• Circular motion in a vertical plane

Consider a pail of water whirling at the end of a rope in a vertical circle. The tension T in the rope varies with the position of the pail.

The figure below shows the forces acting on the pail when the rope is at an angle of θ to the vertical.

From Newton’s 2^nd law: Resultant force F = ma

Net force towards the centre of the circle O is

T – mg cosθ

The net force provides the centripetal acceleration.

Therefore, T – mg cosθ = mv² / r

where v = speed of pail; r = radius of circle

Tension T = mv²/r + mg cosθ

Now, let’s consider how the tension changes as the position of the pail changes.

·         When the pail is at the lowest point (θ = 0),

Tension T = mv²/r + mg        (which is the maximum tension)

·         When the rope is horizontal (θ = π/2 (=90°) or θ = 3π/2 (=270°)), cosθ = 0 and thus

Tension T = mv²/r

·         When the pail is at the highest point (θ = π = 180°), cosθ = –1 and

Tension T = mv²/r – mg        (which is the minimum tension)

Now if the pail reaches the highest point (at a distance equal to the radius r form centre O of the circle), the rope must be taut (there should be a tension in the rope, which is not zero).

( T = mv²/r – mg )      > 0

mv²/r > mg

v > √(gr)

Thus, they velocity of the pail at the highest point must be greater than √(gr).

Now consider the forces acting on the water in the pail when it is at the highest point.

For the water not to fall out (that is, the water is in contact with the pail – there is a contact force on the pail), there should be a reaction R (from Newton’s 3^rd law) on the water by the base of the pail. The resultant force towards the centre of the circle is

F = R + mg = mv² / r

The centripetal force is provided by the reaction R and the weight mg.

R = mv²/r – mg

For the water not to fall, R > 0 (that is, there should be a reaction R on the water).

( R = mv²/r – mg )      > 0

mv² / r > mg

Therefore, water in the pail does not fall out because the centripetal force is greater than the weight of the water. (The weight mg of the water is less than the required force mv²/r towards the centre and so the water stays in. The reaction R of the bucket base on the water provides the rest of the force mv²/r.)

Remember that the weight acts vertically downwards (towards the ground). The centripetal force mv²/r acts towards the centre, causing the bucket to move in a circle.

On the other hand, if the pail is whirled slowly, then mg > mv²/r. The weight is greater than the centripetal force and the water falls out of the bucket. (Here, part of the weight provides the force mv²/r. The rest of the weight causes the water to accelerate downward and hence to leave the bucket.)

References:

1. “Pacific Physics A Level,” by POH LIONG YONG, pg 154 – 155 ‘Motion in a vertical circle’

2. “Advanced Level Physics,” by Nelkon & Parker, 7^th edition, pg 63 – 64

• Physics 9702 Doubts | Help Page 230

Question 1085: [Kinetic model of Matter]

The Brownian motion of smoke particles in air may be observed using the apparatus shown in Fig. 2.1.

Fig. 2.1

(a) Describe what is seen when viewing a smoke particle through the microscope.

(b) Suggest and explain what difference, if any, would be observed in the movement of smoke particles when larger smoke particles than those observed in (a) are viewed through the microscope.

Reference: Past Exam Paper – June 2005 Paper 2 Q2

Solution 1085:

(a) A speck of light {not the smoke particles themselves} that moves haphazardly/randomly/jerkily/etc. is seen

{We are asked to describe what is OBSERVED, not ‘why’ this is happening.}

(b) The randomness of collisions would be ‘averaged out’ {the motion would still be random, but slower}. So there would be less (haphazard) movement.

{There is now a greater surface area and, because of the random distribution of velocities of air molecules, the effects of collisions of the smoke particle with air molecules would tend to average out. This greater rate of collision would lead to a smaller, rather than a larger, randomness of collision and hence motion of the smoke particle.}

Question 1086: [Measurement]

The diagrams show digital voltmeter and analogue ammeter readings from a circuit in which electrical heating is occurring.

What is the electrical power of the heater?

A 0.53 W                    B 0.58 W                     C 530 W                      D 580 W

Reference: Past Exam Paper – June 2009 Paper 1 Q4

Solution 1086:

Answer: B.

Power = VI

The values should be given in SI units.

p.d. V = 1200 mV = 1.2 V

Current I = 0.48 A

Power = VI = 1.2 × 0.48 = 0.576 W = 0.58 W

Question 1087: [Work, Energy and Power]

(a) Explain the concept of work.

(b) A table tennis ball falls vertically through air. Fig. 8.1 shows the variation of the kinetic energy E_K of the ball with distance h fallen. The ball reaches the ground after falling through a distance h₀.

Fig. 8.1

(i) Describe the motion of the ball.

(ii) On Fig. 8.1, draw a line to show the variation with h of the gravitational potential energy E_P of the ball. At h = h₀, the potential energy is zero.

Reference: Past Exam Paper – November 2005 Paper 2 Q8

Solution 1087:

(a) Work is the product of force and the distance moved in the direction of the force.

(b)

(i) The ball falls from rest with a decreasing acceleration until it reaches a constant (terminal) speed.

(ii)

The graph is a straight line with a negative gradient.

It intercepts the y-axis at a point which is above the maximum value for E_k.

The gradient should be reasonable (same magnitude as that for E_k initially).

{Gravitational potential energy = mgx. That is, it depends linearly on height x (remember that the equation for a straight line is of the form y = mx + c), so graph is a straight line. But here, h is the distance fallen, not height. So, gradient is negative.

The straight line should be well above the point for maximum kinetic energy in order to reflect the energy expended in doing work to overcome the air resistance opposing the motion.}