Questions on motion in a vertical circle

Question 1

Explain why conservation of energy can be used for a particle on a string in a vertical circle, even though the tension varies.

Accepted Answer

The tension always acts along the string, which is perpendicular to the direction of motion at every instant, so it does no work. Only gravity does work, and it is conservative, so the total of kinetic and potential energy is constant. B1 for the tension doing no work, B1 for gravity being the only force that does work.

Question 2

A particle on a string of length 1.2 m is to complete a vertical circle. Find the least speed it can have at the top, taking g = 9.8 m/s².

Accepted Answer

At the critical speed the string is just taut, so T = 0 and the only radial force is the weight: mg = mv²/r. Hence v² = gr = 9.8 × 1.2 = 11.76, so v = 3.43 m/s. M1 for setting T = 0, A1 for mg = mv²/r, A1 for the least speed.

Question 3

A particle hangs from a fixed point by a light inextensible string of length 0.8 m and is projected horizontally from the lowest point with speed 4 m/s. Find its speed at the instant the string is horizontal, taking g = 9.8 m/s².

Accepted Answer

The particle rises a height equal to the radius, 0.8 m. Conserving energy: ½v² = ½(16) − 9.8 × 0.8, so v² = 16 − 15.68 = 0.32 and v = 0.566 m/s. M1 for the energy equation over a rise of r, A1 for v² = 0.32, A1 for 0.566 m/s. The particle only just gets there: 4² = 16 is barely more than 2gr = 15.68, so it reaches the horizontal with the string still taut and almost no speed left, and it will swing back rather than go over the top.

Question 4

A particle of mass 0.3 kg moves in a vertical circle on the end of a light inextensible string of length 0.5 m. Its speed at the lowest point is 3 m/s. Find the tension in the string at that point, taking g = 9.8 m/s².

Accepted Answer

At the lowest point the tension acts upward, towards the centre, and the weight downward, so the radial equation is T − mg = mv²/r. Then T = 0.3 × 9.8 + 0.3 × 9/0.5 = 2.94 + 5.4 = 8.34 N. M1 for T − mg = mv²/r, A1 for 5.4 N as the mv²/r term, A1 for 8.34 N. The tension exceeds the weight by the amount needed to bend the path upward; writing T = mv²/r alone leaves the weight out.

Question 5

A particle on a string of length 1.2 m is to complete a vertical circle. Taking g = 9.8 m/s², find the least speed it can have at the lowest point.

Accepted Answer

Conserving energy from the bottom to the top, a rise of 2r, gives ½u² = ½v² + g(2r), so u² = v² + 4gr. The critical case has v² = gr, so u² = 5gr = 5 × 9.8 × 1.2 = 58.8 and u = 7.67 m/s. M1 for conserving energy over a rise of 2r, A1 for u² = v² + 4gr, M1 for using v² = gr, A1 for the least speed.

Question 6

A bead slides from rest on the inside of a smooth circular track of radius 2 m, starting level with the centre. Find its speed at the lowest point and the reaction there, per unit mass.

Accepted Answer

It falls a height equal to the radius, so v² = 2gr = 2(9.8)(2) = 39.2 and v = 6.26 m/s. At the lowest point the radial equation is R − mg = mv²/r, so per unit mass R = 9.8 + 39.2/2 = 29.4 N per kg, three times g. M1 for the energy equation over a drop of r, A1 for the speed, M1 for the radial equation at the lowest point, A1 for the reaction.

Question 7

State the condition for a complete circle when the particle is attached to a light rod rather than a string, and explain the difference.

Accepted Answer

A rod can push outwards as well as pull inwards, so the constraint never fails and the particle only has to reach the top still moving. That needs v > 0 at the top and so u² > 4gr at the bottom. A string can only pull, so its tension must stay non-negative all the way round, forcing v² ≥ gr at the top and u² ≥ 5gr at the bottom. B1 for the rod being able to push, B1 for needing only v > 0 at the top, B1 for the string needing u² ≥ 5gr. A bead threaded on a wire or inside a smooth tube behaves like the rod, since the tube can push inwards or outwards.

Question 8

A particle hangs from a fixed point by a light inextensible string of length 0.9 m and is set moving horizontally from the lowest point with speed u. Taking g = 9.8 m/s², determine with justification what happens when u = 3 m/s, when u = 5 m/s and when u = 7 m/s.

