R F ? mg @ @ @ @ @ I f θ &% '$
Let the angle between the radius to the point of contact with the step and the downward vertical be θ, as shown. At the point of contact, let the frictional force, which is tangent to the sphere, be f and the reaction R (along the radius of the sphere).
To investigate F , we take moments about a point cunningly chosen to eliminate other unknown forces. Since the lines of action both R and f pass through the point of contact of the sphere with the step, we can eliminate both forces by taking moments about this point:
F a(1 + cos θ) = mga sin θ ,
so
F = mg sin θ
1 + cos θ = mg tan
1 2θ .
This takes its maximum value when θ is largest, i.e. when the sphere just touches the horizontal ground. At this position, cos θ = 1
2and sin θ = 1 2 √ 3and Fmax= mg/ √ 3.
Taking moments about the centre of the sphere (to eliminate the normal reaction and weight) gives
F = f, and resolving forces parallel to the radius at the point of contact gives
R = F sin θ + mg cos θ. Now F sin θ = mg sin 2 θ 1 + cos θ = mg 1− cos2θ 1 + cos θ = mg(1− cos θ) so R = mg. We therefore need µmg > Fmax, as required.
Post-mortem
Should we have known that R = mg (in the last line of the solution)? It doesn’t look obvious from the diagram. However, if we follow the principle, used twice already, of taking moments about a point that eliminates unwanted forces, the result drops out. To relate mg and R directly, we need to eliminate
F and f . We therefore need to take moments about the intersection of the lines of action of these two forces, which is shown in the diagram (call it P ). Moments must balance about any point: it doesn’t matter whether the point is inside the body, or outside it as in the case of P . It is easy to see from the geometry (f and F act along tangents to the sphere) that the distances from P to the lines of action of
mgand R are equal, which means that the two forces are equal.
To be precise about the meaning of slow in this context, you have to compare the dynamic forces con- nected with the motion of the sphere with the static forces. The motion of the sphere is rotation about the fixed point of contact with the step. If we assume that the centre moves with constant speed, the only extra force due to the motion is an additional reaction at the point of contact with the step. This reaction is centrifugal in nature and so is roughly of the form mv2/a, which is to be small compared
2001 Paper I
A smooth cylinder with circular cross-section of radius a is held with its axis horizontal. A light elastic band of unstretched length 2πa and modulus of elasticity λ is wrapped round the circum- ference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass m is then attached to the rubber band at its lowest point and released from rest.
(i) Given that the particle falls to a distance 2a below the axis of the cylinder, but no further, show that
λ = 9πmg
(3√3− π)2 .
(ii) Given instead that the particle reaches its maximum speed at a distance 2a below the axis of the cylinder, find a similar expression for λ .
Comments
This question uses the most basic ideas in mechanics, such as conservation of energy. I included it in this selection of questions without realising that the properties of stretched strings are not in the syllabus (which is given in the appendix). However, I didn’t throw it out: it is a nice question and the only two things you need to know about stretched strings are, for a stretched string of natural (i.e. unstretched) length l and extended length l + x with modulus of elasticity λ:
(i) the potential energy stored in the stretched string is λx
2
2l ; (ii) the tension in the stretched string, by Hooke’s law, is λx
Solution to problem 64
The diagram below shows the system when the particle has fallen a distance a from its initial position.
2a
a a
b
A bit of geometry (including Pythagoras) on the above diagram shows that the length of the extended band is 2πa− 2a arccos1
2+ 2
√
3 aso the extension of the band is
−2a arccos1 2+ 2
√
3 a , i.e. 2a(√3−13π) .
(i) At the lowest point, the speed is zero. That suggests using an energy equation, which will involve speed and displacement (for the potential energy) but not acceleration.
Conserving energy (taking the initial potential energy to be zero), we find that when the particle has fallen a distance a and has speed v,
0 = 12mv2+12λ[2a( √
3−13π)]2
2πa − mga .
At the lowest point, v = 0, which gives the required answer.
(ii) Now we are interested in the point where the speed is greatest, i.e. where the acceleration is zero, so this time we should use the equation of motion of the particle.
The component of the tension T in the band in the vertical direction, acting upwards on the particle, when the particle has fallen a distance y is 2T cos θ where sin θ = a/(a + y). Applying Newton’s second law to the motion of the particle gives
m¨y = 2T cos θ− mg , (∗)
where, by Hooke’s law, T = λ× extension/2πa . At the maximum speed, ¨y = 0. This occurs at y = a, so (∗) becomes 0 = 2λ2a( √ 3−1 3π) 2πa √ 3 2 − mg . In this case λ = √ 3 πmg 3√3− π.
Post-mortem
The question was relatively simple because you only needed to evaluate energy or tension at very special points: maximum extension or maximum speed. You could of course have worked from the general equation of motion (∗), with cos θ =√2ay + y2/(a+y)and extension 2√2ay + y2−2a arcsin(a/(a+y)),
but you can’t solve the differential equation to find the general motion. It would be possible for small oscillations.
1987 Specimen Paper II
A tennis tournament is arranged for 2nplayers. It is organised as a knockout tournament, so that only the winners in any given round proceed to the next round. Opponents in each round except the final are drawn at random, and in any match either player has a probability 1
2 of winning.
Two players are chosen at random before the start of the first round. Find the probabilities that they play each other:
(i) in the first round;
(ii) in the final round;
(iii) in the tournament.
Comments
Note that the set-up is not the usual one for a tennis tournament, where the only random element is in the first round line-up. Two players cannot then meet in the final if they are in the same half of the draw. Part (i) is straightforward, but parts (ii) and (iii) need a bit of thought. There is a short way and a long way of tackling these parts, and both have merits.
It is a good plan to check your answers, if possible, by reference to simple special cases where you can see what the answers should be; n = 1 or n = 2, for example.
Interestingly, the answers are independent of the probability that the players have of winning a match; the 2s in the answers represent the number of players in each match rather than (the reciprocal of) the probability that each player has of winning a match. It also does not matter how the draw for each round is made. This is clear if you use the short method mentioned above.