain V2 < arg/h

i.e. V2/r < ag/h ( V is large, and r is small )

Therefore, a must be large, and h must be small, and the tyres are as wide as possible in order to get enough friction for turning.

(iii) Motion of a car (or train) Round Banked Track

Suppose a car is moving round a banked track in a circular path of radius r, if the road is very slippy, the only forces acting on the wheels are the normal reactions R1 and R2. The centripetal force is provided by the component of (R1 + R2)

∴ (R1 + R2)Sinθ = mV2/r (1)

For vertical equilibrium to occur:

(R1 + R2)Cosθ = mg (2)

(1)/(2)⇒ tanθ = V2/rg

Note: - (1) the banking angle θ depends on V2 and r if V2 is large and r is small (sharp corner) then the value of θ should be large.

(2) The design of a racing track is not linear but curved upwards with θ increasing outwards because as the car is moving too fast and turnning too sharp, the car will skid outwards, the larger value of θ outwards provides a larger centripetal force.

(iv) Aircraft turning in the flight

When an aeroplane is about to turn about O, it must be tilted at an angle θ to the vertical because, the centripetal force is provided by the horizontal component of the uplifting force N.i.e.

Centripetal force⇒ Nsinθ = mV2/r (1)

Vertical equilibrium⇒ Ncosθ = mg (2)

(2)/(1)⇒ tanθ = V2/rg

Note:- the angle of tilt depends on the value V2/rg .

(v) Looping the loop

A bucket of water is looping in the loop, at its highest point A, the centripetal force is provided by mg, i.e.

mg = mV2/r

g = V2/r

or V = (gr)1/2

Note: (1) this is the condition for the waterto remain in the bucket.

(2) If the velocity is too small that

V < (gr)1/2, then mg > mV2/r i.e. the weight is not completely used in the centripetal force, the used part of the weight causes the water to leave the bucket.

(3) If V > (gr)1/2 i.e. mg < mV2/r, that means mg is not large enough to provide the required centripetal force, hence, water will press the bottom of the bucket. Then the bucket will provide a normal reaction N, so that

N + mg = mV2/r

Examples of essay type question(87 II A /1)

1. (a) Write down the Newton's law of motion. Apply to the situation where a body, initially at rest, is subject to a constant force, and describe the subsequent motion.

(b) Clearly distinguish the difference between "mass' and 'weight' of a body and explain why a passenger sometimes has the feeling of 'weightlessness' in a lift.

(c) Without giving any mathematical derivations, explain how it is possible for a body to move with constant speed in a horizontal circular path.

(d) Describe an experiment demonstrate the relation between the angular velocity of the body in (c) and the radius of the path, for a constant acting force, explain any source of error.

Solution: -

(a) The rate of change of momentum produce in a body is proportional to the resultant force acting on it and occurs in the direction of the force.

∴When a body, initially at rest is subject to a constant force, the body will move with constant acceleration in the direction of the force.

(b) The mass of a body 'm' is a measure of the inertia of the body and is a constant. The weight 'W' of a body is the force due to the gravitational pull, which is variable from place to place.

In a lift, a body will experience less reaction force if the lift is moving with downward acceleration. Our usual balance usually measures the normal reaction. If the downward acceleration reaches the value of 'g', the reaction becomes zero and the passenger feels weightlessness.

(c) A constant centripetal force acts towards the centre O of a circle provides an acceleration in this direction for a rotating body. The centripetal force changes only the direction of the motion but not it's magnitude, therefore, the speed V is constant.

(d)

(1) One student holds the glass tube vertically and whirls the bung around above his head, and the speed of the bung is increased until the marker is just below the tube.

(2) Another student times a certain number of revolutions of the bung, 50 say.

- By moving the marker, the length l of the string can be varied and the

relation between l and ω (angular velocity) can be obtained as follow:

T cosθ = mg

T sinθ = mrω2

Since, T and mg are constant, therefore, θ is also a constant.

i.e. ω ∝ r1/2

- By plotting a graph of ω against r1/2 a straight line should be obtained.

- The main source of error: there is friction between the string and the glass rod, which vary throughout the experiment.

94 I /1(a)

The figure below shows a car travelling over a hump, which is an arc of a vertical circle. Compared with travelling on a level road, would a passenger feel heavier, lighter or same as usual when the car passes the top of the hump? Explain your answer. (Assume that the passenger remains in contact with the seat).

Solution:-

The car and the passenger is undergoing circular motion, so there exists a centripetal acceleration a = V2/r. There are two forces acting on the passengers:(i) the weight of the passenger (ii) the normal reaction. The centripetal force is the resultant of the two forces, i.e. mg - R = mV2/r. This means that part of the weight is used as the centripetal force. Thus, the force of seat R is smaller than the weight and the passenger feels lighter

## Centrifuge

A Centrifuge consists of a test tube in a casing, which is freely hinged at a point B and rotates rapidly at angular velocity ω about the axle C.

As the speed of rotation increases, the tube rises and becomes horizontal. A liquid of higher density will 'sink' to the bottom of the tube and a liquid of lower density will 'float' to the top of the tube.

From the diagrams (a) below:

- Consider a part of the liquid extending from r to r+Δr from the axle, the centripetal force is provided by the pressure difference, ΔP, on the left and the right side of the liquid. Then the net inward force on this part of the liquid is

F = ΔPA

where A is the cross-sectional area of the tube.

- By Newton's 2nd Law:

F = ma ⇒ ΔPA = mrω2 = [ρΔrA]ω2r

where m = ρΔrA is the mass of this small volume of liquid. So,

ΔP = ρω2rΔr

This shows that when the tube is rotating the pressure increases towards the bottom of the tube.

Now consider diagram (b):

- Suppose there is a small block of another types of liquid, with density ρ' (ρ' < ρ), which extending from r to r + Δr. The net force acting on this block of liquid is calculated in the same way as above:

F = ΔPA' =[ρω2rΔr]A'

where A' is the cross-sectional area of this liquid block.

- Since the centripetal force needed to keep this liquid block in circular motion is

F' = m'ω2r = [ρ'ΔrA]ω2r

- So the excess force

Fe = F - F' =[ρ-ρ']ω2rΔrA'>0.

This excess force has 2 properties:

- push the liquid of smaller density (ρ' < ρ) to the top
- push the liquid of higher density (ρ' > ρ) to the bottom.

We know that the liquid of lower density will float naturally to the top and that of higher density will sink to the bottom, but the process is very slow. The excess force in the centrifuge will speed up this process.

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.