Accepted Answer

Work out the two boundaries first: 2gr = 2(9.8)(0.9) = 17.64 and 5gr = 44.1, both in m²/s².
With u = 3, u² = 9 ≤ 17.64, so the particle never rises past the horizontal through the centre. The string stays taut throughout and the particle swings back and forth like a pendulum.
With u = 5, u² = 25 lies between the boundaries, so the string goes slack at a point above the horizontal. The particle leaves the circle there and moves as a projectile under gravity until the string tightens again.
With u = 7, u² = 49 ≥ 44.1, so the particle completes the circle with the string taut all the way round.
B1 for computing the boundaries 2gr and 5gr, B1 for the pendulum case with its reason, B1 for the slack string and the projectile motion, B1 for the completed circle.

Question 9

A particle of mass m on a light inextensible string of length r completes vertical circles, with speed u at the lowest point. Find the tension in the string at the lowest point and at the highest point, and show that the difference between them is 6mg, whatever the value of u.

Accepted Answer

At the lowest point the resultant upward force supplies the radial acceleration: T − mg = mu²/r, so the tension there is m(u²/r + g).
Energy from bottom to top, a rise of 2r: v² = u² − 4gr.
At the highest point both the tension and the weight act towards the centre: T + mg = mv²/r, so the tension there is m(v²/r − g) = m(u²/r − 5g).
Subtracting, the difference is m(u²/r + g) − m(u²/r − 5g) = 6mg. The u² terms cancel, so the difference is six weights however fast the circle is traversed.
M1 for the radial equation at the lowest point, A1 for m(u²/r + g), M1 for the energy equation over a rise of 2r, A1 for m(u²/r − 5g), A1 for the difference 6mg with the u² terms seen to cancel. The 2mg from the weight changing sides and the 4mg from the energy loss together make the 6mg.

Question 10

A particle of mass 0.2 kg hangs from a fixed point by a light inextensible string of length 1 m and is projected horizontally from the lowest point with speed 5 m/s. Find the tension in the string when it makes an angle of 60° with the downward vertical, taking g = 9.8 m/s².

Accepted Answer

At 60° from the downward vertical the particle has risen r(1 − cos 60°) = 0.5 m. Energy: v² = 25 − 2 × 9.8 × 0.5 = 15.2. Resolving along the string towards the centre, the weight has a component mg cos 60° away from the centre: T − mg cos 60° = mv²/r, so T = 0.2 × 15.2/1 + 0.2 × 9.8 × 0.5 = 3.04 + 0.98 = 4.02 N. M1 for the energy equation with the rise r(1 − cos θ), A1 for v² = 15.2, M1 for the radial equation with mg cos θ, A1 for 4.02 N.

Question 11

A small particle moves inside a smooth hollow tube bent into a vertical circle of radius 0.6 m. It passes the lowest point with speed 5 m/s. Show that the particle reaches the highest point, and find the force the tube exerts on it there, per unit mass, stating its direction. Take g = 9.8 m/s².

Accepted Answer

Energy from bottom to top, a rise of 2r = 1.2 m: v² = 25 − 4 × 9.8 × 0.6 = 25 − 23.52 = 1.48, which is positive, so the particle is still moving at the top and does reach it. A tube can push either way, so unlike a string it needs nothing more than v > 0. At the top the required acceleration towards the centre is v²/r = 1.48/0.6 = 2.47 m/s², which is less than g, so gravity alone would be too much: the tube must push outward (upward) with a force per unit mass of 9.8 − 2.47 = 7.33 N per kg. M1 for the energy equation, A1 for v² = 1.48 > 0 with the conclusion, M1 for the radial equation at the top, A1 for 7.33 N per kg outward. On a string this speed would not do, since v² = 1.48 is below gr = 5.88: the string would have gone slack on the way up.

Question 12

A particle slides from rest at the top of a smooth sphere of radius r. Show that it leaves the surface where the radius makes an angle θ with the upward vertical satisfying cos θ = 2/3, and find its speed at that instant in terms of g and r.

Accepted Answer

At angle θ the radial equation is mg cos θ − R = mv²/r. Energy from the top, having dropped r(1 − cos θ): v² = 2gr(1 − cos θ). The particle leaves when R = 0, so g cos θ = v²/r = 2g(1 − cos θ). Hence 3cos θ = 2 and cos θ = 2/3, that is θ = 48.2°.
Substituting back, v² = 2gr(1 − 2/3) = 2gr/3, so the speed is √(2gr/3).
M1 for the radial equation, A1 for mg cos θ − R = mv²/r, M1 for the energy equation, A1 for v² = 2gr(1 − cos θ), M1 for setting R = 0, A1 for cos θ = 2/3, A1 for the speed.
The result is independent of r and of the mass, so every smooth sphere sheds a particle at the same place.

