3.3 - Vertical circular motion

3.3 - Vertical circular motion

A particle usually slows down as it rises around a vertical circle and speeds up as it falls. Use energy to find its speed, resolve the actual forces towards the centre, and decide whether a string or surface can maintain the assumed circular path. A zero-force calculation is useful only when the particle can physically reach that position.

Speed changes around a circle

Can a particle have enough energy to rise above the centre but still fail to stay on its circle? We need two different checks: energy determines its speed, while the forces determine whether the constraint can keep it on the path.

Start with energy. A light inextensible string has no elastic energy. Its tension is perpendicular to the particle's displacement while the string is taut, so tension does no work. The reaction from a smooth fixed circular track is also perpendicular to the motion. With no resistance, kinetic energy plus gravitational potential energy is conserved.

You already know

Kinetic energy is 12mv2\tfrac12 mv^2, and a rise through vertical height hh increases gravitational potential energy by mghmgh. Choose one height datum and use the actual vertical rise, not the distance along the path.

Let a circle have radius aa, and let θ\theta be the angle from the downward vertical to the radius through the particle. At the bottom θ=0\theta=0; at the top θ=π\theta=\pi. The downward projection of the radius is acos⁡θa\cos\theta, so the height above the bottom is a−acos⁡θa-a\cos\theta.

[DIAGRAM: asset_name: The height used in vertical-circle energy; asset_slug: cam9231-vertical-circle-height; caption: Measuring the angle from the downward vertical gives a rise of a times one minus cos theta; description: Exact vertical circle of radius a centred at O with bottom B, top A and a particle P in the upper right quadrant. The angle theta is measured from OB to OP. A horizontal level through P and a vertical height bracket from the bottom show h equals a minus a cos theta. The projection a cos theta is signed: it is negative at the displayed upper-half position. A side panel gives bottom, side and top heights zero, a and 2a.]
Measuring the angle from the downward vertical gives a rise of a times one minus cos theta

Energy from the bottom

If the speed at the bottom is uu, then, while the circular constraint remains valid,

12mu2=12mv2+mga(1−cos⁡θ),\frac12 mu^2=\frac12 mv^2+mga(1-\cos\theta),

so

v2=u2−2ga(1−cos⁡θ).v^2=u^2-2ga(1-\cos\theta).

This vertical-circle relation is not printed in the formula booklet: derive it from conservation of energy and the height geometry.

A negative value of v2v^2 means the particle cannot reach that position with the given energy. A nonnegative value alone does not guarantee that a string stays taut or that a one-sided track retains contact.

Worked example 1: rising from the bottom

Question. A bead threaded on a smooth fixed vertical circular wire of radius 1.2 m1.2\,\mathrm{m} passes the bottom with speed 8 m s−18\,\mathrm{m\,s^{-1}}. Find its speed when θ=60∘\theta=60^\circ and at the top. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}.

Plan. Calculate each height above the bottom, then subtract the corresponding gain in potential energy.

At 60∘60^\circ, the rise is 1.2(1−1/2)=0.6 m1.2(1-1/2)=0.6\,\mathrm{m}:

v2=64−2(10)(0.6)=52.v^2=64-2(10)(0.6)=52.

At the top, the rise is 2(1.2)=2.4 m2(1.2)=2.4\,\mathrm{m}:

vtop2=64−2(10)(2.4)=16.v_{\text{top}}^2=64-2(10)(2.4)=16.

Answer. The speeds are 213 m s−12\sqrt{13}\,\mathrm{m\,s^{-1}} at 60∘60^\circ and 4 m s−14\,\mathrm{m\,s^{-1}} at the top.

Check. Per unit mass, the initial energy is 32 J kg−132\,\mathrm{J\,kg^{-1}}. The two later totals are 26+6=3226+6=32 and 8+24=328+24=32, respectively. A threaded bead can be constrained from either side of the wire, so no one-sided contact condition has been assumed.

