Mr4 - Horizontal circles at constant speed

Mr4 - Horizontal circles at constant speed

Mr4 brings circular acceleration into familiar force models: a car on a flat bend, a conical pendulum and a banked track. You will use zero vertical acceleration alongside a non-zero horizontal radial acceleration, choose the correct circle radius and test whether friction or contact can sustain the motion.

A level circle needs horizontal grip

Can an 800 kg800\,\mathrm{kg} car take a level circular bend of radius 40 m40\,\mathrm{m} at 14 m s−114\,\mathrm{m\,s^{-1}}, if the coefficient of friction is 0.60.6? The car is neither rising nor falling, but it is accelerating. One force equation will not describe both directions.

Model the car as a particle, with no aerodynamic forces and with horizontal tyre friction providing the inward resultant. At constant speed there is no tangential acceleration. In this ideal model the friction is sideways, opposing the tendency to slide outwards; it need not oppose the forward velocity.

You already know

For a circle, ain=v2/ra_{\mathrm{in}}=v^2/r. Newton's law uses the actual inward force components. At a rough contact, static friction satisfies ∣F∣≤μR|F|\le\mu R, with equality only at limiting grip. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Vertical balance gives the normal force. The radial equation gives the friction actually required. Compare that value with the available maximum only after finding it.

A level circular track

For a particle of mass mm travelling uniformly on a level circle of radius rr, with friction as the only horizontal force,

R=mg,F=mv2rR=mg,\qquad F=\frac{mv^2}{r}

The motion is possible within this friction model if

F≤μR,v2≤μgrF\le\mu R,\qquad v^2\le\mu gr

The limiting speed is μgr\sqrt{\mu gr}.

Not in the formula booklet: learn it.

Mass cancels from the limiting-speed condition. Doubling the car's modelled mass doubles both required sideways force and available grip, so it leaves the predicted limiting speed unchanged.

Worked example 1: enough friction

Question. An 800 kg800\,\mathrm{kg} car travels at 14 m s−114\,\mathrm{m\,s^{-1}} on a level circular track of radius 40 m40\,\mathrm{m}. The coefficient of friction is 0.60.6. Determine whether the motion is possible.

Plan. Find the required inward friction and compare it with μR\mu R.

R=800(9.8)=7840 Nvertical balanceF=800(142)40=3920 Nrequired inward forceFmax⁡=0.6(7840)=4704 Navailable grip3920<4704requirement is below the limit\begin{aligned} R&=800(9.8)=7840\,\mathrm{N}&&\text{vertical balance}\\ F&=\frac{800(14^2)}{40}=3920\,\mathrm{N}&&\text{required inward force}\\ F_{\max}&=0.6(7840)=4704\,\mathrm{N}&&\text{available grip}\\ 3920&<4704&&\text{requirement is below the limit} \end{aligned}

Why this step? Using F=μRF=\mu R immediately would describe impending sliding, which has not been stated.

Answer. The motion is possible in this model. The friction is 3920 N3920\,\mathrm{N} towards the centre.

Check. The minimum coefficient needed is 3920/7840=0.53920/7840=0.5, below 0.60.6.

Change only the speed to 16 m s−116\,\mathrm{m\,s^{-1}}. The available grip is unchanged, but the required force grows as the square of speed.

Worked example 2: the model cannot sustain the turn

Question. Determine whether the same car can follow the same circle at 16 m s−116\,\mathrm{m\,s^{-1}}.

Plan. Recalculate the radial force and compare it with the same maximum friction.

F=800(162)40=5120 Nforce required by the pathFmax⁡=4704 Nsame normal force and coefficient5120>4704insufficient grip\begin{aligned} F&=\frac{800(16^2)}{40}=5120\,\mathrm{N}&&\text{force required by the path}\\ F_{\max}&=4704\,\mathrm{N}&&\text{same normal force and coefficient}\\ 5120&>4704&&\text{insufficient grip} \end{aligned}

Why this step? Newton's law calculates what maintaining the specified circle would require. The contact inequality decides whether the surface can supply it.

Answer. The car cannot maintain this circular path at 16 m s−116\,\mathrm{m\,s^{-1}} under the stated model.

Check. The limiting speed is 0.6(9.8)(40)=15.3 m s−1\sqrt{0.6(9.8)(40)}=15.3\,\mathrm{m\,s^{-1}} to 33 significant figures, between the two trial speeds.

Once grip is lost, the original circular-motion equations no longer determine the path. They diagnose failure of the proposed motion, rather than calculating the later skid.

Watch out

Wrong: friction always has magnitude μR\mu R. Right: find the friction needed and test ∣F∣≤μR|F|\le\mu R. Equality is the limiting case.

The next speed lies between the two worked examples.

The circle radius is not the string length

In a conical pendulum, a particle attached to a fixed point by a light inextensible string travels uniformly in a horizontal circle. The string sweeps out a cone. Its length is a sloping distance; the circle's radius is the horizontal projection.

Let the string have length ℓ\ell and make an angle θ\theta, where 0<θ<90∘0<\theta<90^\circ, with the downward vertical. The circle radius is r=ℓsin⁡θr=\ell\sin\theta. Tension supplies both an upward component to balance weight and a horizontal component to turn the particle.

