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A-Level · Further Mechanics 3 (Circular Motion)

FP.FM3 Further Mechanics: Circular Motion

AQA 7367 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Formulae and modelling assumptions used in this worksheet

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Unless a question states otherwise, take g = 9.8 m/s^2 and model each object as a particle. Strings are light and inextensible; rods are light and rigid; all surfaces, tracks, hemispheres and hinges are smooth (frictionless) unless friction is explicitly stated. For a particle moving on a circle of radius r with angular speed omega, the linear speed is v = r*omega, and the magnitude of the centripetal (radial) acceleration, directed towards the centre of the circle, is a = v^2/r = r*omega^2. For a conical pendulum and for a vehicle on a banked track, resolve the forces vertically and horizontally (or perpendicular and parallel to the surface) to form simultaneous equations linking the tension or normal reaction to the speed. For vertical circular motion, combine Newton's second law in the radial direction with conservation of energy; the critical condition for a particle on a string, or moving on the inside of a smooth circular track or the outside of a smooth sphere, occurs where the tension or normal reaction is zero, whereas a particle threaded on a rigid rod can maintain circular motion even when its speed is instantaneously zero.

At a funfair in Brighton, a big wheel of diameter 24 m rotates at a constant angular speed, completing one full revolution every 30 seconds. Sam sits in a pod fixed to the rim of the wheel.

Figure (to be drawn): Diagram: circular wheel of radius 12 m, centre O, with a pod (marked S for Sam) on the rim; an arrow shows the direction of rotation and the radius OS = 12 m is labelled.

(a)Show that the angular speed of the wheel is π/15 rad/s.(2)

(b)Find the speed of Sam's pod, giving your answer in m/s to 3 significant figures.(2)

(c)Calculate the magnitude of Sam's centripetal acceleration using the formula a = v²/r, and also using the formula a = r*ω², showing that the two methods give the same value to 3 significant figures.(4)

(Total for Question 1 is 8 marks)

In a school physics lab, Aisha sets up an experiment in which a particle P of mass 0.4 kg lies on a smooth horizontal table and is attached to a light inextensible string. The string passes through a smooth hole at the centre of the table and hangs vertically below the table, where it is attached to a particle Q of mass 1.2 kg. Q hangs in equilibrium, and P moves in a horizontal circle of radius 0.75 m at constant angular speed.

Figure (to be drawn): Diagram: horizontal table with particle P moving in a circle of radius 0.75 m around a central hole; the string passes vertically down through the hole to hanging particle Q, which is in equilibrium under gravity and the tension in the string.

(a)Find the tension in the string.(2)

(b)Find the angular speed of P.(3)

(c)Find the speed of P.(2)

(d)Find the time taken for P to complete one full revolution.(2)

(Total for Question 2 is 9 marks)

An engineering student, Kwame, models a fairground ride using a conical pendulum: a light inextensible string of length 0.9 m has one end fixed at a point O, and a small ball of mass 0.25 kg is attached to the other end. The ball moves in a horizontal circle at constant speed, with the string making a constant angle of 25 degrees with the vertical.

Figure (to be drawn): Diagram: fixed point O with a string of length 0.9 m making an angle of 25 degrees with the downward vertical; the ball is attached at the lower end of the string and moves in a horizontal circle of radius r below and to the side of O, with tension T acting along the string and weight mg acting vertically down on the ball.

(a)Show that the tension in the string is awrt 2.70 N.(3)

(b)Find the radius of the circle in which the ball moves.(2)

(c)Find the speed of the ball.(4)

(d)Find the time taken for the ball to complete one full revolution.(3)

(Total for Question 3 is 12 marks)

A go-kart track designer, Olivia, is designing a bend of radius of curvature 120 m, banked at an angle θ to the horizontal. The bend is designed so that a car travelling at a constant 108 km/h experiences no sideways friction force from the track surface.

Figure (to be drawn): Diagram: car on a banked circular bend of radius 120 m; the track surface is inclined at angle θ to the horizontal, with the normal reaction N perpendicular to the surface and the weight mg acting vertically down; for part (b) a friction force F is also shown acting along the slope.

