Mr5–Mr8 - Varying speed and vertical circles

Mr5–Mr8 - Varying speed and vertical circles

A changing speed adds tangential acceleration to the inward radial component. Mr5 and Mr6 separate those jobs; Mr7 combines forces and energy in vertical circles; Mr8 checks when a string or surface can no longer maintain the circle and follows the motion after departure.

Turning and speeding up are different components

Can a particle leave a circular path while it still has plenty of speed? Yes. Its speed may be large enough to keep moving, yet the string or surface may be unable to supply the force that would keep it on that particular circle.

First separate changes of direction from changes of speed. A particle on a 0.5 m0.5\,\mathrm{m} radius circle with angular position θ=t2\theta=t^2 has θ˙=2t\dot{\theta}=2t: its speed increases, so its acceleration cannot be wholly radial.

You already know

For fixed radius, v=rθ˙v=r\dot{\theta} and the inward acceleration is rθ˙ 2=v2/rr\dot{\theta}^{\,2}=v^2/r. Energy conservation connects speed to vertical height when gravity is the only force doing work. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Choose the positive tangent to match increasing θ\theta. Differentiating the signed tangential velocity gives atan=rθ¨a_{\mathrm{tan}}=r\ddot{\theta}. This component changes the speed; the perpendicular inward component changes the velocity's direction.

Acceleration on a fixed circle

With θ\theta in radians and the positive tangent in the sense of increasing θ\theta,

ain=rθ˙ 2,atan=rθ¨a_{\mathrm{in}}=r\dot{\theta}^{\,2},\qquad a_{\mathrm{tan}}=r\ddot{\theta}

The inward component is non-negative, while the tangential component is signed.

In the formula booklet.

Newton's second law applies separately in those perpendicular directions:

∑Fin=mrθ˙ 2,∑Ftan=mrθ¨\sum F_{\mathrm{in}}=mr\dot{\theta}^{\,2},\qquad \sum F_{\mathrm{tan}}=mr\ddot{\theta}

The acceleration magnitude is ain2+atan2\sqrt{a_{\mathrm{in}}^2+a_{\mathrm{tan}}^2}. These force and component-combination equations are not in the formula booklet: learn them. When atan≠0a_{\mathrm{tan}}\ne0, the whole acceleration does not point at the centre.

Worked example 1: a prescribed angular position

Question. A 2 kg2\,\mathrm{kg} particle follows a horizontal circle of radius 0.5 m0.5\,\mathrm{m}. Its angular position is θ=t2\theta=t^2 radians, with tt in seconds. At t=2t=2, find the radial and tangential acceleration components, the magnitude of acceleration and the corresponding horizontal resultant-force components.

Plan. Differentiate θ\theta twice, then use each derivative in its own component formula.

θ˙=2t,θ¨=2differentiate in timeθ˙(2)=4 rad s−1evaluate at the instantain=0.5(42)=8 m s−2turning componentatan=0.5(2)=1 m s−2speed-changing component∣a∣=82+12=65=8.06 m s−2perpendicular componentsFin=2(8)=16 N,Ftan=2(1)=2 NNewton’s second law\begin{aligned} \dot{\theta}&=2t,\qquad\ddot{\theta}=2&&\text{differentiate in time}\\ \dot{\theta}(2)&=4\,\mathrm{rad\,s^{-1}}&&\text{evaluate at the instant}\\ a_{\mathrm{in}}&=0.5(4^2)=8\,\mathrm{m\,s^{-2}}&&\text{turning component}\\ a_{\mathrm{tan}}&=0.5(2)=1\,\mathrm{m\,s^{-2}}&&\text{speed-changing component}\\ |\mathbf a|&=\sqrt{8^2+1^2}=\sqrt{65}=8.06\,\mathrm{m\,s^{-2}}&&\text{perpendicular components}\\ F_{\mathrm{in}}&=2(8)=16\,\mathrm{N},\quad F_{\mathrm{tan}}=2(1)=2\,\mathrm{N}&&\text{Newton's second law} \end{aligned}

Why this step? The first derivative is squared for radial acceleration; the second derivative is used directly for tangential acceleration. They describe different changes of velocity.

Answer. Acceleration components 8 m s−28\,\mathrm{m\,s^{-2}} inward and 1 m s−21\,\mathrm{m\,s^{-2}} along increasing θ\theta; magnitude 8.06 m s−28.06\,\mathrm{m\,s^{-2}}. Resultant-force components are 16 N16\,\mathrm{N} inward and 2 N2\,\mathrm{N} tangentially.