[DIAGRAM: asset_name: Conical pendulum geometry and actual forces; asset_slug: h645_b072_conical; caption: The horizontal circle has radius ell sin theta; tension's vertical and horizontal components have different jobs; description: A fixed pivot is above the centre of a horizontal circle. A sloping string of length ell makes angle theta with the downward vertical, ending at a particle. The horizontal radius is labelled r equals ell sin theta and the vertical depth ell cos theta. A separate force panel shows tension along the string towards the pivot, weight vertically down, and the inward horizontal direction.]
The horizontal circle has radius ell sin theta

Conical pendulum equations

For a particle of mass mm on a taut light string of length ℓ\ell, rotating uniformly at angle θ\theta to the vertical,

r=ℓsin⁡θ,Tcos⁡θ=mg,Tsin⁡θ=mrω2r=\ell\sin\theta,\qquad T\cos\theta=mg,\qquad T\sin\theta=mr\omega^2

Eliminating TT, for 0<θ<90∘0<\theta<90^\circ,

ω2=gℓcos⁡θ\omega^2=\frac{g}{\ell\cos\theta}

The speed is r∣ω∣r|\omega|, and the period is 2π/∣ω∣2\pi/|\omega|.

Not in the formula booklet: learn it.

The angular-speed equation follows by dividing the horizontal equation by the vertical equation, then substituting r=ℓsin⁡θr=\ell\sin\theta. It is specific to this one-string force model. If a second string or surface acts, return to the actual forces.

Worked example 3: a conical pendulum at a given angle

Question. A 0.3 kg0.3\,\mathrm{kg} particle is attached to a light string of length 1.25 m1.25\,\mathrm{m}. It rotates uniformly with the string at 60∘60^\circ to the downward vertical. Find the tension, circle radius, angular speed and period.

Plan. Use vertical balance for tension and the horizontal geometry for the circular motion.

Tcos⁡60∘=0.3(9.8)zero vertical accelerationT=5.88 Ndivide by cos⁡60∘r=1.25sin⁡60∘=1.083 mhorizontal radiusω2=9.81.25cos⁡60∘=15.68eliminate tension∣ω∣=3.96 rad s−1positive angular speedP=2π15.68=1.59 sone revolution\begin{aligned} T\cos60^\circ&=0.3(9.8)&&\text{zero vertical acceleration}\\ T&=5.88\,\mathrm{N}&&\text{divide by }\cos60^\circ\\ r&=1.25\sin60^\circ=1.083\,\mathrm{m}&&\text{horizontal radius}\\ \omega^2&=\frac{9.8}{1.25\cos60^\circ}=15.68&&\text{eliminate tension}\\ |\omega|&=3.96\,\mathrm{rad\,s^{-1}}&&\text{positive angular speed}\\ P&=\frac{2\pi}{\sqrt{15.68}}=1.59\,\mathrm{s}&&\text{one revolution} \end{aligned}

Why this step? The particle circles the vertical axis, not the pivot. Substituting 1.251.25 as the horizontal radius would give the wrong radial equation.

Answer. T=5.88 NT=5.88\,\mathrm{N}, r=1.08 mr=1.08\,\mathrm{m}, ∣ω∣=3.96 rad s−1|\omega|=3.96\,\mathrm{rad\,s^{-1}}, P=1.59 sP=1.59\,\mathrm{s}, to 33 significant figures where needed.

Check. Tsin⁡60∘=5.09 NT\sin60^\circ=5.09\,\mathrm{N}, while 0.3(1.25sin⁡60∘)(15.68)0.3(1.25\sin60^\circ)(15.68) gives the same inward force.

Change only the angle to 45∘45^\circ. The string length and particle mass remain the same.

Worked example 4: a narrower cone

Question. Find the tension, radius, angular speed and period for the same pendulum when the string is at 45∘45^\circ to the vertical.

Plan. Recalculate both string projections before applying the same force equations.

T=0.3(9.8)cos⁡45∘=4.16 Nvertical balancer=1.25sin⁡45∘=0.884 mnew circle radiusω2=9.81.25cos⁡45∘=11.087…radial equation∣ω∣=3.33 rad s−1square rootP=2π∣ω∣=1.89 suse unrounded speed\begin{aligned} T&=\frac{0.3(9.8)}{\cos45^\circ}=4.16\,\mathrm{N}&&\text{vertical balance}\\ r&=1.25\sin45^\circ=0.884\,\mathrm{m}&&\text{new circle radius}\\ \omega^2&=\frac{9.8}{1.25\cos45^\circ}=11.087\ldots&&\text{radial equation}\\ |\omega|&=3.33\,\mathrm{rad\,s^{-1}}&&\text{square root}\\ P&=\frac{2\pi}{|\omega|}=1.89\,\mathrm{s}&&\text{use unrounded speed} \end{aligned}

Why this step? Changing the angle changes the circle radius as well as the force components. Keeping the old radius would mix two different geometries.

Answer. T=4.16 NT=4.16\,\mathrm{N}, r=0.884 mr=0.884\,\mathrm{m}, ∣ω∣=3.33 rad s−1|\omega|=3.33\,\mathrm{rad\,s^{-1}}, P=1.89 sP=1.89\,\mathrm{s}, to 33 significant figures.

Check. At 45∘45^\circ, the tension's horizontal and vertical components are equal. The inward force is therefore mg=2.94 Nmg=2.94\,\mathrm{N}, agreeing with mrω2mr\omega^2.

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