(a)Show that tan(θ) = v²/(r*g), where v is the car's speed and r is the radius of curvature, and hence find θ to 1 decimal place. (Use v = 108 km/h = 30 m/s.)(4)

(b)The bend is banked at the angle θ found in part (a). Given that the coefficient of friction between a car's tyres and the track surface is 0.3, find the range of speeds at which a car can travel around the bend without sliding up or down the slope.(8)

(Total for Question 4 is 12 marks)

A particle P of mass 0.6 kg is attached to one end of a light inextensible string of length 1.5 m. The other end of the string is fixed at a point O, and P moves in a vertical circle of radius 1.5 m with centre O.

Figure (to be drawn): Diagram: vertical circle of radius 1.5 m, centre O; P is shown at the top of the circle with tension T (if any) acting downward along the string towards O and weight mg acting vertically down, and separately at the bottom of the circle with tension T acting upward along the string towards O and weight mg acting down.

(a)Given that P is on the point of the string becoming slack when P is at the top of the circle, show that the speed of P at the top of the circle, v_top, satisfies v_top² = g*r.(3)

(b)Find the speed of P at the top of the circle in this case.(2)

(c)Using conservation of energy, find the speed of P at the bottom of the circle in this case.(4)

(d)Find the tension in the string when P is at the bottom of the circle.(3)

(Total for Question 5 is 12 marks)

In a mechanics practical, Ravi compares two versions of the same apparatus. In each version, a small ball of mass 0.3 kg is attached to the end of a light connecting piece of length 0.8 m, the other end of which is fixed at a point O, and the ball is projected from the lowest point of a vertical circle of radius 0.8 m with speed u. In version 1, the connecting piece is a light rigid rod; in version 2, it is a light inextensible string.

Figure (to be drawn): Diagram: vertical circle of radius 0.8 m, centre O, with the ball shown at the lowest point moving with speed u and at the top of the circle moving with speed v_top; version 1 shows a rigid rod from O to the ball, version 2 shows a string from O to the ball.

(a)Explain why the minimum value of u needed for the ball to complete full circles is different for the rod (version 1) and the string (version 2).(2)

(b)For version 1 (the rigid rod), find the minimum speed of projection, u, at the lowest point for the ball to complete full circles.(4)

(c)For version 2 (the string), find the minimum speed of projection, u, at the lowest point for the ball to complete full circles.(4)

(d)State the ratio of the minimum value of u² for the rod to the minimum value of u² for the string, giving your answer in its simplest form.(1)

(Total for Question 6 is 11 marks)

At a funfair in Blackpool, a small 50p coin of mass 0.05 kg rests on a rotating horizontal disc at a distance of 0.4 m from the centre. The coefficient of friction between the coin and the disc is 0.25. The disc's angular speed is slowly increased from rest.

Figure (to be drawn): Diagram: horizontal disc viewed from above, rotating about a vertical axis through its centre; the coin is shown at radius 0.4 m from the centre, with the friction force F acting towards the centre providing the centripetal force.

(a)Find the maximum angular speed of the disc for which the coin does not slide.(4)

(b)Find the corresponding linear speed of the coin at this angular speed.(2)

(c)The coin is now moved to a distance of 0.6 m from the centre of the disc, which continues to rotate at the angular speed found in part (a). Determine, with justification, whether the coin remains in place at this new radius.(4)

(Total for Question 7 is 10 marks)

A smooth solid hemisphere of radius 0.6 m rests with its flat face on horizontal ground. A small particle is placed at the highest point of the hemisphere and given a negligible speed to disturb it from rest, after which it slides on the smooth outer curved surface of the hemisphere. Let θ be the angle between the radius from the centre of the hemisphere to the particle and the upward vertical, and let v be the particle's speed when it has turned through angle θ.

Figure (to be drawn): Diagram: hemisphere of radius 0.6 m with flat face on the ground and curved surface uppermost; the particle starts at the top and slides down the outside, shown at a general angle θ from the upward vertical through the centre, with weight mg acting vertically down and normal reaction N acting radially outward (away from the centre) on the particle.