Check. v=0.5(2t)=tv=0.5(2t)=t, so dv/dt=1 m s−2dv/dt=1\,\mathrm{m\,s^{-2}}. At t=2t=2, v2/r=22/0.5=8 m s−2v^2/r=2^2/0.5=8\,\mathrm{m\,s^{-2}}.

Change only the angular-position law to θ=2t2\theta=2t^2. The angular acceleration doubles, but the radial acceleration at the same time quadruples.

Worked example 2: the two components scale differently

Question. For the same particle and circle, let θ=2t2\theta=2t^2. Find the same quantities at t=2t=2.

Plan. Differentiate the new angular law and retain the distinction between θ˙ 2\dot{\theta}^{\,2} and θ¨\ddot{\theta}.

θ˙=4t,θ¨=4new derivativesain=0.5(82)=32 m s−2evaluate at t=2atan=0.5(4)=2 m s−2tangential component∣a∣=322+22=1028=32.1 m s−2resultant magnitudeFin=64 N,Ftan=4 Nmultiply by the mass\begin{aligned} \dot{\theta}&=4t,\qquad\ddot{\theta}=4&&\text{new derivatives}\\ a_{\mathrm{in}}&=0.5(8^2)=32\,\mathrm{m\,s^{-2}}&&\text{evaluate at }t=2\\ a_{\mathrm{tan}}&=0.5(4)=2\,\mathrm{m\,s^{-2}}&&\text{tangential component}\\ |\mathbf a|&=\sqrt{32^2+2^2}=\sqrt{1028}=32.1\,\mathrm{m\,s^{-2}}&&\text{resultant magnitude}\\ F_{\mathrm{in}}&=64\,\mathrm{N},\quad F_{\mathrm{tan}}=4\,\mathrm{N}&&\text{multiply by the mass} \end{aligned}

Why this step? Squaring θ¨\ddot{\theta}, or omitting the square on θ˙\dot{\theta}, confuses changing speed with changing direction.

Answer. Acceleration components 32 m s−232\,\mathrm{m\,s^{-2}} inward and 2 m s−22\,\mathrm{m\,s^{-2}} along increasing θ\theta, with magnitude 32.1 m s−232.1\,\mathrm{m\,s^{-2}}. The force components are 64 N64\,\mathrm{N} and 4 N4\,\mathrm{N}, respectively.

Check. v=2tv=2t, so at t=2t=2, v=4v=4 and dv/dt=2dv/dt=2, reproducing both components.

The force components describe the resultant of actual interactions. A radial guide force cannot, by itself, produce the changing speed in these examples; a tangential force is needed too.

Watch out

Wrong: call v2/rv^2/r “the acceleration” when the speed is changing. Right: it is the inward component. Find the tangential component as well before giving a magnitude or a direction.

The angular law now changes its coefficient once more.

Energy gives speed; force equations give acceleration and contact

For a fixed smooth vertical circle, weight changes the speed while the radial constraint turns the velocity. The constraint force does no work because it is perpendicular to the motion.

Measure θ\theta from the downward vertical, increasing as the particle rises on the right-hand side. Its height above the lowest point is r(1−cos⁡θ)r(1-\cos\theta). The inward direction is towards the centre; the positive tangent points in the sense of increasing θ\theta.

[DIAGRAM: asset_name: Forces in vertical circular motion and surface contact; asset_slug: h645_b073_constraints; caption: Fix the angle convention before resolving; an interior surface pushes inward while an exterior surface pushes outward; description: The left panel shows a particle on the lower right of a vertical circle, angle theta measured from downward vertical, height r times one minus cos theta, weight down and interior reaction inward. A small tangent arrow points up-right. The right panel shows a particle on the upper exterior of a sphere, angle phi measured from upward vertical, reaction outward, weight down and a downward-right tangent velocity. Each panel names its angle convention explicitly.]
Fix the angle convention before resolving

A smooth vertical circle from the lowest point

Let uu be speed at the lowest point of a fixed smooth circle of radius rr. While the circular constraint is maintained and gravity is the only force doing work,

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

If SS denotes a signed inward constraint force,

S−mgcos⁡θ=mv2r,mrθ¨=−mgsin⁡θS-mg\cos\theta=\frac{mv^2}{r},\qquad mr\ddot{\theta}=-mg\sin\theta

For an interior surface, S=R≥0S=R\ge0; for a taut string, S=T≥0S=T\ge0.