Now specify the speed at a higher point. The same energy principle works between any two positions, including on a descending arc.

Worked example 2: starting from a higher position

Question. A bead on the same smooth fixed circular wire has speed 4 m s−14\,\mathrm{m\,s^{-1}} at θ=120∘\theta=120^\circ. Find its speed at the bottom and when it next descends through θ=60∘\theta=60^\circ. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}.

Plan. Find the initial height, then use the energy released in each fall.

The initial height is 1.2(1−(−1/2))=1.8 m1.2(1-(-1/2))=1.8\,\mathrm{m}. Therefore the bottom speed uu satisfies

u2=16+2(10)(1.8)=52.u^2=16+2(10)(1.8)=52.

The 60∘60^\circ position is 0.6 m0.6\,\mathrm{m} above the bottom, so the fall from the given point is 1.2 m1.2\,\mathrm{m}:

v2=16+2(10)(1.2)=40.v^2=16+2(10)(1.2)=40.

Answer. The bottom speed is 213 m s−12\sqrt{13}\,\mathrm{m\,s^{-1}}, and the speed at 60∘60^\circ is 210 m s−12\sqrt{10}\,\mathrm{m\,s^{-1}}.

Check. Energy referred to the bottom is initially 16/2+10(1.8)=2616/2+10(1.8)=26 per unit mass. At 60∘60^\circ it is 40/2+10(0.6)=2640/2+10(0.6)=26. The direction of travel does not change this energy total.

The angle convention fixes the height formula. If an angle ϕ\phi is instead measured from the upward vertical, the height above the bottom is a(1+cos⁡ϕ)a(1+\cos\phi). Draw the geometry before choosing either expression.

Watch out

Using mgaθmga\theta for the gain in potential energy uses an arc length as a height. Use mga(1−cos⁡θ)mga(1-\cos\theta) when θ\theta is measured from the downward vertical.

Practise that height calculation without introducing a force equation yet.

Tension or normal reaction

Energy gives the speed, but it does not give the force maintaining the circle. At each position, resolve towards the centre and set that resultant equal to mv2/amv^2/a. Although the speed may be changing, this remains the radial component of Newton's second law.

For a string, tension points inwards. With θ\theta measured from the downward vertical, the inward component of weight is −mgcos⁡θ-mg\cos\theta. At the bottom weight points outwards; at the top it points inwards.

Radial force for a string or an inside track

While a string is taut,

T−mgcos⁡θ=mv2a.T-mg\cos\theta=\frac{mv^2}{a}.

For a particle in contact with the inside of a smooth circular track, replace TT by its inward normal reaction NN. These equations are not printed in the formula booklet: form them from the actual forces.

This sign convention produces the familiar extremes: T−mg=mv2/aT-mg=mv^2/a at the bottom and T+mg=mv2/aT+mg=mv^2/a at the top. It also gives T=mv2/aT=mv^2/a at the side, where gravity is tangential.

[DIAGRAM: asset_name: Inward resolution depends on the side of the surface; asset_slug: cam9231-radial-force-signs; caption: Weight may help the inward resultant, while a surface reaction can point either inwards or outwards; description: Two exact force diagrams. The left shows a string particle in the upper right of a vertical circle with angle theta from the downward vertical, inward tension T and downward weight mg. The right shows a particle on the outside of a smooth dome at angle phi from the upward vertical, with outward normal reaction N and downward weight mg. Text below gives T minus mg cos theta equals mv squared over a on the left and mg cos phi minus N equals mv squared over a on the right.]
Weight may help the inward resultant, while a surface reaction can point either inwards or outwards

The force diagram changes for the outside of a convex circular surface. Its reaction pushes away from the centre. If ϕ\phi is measured from the upward vertical, the inward equation is

mgcos⁡ϕ−N=mv2a.mg\cos\phi-N=\frac{mv^2}{a}.

Changing the angle convention does not change the forces; it changes which trigonometric component represents the weight.

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