Further Mechanics: Circular Motion
In A-level Further Mechanics, a particle moving on a circular path of radius r at angular speed omega (in radians per second) has speed v = r*omega and accelerates toward the centre of the circle at v^2/r (equivalently r*omega^2), even while its speed is constant, because its direction is always changing. This centripetal acceleration requires a resultant force of magnitude mv^2/r directed toward the centre, provided in practice by tension, friction, a normal reaction, or a combination. For a particle moving in a vertical circle on a string (or the inside of a smooth track), gravity's contribution to this centripetal force changes with position, and there is a critical minimum speed at the top of the circle, v^2 = gr, below which the string goes slack (or the particle leaves the track) before reaching the top.
Before you start
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Method
- Identify every force acting on the particle at the position being considered: weight mg always acts vertically down; tension, thrust or normal reaction acts along the radius (toward or away from the centre, depending on the situation).
- Resolve the forces toward the centre of the circle and apply Newton's second law for circular motion: resultant force toward the centre = mv^2/r (or m*r*omega^2).
- For horizontal circular motion at a fixed height (for example a conical pendulum), also resolve vertically and set the resultant to zero, since there is no vertical acceleration; solve the vertical and horizontal (centripetal) equations together.
- For vertical circular motion, the component of weight toward the centre changes with position, so write the centripetal equation separately at each position of interest, being careful with the sign of mg in each.
- The critical (minimum) condition for a particle on a string, or on the inside of a smooth track, to maintain circular motion at the top of a vertical circle is that the tension or normal reaction there is exactly zero, giving the minimum v_top^2 = gr; below this the string would go slack, or the particle would leave the track, before reaching the top.
- Combine the critical speed condition with conservation of energy (since tension and normal reaction do no work, as they always act perpendicular to the velocity) to relate the speed at the top to the speed at another point, accounting for the height change between them.
- State clearly whether a question is asking for the minimum (critical) case or a general speed above it, since 'just completes the circle' specifically means v_top^2 = gr, not merely v_top^2 >= gr.
Worked example
A particle of mass 0.5 kg is attached to one end of a light inextensible string of length 0.4 m; the other end is fixed at a point O. The particle moves in a vertical circle of radius 0.4 m, centre O. Find the minimum speed of the particle at the lowest point of the circle for it to complete a full circle without the string going slack. (Take g = 9.8 m/s^2.)
- At the top of the circle, the minimum condition for the string to remain taut is tension = 0, so the weight alone provides the centripetal force: mg = mv_top^2/r.
- Solve for v_top^2: v_top^2 = gr = 9.8 x 0.4 = 3.92 (m/s)^2.
- Use conservation of energy between the lowest and highest points; the particle rises a height of 2r = 0.8 m, and the tension does no work, since it always acts perpendicular to the velocity: (1/2)m*v_bottom^2 = (1/2)m*v_top^2 + mg(2r).
- The mass m cancels: v_bottom^2 = v_top^2 + 4gr = 3.92 + 4(9.8)(0.4) = 3.92 + 15.68 = 19.6.
- Solve: v_bottom = sqrt(19.6) = 7sqrt(10)/5 m/s exactly, approximately 4.43 m/s (3 s.f.).
Practice questions
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Q1State the formula for centripetal acceleration in terms of v and r, and in terms of omega and r.Show answer
Answer: a = v^2/r = r*omega^2.
Q2A particle moves in a horizontal circle of radius 2 m at angular speed 3 rad/s. Find its speed and its centripetal acceleration.Show answer
Answer: v = r*omega = 2 x 3 = 6 m/s; a = r*omega^2 = 2 x 9 = 18 m/s^2.
Q3A car of mass 900 kg travels round a flat circular track of radius 40 m at a constant 14 m/s. Find the friction force required to keep it on the circular path.Show answer
Answer: F = mv^2/r = 900 x 196/40 = 900 x 4.9 = 4410 N.
Q4State the condition, in terms of tension, for a string attached to a particle moving in a vertical circle to remain taut throughout the motion.Show answer
Answer: The tension T must satisfy T >= 0 at every point of the circle, since a string cannot push, only pull.
Q5A particle of mass 0.2 kg moves on the inside of a smooth vertical circular track of radius 0.5 m. Find the minimum speed at the top of the track for the particle to maintain contact with the track.Show answer
Answer: mg = mv^2/r at the critical case, so v^2 = gr = 9.8 x 0.5 = 4.9, giving v = sqrt(4.9) = 7sqrt(10)/10, approximately 2.21 m/s (3 s.f.).
Q6A particle of mass 0.3 kg on a light string of length 1 m moves as a conical pendulum in a horizontal circle of radius 0.6 m, with the string making a constant angle to the vertical such that cos(theta) = 0.8. Find the tension in the string and the speed of the particle (take g = 9.8 m/s^2).Show answer
Answer: Vertically, T cos(theta) = mg, so T(0.8) = 0.3(9.8) = 2.94, giving T = 3.675 N. Horizontally, T sin(theta) = mv^2/r with sin(theta)=0.6, so 3.675(0.6) = 0.3v^2/0.6, giving v^2 = 4.41 and v = 2.1 m/s.
Q7Explain why the tension in the string of a conical pendulum must be greater than the weight of the particle.Show answer
Answer: Since T cos(theta) = mg and cos(theta) < 1 for any string that is not vertical, T = mg/cos(theta) > mg, so the tension must exceed the weight to supply both the vertical support and the horizontal centripetal component.
Q8A particle moving in a vertical circle on the inside of a smooth track has speed exactly sqrt(gr) at the top. State the normal reaction from the track at this instant, and explain what would happen if the particle's speed at the top were slightly less than this value.Show answer
Answer: The normal reaction is 0 at this instant, since this is precisely the critical condition. If the speed were slightly less, the centripetal force needed at the top would be less than mg, which would require the track to pull the particle inward; since a normal reaction can only push, this is impossible, so the particle would leave the track before reaching the top.
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
A particle of mass 0.6 kg is attached to a light inextensible string of length 0.5 m, the other end fixed at a point O, and moves in a vertical circle of radius 0.5 m about O. The particle passes through the lowest point of the circle with speed 6 m/s. (a) Show that the string does not go slack, so the particle completes a full circle. (b) Find the tension in the string when the particle is at the lowest point. (c) Find the speed of the particle at the point level with O, where the string is horizontal. (Take g = 9.8 m/s^2.)
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A car of mass 1000 kg travels around an unbanked circular bend of radius 62.5 m at constant speed. The coefficient of friction between the car's tyres and the road is 0.4, and friction provides the full centripetal force needed. (a) Find the maximum speed at which the car can travel around the bend without skidding. (b) State one modelling assumption you have made about the car. (Take g = 9.8 m/s^2.)
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