Not in the formula booklet: learn it.

The tangential equation is rθ¨=−gsin⁡θr\ddot{\theta}=-g\sin\theta. It is not exact SHM, because it involves sin⁡θ\sin\theta, rather than θ\theta. A bead threaded on a complete wire can have a constraint force in either radial sense, so its signed SS may be negative; it does not detach just because the chosen sign changes.

Worked example 3: motion on an interior surface

Question. A 0.2 kg0.2\,\mathrm{kg} particle moves in a vertical plane on the inside of a fixed smooth sphere of radius r=0.5 mr=0.5\,\mathrm{m}. Its speed at the lowest point is u=2gru=\sqrt{2gr}. When θ=60∘\theta=60^\circ, find its speed, normal reaction, tangential acceleration and vertical acceleration.

Plan. Use energy for speed, then resolve inward and along the tangent. Finally project both acceleration components vertically.

v2=2gr−2gr(1−cos⁡60∘)=gr=4.9energyv=4.9=2.21 m s−1positive speedR−mgcos⁡60∘=m(gr)rinward Newton’s lawR=32mg=2.94 Ncontact forceatan=−gsin⁡60∘=−32gtangential Newton’s laway=v2rcos⁡60∘+atansin⁡60∘upward components=12g−34g=−14gcombine both components\begin{aligned} v^2&=2gr-2gr(1-\cos60^\circ)=gr=4.9&&\text{energy}\\ v&=\sqrt{4.9}=2.21\,\mathrm{m\,s^{-1}}&&\text{positive speed}\\ R-mg\cos60^\circ&=\frac{m(gr)}r&&\text{inward Newton's law}\\ R&=\frac32mg=2.94\,\mathrm{N}&&\text{contact force}\\ a_{\mathrm{tan}}&=-g\sin60^\circ=-\frac{\sqrt3}{2}g&&\text{tangential Newton's law}\\ a_y&=\frac{v^2}{r}\cos60^\circ+a_{\mathrm{tan}}\sin60^\circ&&\text{upward components}\\ &=\frac12g-\frac34g=-\frac14g&&\text{combine both components} \end{aligned}

Why this step? The inward direction is inclined, so v2/rv^2/r is not the vertical acceleration. The tangential component has a vertical projection too.

Answer. Speed 2.21 m s−12.21\,\mathrm{m\,s^{-1}}, reaction 2.94 N2.94\,\mathrm{N}, tangential acceleration −8.49 m s−2-8.49\,\mathrm{m\,s^{-2}}, and vertical acceleration 2.45 m s−22.45\,\mathrm{m\,s^{-2}} downwards.

Check. R>0R>0, so contact is feasible at this point. Direct vertical Newton's law gives Rcos⁡60∘−mg=1.47−1.96=−0.49 NR\cos60^\circ-mg=1.47-1.96=-0.49\,\mathrm{N}, which divided by 0.20.2 gives −2.45 m s−2-2.45\,\mathrm{m\,s^{-2}}.

Change only the initial speed to 3gr\sqrt{3gr}. The geometry and tangential acceleration at the specified angle stay the same, but the radial component increases.

Worked example 4: a greater initial energy

Question. For the same particle and sphere, let the speed at the lowest point be 3gr\sqrt{3gr}. Find the same quantities at θ=60∘\theta=60^\circ.

Plan. Update the energy equation, then use the same force directions and projections.

v2=3gr−2gr(1−cos⁡60∘)=2gr=9.8energyv=9.8=3.13 m s−1speedR=m(2gr)r+mgcos⁡60∘radial equation=52mg=4.90 Nreactionatan=−32gsame angle and gravityay=2gcos⁡60∘−gsin⁡260∘=14gvertical component\begin{aligned} v^2&=3gr-2gr(1-\cos60^\circ)=2gr=9.8&&\text{energy}\\ v&=\sqrt{9.8}=3.13\,\mathrm{m\,s^{-1}}&&\text{speed}\\ R&=\frac{m(2gr)}r+mg\cos60^\circ&&\text{radial equation}\\ &=\frac52mg=4.90\,\mathrm{N}&&\text{reaction}\\ a_{\mathrm{tan}}&=-\frac{\sqrt3}{2}g&&\text{same angle and gravity}\\ a_y&=2g\cos60^\circ-g\sin^2 60^\circ=\frac14g&&\text{vertical component} \end{aligned}

Why this step? Gravity's tangential component depends on the position, not the speed. The normal force adjusts to supply the larger radial acceleration.