Question 13

A particle is attached to a fixed point by a light inextensible string of length r and is projected horizontally from the lowest point with speed √(3.5gr). Show that the string goes slack when it makes an angle of 60° with the upward vertical, find the speed of the particle at that instant in terms of g and r, and state how the particle moves immediately afterwards.

Accepted Answer

Since 2gr < u² = 3.5gr < 5gr, the particle must leave the circle somewhere above the horizontal, so let the string make angle θ with the upward vertical there. Resolving towards the centre, the weight now has an inward component: T = mv²/r − mg cos θ.
Energy from the lowest point, a rise of r(1 + cos θ): v² = 3.5gr − 2gr(1 + cos θ).
The string goes slack when T = 0, which gives v² = gr cos θ. Substituting into the energy equation: gr cos θ = 3.5gr − 2gr − 2gr cos θ, so 3 cos θ = 1.5 and cos θ = ½, that is θ = 60°.
Then v² = gr cos 60° = gr/2, so the speed is √(gr/2).
With the string slack the only force is gravity, so the particle moves as a projectile from that point until the string snaps taut again.
M1 for the radial equation at a general angle, A1 for T = mv²/r − mg cos θ, M1 for the energy equation, A1 for v² = 3.5gr − 2gr(1 + cos θ), M1 for setting T = 0, A1 for cos θ = ½, A1 for the speed √(gr/2), B1 for the projectile statement.

Question 14

A particle is placed at the highest point of a fixed smooth sphere of radius r and given a horizontal speed u. Show that if u² ≥ gr the particle leaves the surface immediately, and that otherwise it leaves where the radius to it makes an angle θ with the upward vertical given by cos θ = (2 + u²/(gr))/3. Find θ when u² = gr/2.

Accepted Answer

At angle θ the particle has dropped r(1 − cos θ), so by energy v² = u² + 2gr(1 − cos θ). Resolving towards the centre, mg cos θ − R = mv²/r. The particle leaves when R = 0, that is g cos θ = v²/r = u²/r + 2g(1 − cos θ), so 3g cos θ = 2g + u²/r and cos θ = (2 + u²/(gr))/3.
At the top itself, θ = 0 and cos θ = 1, so the reaction is already zero when u² = gr; for u² > gr the formula would need cos θ > 1, which means no contact position exists and the particle leaves at once, since gravity cannot supply the acceleration u²/r the circle would demand.
With u² = gr/2: cos θ = (2 + ½)/3 = 5/6, so θ = 33.6°.
M1 for the energy equation, M1 for the radial equation, A1 for the general cos θ, B1 for the u² ≥ gr case with the reason, A1 for cos θ = 5/6, A1 for 33.6°. Placing u = 0 recovers cos θ = 2/3, the result for a particle that starts from rest, and a larger u always makes the particle leave sooner.

Question 15

A particle hangs from a fixed point by a light inextensible string of length r and is projected horizontally from the lowest point with speed u, where 2gr < u² < 5gr. Let θ be the angle the string makes with the upward vertical, measured from that upward vertical, so that θ = 180° at the lowest point. Find, in terms of u, g and r, the value of cos θ at which the string would go slack and the value at which the particle would come to rest, and prove that the string goes slack before the particle stops.

Accepted Answer

Measured from the upward vertical, the particle rises r(1 + cos θ) from the lowest point, so by energy v² = u² − 2gr(1 + cos θ). Resolving towards the centre, T + mg cos θ = mv²/r.
Slack: T = 0 gives v² = gr cos θ, so gr cos θ = u² − 2gr − 2gr cos θ and cos θ = (u² − 2gr)/(3gr).
Rest: v = 0 gives 2gr(1 + cos θ) = u², so cos θ = (u² − 2gr)/(2gr).
As the particle rises from the bottom, θ decreases from 180° and cos θ increases steadily from −1. Since u² − 2gr > 0, both values are positive, and the slack value has the larger denominator, so it is the smaller of the two and is reached first. The string therefore goes slack while the particle is still moving, and it leaves the circle as a projectile rather than stopping on it.
M1 for the energy equation with the rise r(1 + cos θ), M1 for the radial equation with T + mg cos θ, A1 for the slack value, A1 for the rest value, M1 for comparing the two with cos θ increasing on the way up, A1 for the conclusion. When u² = 5gr the slack value reaches 1, at the top, which is the complete-circle limit; when u² = 2gr both are zero, at the horizontal, which is the pendulum limit.