(a)Using conservation of energy, show that v² = 2*g*r*(1-cos(θ)), where r is the radius of the hemisphere.(3)

(b)By applying Newton's second law in the radial direction, show that the particle leaves the surface of the hemisphere when cos(θ) = v²/(g*r).(3)

(c)Hence show that, at the point where the particle leaves the surface, cos(θ) = 2/3.(2)

(d)Find the value of θ, to 1 decimal place, at which the particle leaves the surface.(2)

(e)Find the speed of the particle at the point where it leaves the surface.(3)

(Total for Question 8 is 13 marks)

A wheel rotates at a constant angular speed of 8 rad/s.

(Total for Question 9 is 3 marks)

A small ball of mass 0.2 kg lies on a smooth horizontal table and is attached to a fixed point on the table by a light inextensible string of length 0.5 m. The ball moves in a horizontal circle of radius 0.5 m at a constant speed of 3 m/s.

(a)Find the angular speed of the ball.(2)

(b)Find the tension in the string.(2)

(Total for Question 10 is 4 marks)

During a practical, Sofia sets up a conical pendulum using a string of length 1.2 m, with the string making a constant angle of 30 degrees with the vertical. She measures the time for one complete revolution and records a period of 1.6 seconds.

Figure (to be drawn): Diagram: conical pendulum with string of length 1.2 m from a fixed point, making an angle of 30 degrees with the vertical, ball moving in a horizontal circle at the lower end of the string.

(Total for Question 11 is 4 marks)

Mark scheme · FP.FM3 Further Mechanics: Circular Motion

Question 1

(a) M1 uses ω = 2*π/T with T = 30 (s)
(a) A1 ω = 2*π/30 = π/15 rad/s, cso
(a) Answer: ω = π/15 rad/s (exact), = awrt 0.209 rad/s
(b) M1 uses v = r*ω with r = 12 (m), ft (a)
(b) A1 v = 2.51 m/s, awrt, cao
(b) Answer: v = awrt 2.51 m/s
(c) M1 substitutes into a = v²/r using their v from (b) and r = 12, ft
(c) A1 a = 0.526 m/s², awrt
(c) M1 substitutes into a = r*ω² using ω = π/15 and r = 12
(c) A1 a = 0.526 m/s² (awrt), matching the value found by the first method, cso
(c) Answer: a = awrt 0.526 m/s² by both methods

Question 2

(a) M1 resolves vertically for Q in equilibrium: T = m_Q * g
(a) A1 T = 11.76 N, cao
(a) Answer: T = 11.76 N
(b) M1 uses the centripetal force equation for P: T = m_P * r * ω², ft (a)
(b) A1 ω² = 11.76/(0.4*0.75) = 39.2, ft
(b) A1 ω = 6.26 rad/s, awrt, cao
(b) Answer: ω = awrt 6.26 rad/s
(c) M1 uses v = r*ω, ft (b)
(c) A1 v = 4.70 m/s, awrt, cao
(c) Answer: v = awrt 4.70 m/s
(d) M1 uses period = 2*π/ω, ft (b)
(d) A1 period = 1.00 s, awrt, cao
(d) Answer: period = awrt 1.00 s

Question 3

(a) M1 resolves vertically for the ball: T*cos(25 degrees) = m*g
(a) A1 correct unsimplified expression, T = (0.25*9.8)/cos(25 degrees)
(a) A1 T = 2.70 N, cso
(a) Answer: T = awrt 2.70 N (shown)
(b) M1 uses r = L*sin(25 degrees) with L = 0.9
(b) A1 r = 0.380 m, awrt, cao
(b) Answer: r = awrt 0.380 m
(c) M1 resolves horizontally for the ball: T*sin(25 degrees) = m*v²/r, ft (a) and (b)
(c) M1 rearranges for v², ft
(c) A1 v² = 1.74, awrt, ft
(c) A1 v = 1.32 m/s, awrt, cao
(c) Answer: v = awrt 1.32 m/s
(d) M1 uses period = 2*π*r/v, ft (b) and (c)
(d) M1 substitutes values correctly
(d) A1 period = 1.81 s, awrt, cao
(d) Answer: period = awrt 1.81 s