Answer. Speed 3.13 m s−13.13\,\mathrm{m\,s^{-1}}, reaction 4.90 N4.90\,\mathrm{N}, tangential acceleration −8.49 m s−2-8.49\,\mathrm{m\,s^{-2}}, and vertical acceleration 2.45 m s−22.45\,\mathrm{m\,s^{-2}} upwards.

Check. (Rcos⁡60∘−mg)/m=(2.45−1.96)/0.2=2.45 m s−2(R\cos60^\circ-mg)/m=(2.45-1.96)/0.2=2.45\,\mathrm{m\,s^{-2}}, matching the component calculation.

At the same point, changing the speed has even reversed the vertical acceleration. The direction of gravity alone does not determine the direction of the total acceleration when a constraint force acts.

Watch out

Wrong: R=mv2/rR=mv^2/r, ignoring the weight component, or ay=(v2/r)cos⁡θa_y=(v^2/r)\cos\theta, ignoring tangential acceleration. Right: resolve all actual forces, or project both acceleration components.

Reduce the initial speed until the particle reaches the specified point with zero speed.

Enough energy is not enough for a taut string

Energy can predict a speed at the top even when the string would already have gone slack. A light inextensible string pulls towards its pivot; it cannot push away from it.

Retain θ\theta from the downward vertical and a lowest-point speed uu. Combining the preceding energy and radial equations gives the tension at each point. Check this alongside the requirement that v2v^2 is non-negative.

String feasibility and a complete circle

For a particle on a light inextensible string of length rr, with only tension and weight acting,

v2=u2−2gr+2grcos⁡θv^2=u^2-2gr+2gr\cos\theta T=m(u2r−2g+3gcos⁡θ)T=m\left(\frac{u^2}{r}-2g+3g\cos\theta\right)

These equations apply while the string is taut. A circular motion requires T≥0T\ge0. At the top,

vtop2=u2−4gr,Ttop=m(u2r−5g)v_{\mathrm{top}}^2=u^2-4gr,\qquad T_{\mathrm{top}}=m\left(\frac{u^2}{r}-5g\right)

The least lowest-point speed for a complete vertical circle is 5gr\sqrt{5gr}, with zero tension at the top in the limiting case.

Not in the formula booklet: learn it.

On an uninterrupted ascent, both speed squared and tension decrease as θ\theta increases from 00 to π\pi. The first event matters: a turning point v=0v=0, slackness T=0T=0, or successful passage through the top. Testing only the top energy misses that ordering.

Worked example 5: completing the circle

Question. A particle on a light inextensible string of length rr starts at the lowest point with speed 6gr\sqrt{6gr}. Show that the string remains taut throughout a complete circle, and find the speed and tension at the top.

Plan. Find the minimum tension, which occurs at the top, and check that the top speed is real.

vtop2=6gr−4gr=2grrise through 2rTtop+mg=m(2gr)rinward is down at the topTtop=mg>0minimum tensionvtop=2grnon-zero top speed\begin{aligned} v_{\mathrm{top}}^2&=6gr-4gr=2gr&&\text{rise through }2r\\ T_{\mathrm{top}}+mg&=\frac{m(2gr)}r&&\text{inward is down at the top}\\ T_{\mathrm{top}}&=mg>0&&\text{minimum tension}\\ v_{\mathrm{top}}&=\sqrt{2gr}&&\text{non-zero top speed} \end{aligned}

Why this step? Positive tension at the highest point is sufficient here because the tension is greater everywhere else on the circle.

Answer. The string stays taut throughout. The top speed is 2gr\sqrt{2gr}, and the top tension is mgmg.

Check. At the bottom, T=6mg+mg=7mgT=6mg+mg=7mg; the general expression T=mg(4+3cos⁡θ)T=mg(4+3\cos\theta) varies between mgmg and 7mg7mg.

Change only the lowest-point speed to 4gr\sqrt{4gr}. Energy would predict zero speed at the top, but the string constraint must be checked before that point.

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