IPhO 2002 · Theory problem 3

A Heavy Vehicle Moving on An Inclined Road

Download PDFIPhO 2002
Figure 1

Figure III-1: A simplified model of a heavy vehicle moving on an inclined road.

The above figure is a simplified model of a heavy vehicle (road roller) with one rear and one front cylinder as its wheels on an inclined road with inclination angle of \(\theta\) (labelled è in the figure) as shown in Figure III-1. Each of the two cylinders has a total mass \(M(m_2=m_3=M)\) and consists of a cylindrical shell of outer radius \(R_o\), inner radius \(R_i=0.8R_o\) and eight number of spokes with total mass \(0.2M\). The mass of the undercarriage supporting the vehicle’s body is negligible. The cylinder can be modeled as shown in Figure III-2. The vehicle is moving down the road under the influence of gravitational and frictional forces. The front and rear cylinder are positioned symmetrically with respect to the vehicle.

Figure 2

Figure III-2: A simplified model of the cylinders.

The static and kinetic friction coefficients between the cylinder and the road are \(\mu_s\) and \(\mu_k\) respectively. The body of the vehicle has a mass of \(5M\), length of \(L\) and thickness of \(t\). The distance between the front and the rear cylinder is \(2l\) while the distance from the center of cylinder to the base of the vehicle’s body is \(h\). Assume that the rolling friction between the cylinder and its axis is negligible.

Questions:

  1. Calculate the moment of inertia of either cylinder [1.5 pts].

  2. Draw all forces that act on the body, the front cylinder, and the rear one. Write down equations of motion for each part of them [2.5 pts].

  3. The vehicle is assummed to move from rest, then freely move under gravitational influence. State all the possible types of motion of the system and derive their accelerations in terms of the given physical quantities [4.0 pts].

  4. Assume that after the vehicle travels a distance \(d\) by pure rolling from rest the vehicle enters a section of the road with all the friction coefficients drop to smaller constant values \(\mu_s'\) and \(\mu_k'\) such that the two cylinders start to slide. Calculate the linear and angular velocities of each cylinder after the vehicle has traveled a total distance of \(s\) meters. Here we assume that \(d\) and \(s\) is much larger than the dimension of vehicle [2.0 pts]

Official English text of the problem.

Answers

Write \(\theta\) for the inclination angle and define \[ R=R_o,\qquad b=h+\frac t2,\qquad \kappa=\frac{262}{375},\qquad q=\frac{5b}{7l}. \] Subscripts \(r,f\) denote rear and front. Linear acceleration is positive downhill; angular acceleration is positive in the direction of downhill rolling.

  1. Moment of inertia of either cylinder \[ \boxed{I=\kappa MR^2=\frac{262}{375}MR^2 \simeq0.699MR^2.} \] Its SI unit is \(\mathrm{kg\,m^2}\).

  2. Equations of motion

Let the axle force on cylinder \(i\) have components \(P_i\) downhill and \(Q_i\) toward the road. Let \(N_i\) be its road-normal force and \(f_i\) its uphill friction force. These are signed reference components.

For the body and massless undercarriage, \[ \boxed{ \begin{aligned} 5Ma&=5Mg\sin\theta-P_r-P_f,\\ Q_r+Q_f&=5Mg\cos\theta,\\ l(Q_f-Q_r)&=b(P_r+P_f). \end{aligned}} \] For each cylinder, \[ \boxed{ Ma=Mg\sin\theta+P_i-f_i,\qquad N_i=Mg\cos\theta+Q_i,\qquad I\alpha_i=Rf_i.} \]

Consequently, \[ \boxed{7Ma=7Mg\sin\theta-f_r-f_f,} \] \[ \boxed{ N_{r,f}(a)=\frac{7M}{2} \left[g\cos\theta\mp q(g\sin\theta-a)\right].} \]

  1. Rolling and sliding regimes

For compactness, define \[ \alpha_i^{\rm sl}(a)=\frac{\mu_kN_i(a)}{\kappa MR}. \]