Question 4

(a) M1 resolves vertically: N*cos(θ) = m*g
(a) M1 resolves horizontally (centripetal direction): N*sin(θ) = m*v²/r
(a) A1 divides the two equations to eliminate N and m, giving tan(θ) = v²/(r*g), cso
(a) A1 θ = 37.4 degrees, awrt, cao
(a) Answer: tan(θ) = v²/(r*g) (shown); θ = awrt 37.4 degrees
(b) M1 for the maximum speed (car about to slide up), resolves vertically with friction acting down the slope: N*cos(θ) - μ*N*sin(θ) = m*g
(b) M1 resolves horizontally for the maximum speed: N*sin(θ) + μ*N*cos(θ) = m*v_max²/r
(b) A1 combines to give v_max² = r*g*(sin(θ)+μ*cos(θ))/(cos(θ)-μ*sin(θ)), oe (e.g. the equivalent tan(θ) form)
(b) A1 v_max = 40.3 m/s, awrt, cao
(b) M1 for the minimum speed (car about to slide down), resolves vertically with friction acting up the slope: N*cos(θ) + μ*N*sin(θ) = m*g
(b) M1 resolves horizontally for the minimum speed: N*sin(θ) - μ*N*cos(θ) = m*v_min²/r
(b) A1 combines to give v_min² = r*g*(sin(θ)-μ*cos(θ))/(cos(θ)+μ*sin(θ)), oe
(b) A1 v_min = 21.1 m/s, awrt, cao
(b) Answer: 21.1 m/s ≤ v ≤ 40.3 m/s (awrt)

Question 5

(a) M1 applies Newton's second law in the radial direction at the top of the circle: T + m*g = m*v_top²/r
(a) M1 uses the critical condition that the string is on the point of becoming slack, i.e. T = 0
(a) A1 v_top² = g*r, cso
(a) Answer: v_top² = g*r (shown)
(b) M1 substitutes r = 1.5 and g = 9.8 into v_top² = g*r, ft (a)
(b) A1 v_top = 3.83 m/s, awrt, cao
(b) Answer: v_top = awrt 3.83 m/s
(c) M1 uses conservation of energy between the bottom and top of the circle, with height gained 2*r
(c) M1 (1/2)*m*v_bottom² = (1/2)*m*v_top² + m*g*(2*r), ft (a)
(c) A1 v_bottom² = v_top² + 4*g*r = 5*g*r = 73.5, ft
(c) A1 v_bottom = 8.57 m/s, awrt, cao
(c) Answer: v_bottom = awrt 8.57 m/s
(d) M1 applies Newton's second law in the radial direction at the bottom of the circle: T - m*g = m*v_bottom²/r, ft
(d) M1 substitutes v_bottom²/r = 73.5/1.5 = 49 (= 5*g)
(d) A1 T = 35.28 N, cao
(d) Answer: T = 35.28 N

Question 6

(a) B1 a rigid rod can push as well as pull (it can act in compression as well as tension), so it can supply the centripetal force required at the top of the circle even when the ball's speed there is very small (or zero), oe
(a) B1 a string can only pull (it can only be in tension), so if the string is to remain taut the weight of the ball alone must not exceed the centripetal force required at the top; this requires a strictly positive minimum speed at the top, oe
(a) Answer: A rod can push or pull, so it can maintain the circular path even when the ball's speed at the top is zero; a string can only pull, so it would go slack unless the ball has enough speed at the top for its weight alone to provide the required centripetal force.
(b) M1 uses conservation of energy between the bottom and top of the circle: (1/2)*u² = (1/2)*v_top² + g*(2*r)
(b) M1 uses the critical condition for a rod, v_top = 0, since the rod can maintain the circular path down to zero speed at the top
(b) A1 u² = 4*g*r = 31.36, ft
(b) A1 u = 5.6 m/s, cao
(b) Answer: u = 5.6 m/s
(c) M1 uses the critical condition for a string, v_top² = g*r (tension zero at the top)
(c) M1 uses conservation of energy: u² = v_top² + 4*g*r
(c) A1 u² = 5*g*r = 39.2, ft
(c) A1 u = 6.26 m/s, awrt, cao
(c) Answer: u = awrt 6.26 m/s
(d) B1 ratio = 4:5, cao (independent of the values of g and r, since the ratio is 4*g*r : 5*g*r)
(d) Answer: 4:5

Question 7

(a) M1 identifies that at maximum angular speed, friction is at its limiting value: F = μ*m*g
(a) M1 sets the limiting friction equal to the required centripetal force: μ*m*g = m*r*ω²
(a) A1 ω² = μ*g/r = 6.125, cao
(a) A1 ω = 2.47 rad/s, awrt, cao
(a) Answer: ω = awrt 2.47 rad/s
(b) M1 uses v = r*ω, ft (a)
(b) A1 v = 0.990 m/s, awrt, cao
(b) Answer: v = awrt 0.990 m/s
(c) M1 finds the friction force required to keep the coin moving in a circle at radius 0.6 m at this angular speed: F_required = m*r*ω², ft
(c) A1 F_required = 0.184 N, awrt, ft
(c) M1 finds the maximum friction force available: F_max = μ*m*g
(c) A1 F_max = 0.1225 N, and since F_required > F_max, the disc cannot supply enough friction, so the coin slides (does not remain in place), cao with correct conclusion
(c) Answer: The coin slides, since the required friction (awrt 0.184 N) exceeds the maximum available friction (0.1225 N).

Question 8

(a) M1 finds the vertical height fallen when the particle has turned through angle θ: h = r*(1-cos(θ))
(a) M1 uses conservation of energy: (1/2)*m*v² - 0 = m*g*r*(1-cos(θ))
(a) A1 v² = 2*g*r*(1-cos(θ)), cso
(a) Answer: v² = 2*g*r*(1-cos(θ)) (shown)
(b) M1 applies Newton's second law towards the centre: m*g*cos(θ) - N = m*v²/r
(b) M1 uses the condition that the particle leaves the surface when the normal reaction N = 0
(b) A1 cos(θ) = v²/(g*r), cso
(b) Answer: cos(θ) = v²/(g*r) (shown)
(c) M1 substitutes v² = 2*g*r*(1-cos(θ)) from part (a) into cos(θ) = v²/(g*r) from part (b)
(c) A1 rearranges 3*cos(θ) = 2 to give cos(θ) = 2/3, cso
(c) Answer: cos(θ) = 2/3 (shown)
(d) M1 uses θ = arccos(2/3), ft (c)
(d) A1 θ = 48.2 degrees, awrt, cao
(d) Answer: θ = awrt 48.2 degrees
(e) M1 substitutes cos(θ) = 2/3 into v² = 2*g*r*(1-cos(θ)), ft (a) and (c)
(e) A1 v² = (2/3)*g*r = 3.92, cao
(e) A1 v = 1.98 m/s, awrt, cao
(e) Answer: v = awrt 1.98 m/s

Question 9

M1 converts angular speed to revolutions per second: revs per second = ω/(2*π)
M1 multiplies by 60 to convert to revolutions per minute
A1 76 revolutions per minute (to the nearest whole number), awrt
Answer: 76 revolutions per minute (to the nearest whole number)

Question 10

(a) M1 uses ω = v/r
(a) A1 ω = 6 rad/s, cao
(a) Answer: ω = 6 rad/s
(b) M1 uses T = m*v²/r
(b) A1 T = 3.6 N, cao
(b) Answer: T = 3.6 N

Question 11

M1 uses the conical pendulum result ω² = g/(L*cos(θ)) with L = 1.2 and θ = 30 degrees
M1 substitutes values correctly and finds the corresponding period, period = 2*π/ω
A1 period = 2.05 s, awrt
A1 compares this calculated period with the recorded value of 1.6 s and correctly concludes that Sofia's measurements are not consistent with a half-angle of 30 degrees, since 2.05 s does not equal 1.6 s, oe
Answer: The theoretical period for a half-angle of 30 degrees is awrt 2.05 s, which does not match Sofia's recorded period of 1.6 s, so the measurements are not consistent.

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