Rear cylinder Front cylinder Common linear acceleration \(a\) \(\alpha_r\) \(\alpha_f\)
Rolls Rolls \(\displaystyle a_0=\frac{7g\sin\theta}{7+2\kappa}\simeq0.8336g\sin\theta\) \(a_0/R\) \(a_0/R\)
Slides Rolls \(\displaystyle a_{\rm SR}=\frac{7g[(2+q\mu_k)\sin\theta-\mu_k\cos\theta]}{14+2\kappa+7q\mu_k}\) \(\alpha_r^{\rm sl}(a_{\rm SR})\) \(a_{\rm SR}/R\)
Rolls Slides \(\displaystyle a_{\rm RS}=\frac{7g[(2-q\mu_k)\sin\theta-\mu_k\cos\theta]}{14+2\kappa-7q\mu_k}\) \(a_{\rm RS}/R\) \(\alpha_f^{\rm sl}(a_{\rm RS})\)
Slides Slides \(\displaystyle a_{\rm SS}=g(\sin\theta-\mu_k\cos\theta)\) \(\alpha_r^{\rm sl}(a_{\rm SS})\) \(\alpha_f^{\rm sl}(a_{\rm SS})\)

These are the four contact-mode equations. Each requires \(N_r,N_f\ge0\), together with \[ \boxed{\kappa Ma\le\mu_sN_i\quad\text{for a rolling cylinder},} \] \[ \boxed{a>R\alpha_i^{\rm sl}\quad\text{for a cylinder sliding downhill from rest}.} \] Equalities describe boundaries between modes.

Both cylinders roll from rest if \[ \boxed{ \tan\theta\le \frac{(7+2\kappa)\mu_s}{2\kappa(1+q\mu_s)}.} \] The rear cylinder reaches its static-friction limit first. Thus loss of grip from the both-rolling branch leads to rear-only sliding or directly to both sliding; front-only sliding is the alternative conditional mixed contact mode, not the first loss of grip of that branch.

  1. Velocities after entering the lower-friction section

For \(s\ge d\), put \[ v_d=\sqrt{2a_0d},\qquad a_1=g(\sin\theta-\mu_k'\cos\theta), \] \[ v=\sqrt{v_d^2+2a_1(s-d)},\qquad \Delta\tau=\frac{2(s-d)}{v+v_d}. \] Then \[ \boxed{v_r=v_f=v,} \] \[ \boxed{ \omega_{r,f}= \frac{v_d}{R} +\frac{7\mu_k'g\cos\theta}{2\kappa R} (1\mp q\mu_k')\,\Delta\tau.} \] The minus sign is for the rear cylinder. Linear velocities have units \(\mathrm{m\,s^{-1}}\), and angular velocities have units \(\mathrm{rad\,s^{-1}}\).

Solution

1. Moment of inertia

Treat the shell as uniform and each spoke as a uniform thin radial rod extending from the axle to \(R_i\).

The shell has mass \(0.8M\), so its moment of inertia about the cylinder axis is \[ I_{\rm shell} =\frac12(0.8M)(R^2+R_i^2). \]

Each spoke has mass \(0.2M/8\). A uniform rod of length \(R_i\), rotating about an axis through one end perpendicular to the rod, has moment of inertia \(mR_i^2/3\). Hence \[ I_{\rm spokes} =8\left[\frac13\left(\frac{0.2M}{8}\right)R_i^2\right] =\frac{0.2M}{3}R_i^2. \]

Using \(R_i=0.8R\), \[ \begin{aligned} I &=MR^2\left[ \frac{0.8}{2}(1+0.8^2) +\frac{0.2}{3}(0.8^2) \right]\\ &=MR^2\left(\frac{82}{125}+\frac{16}{375}\right). \end{aligned} \] Therefore \[ \boxed{I=\frac{262}{375}MR^2=\kappa MR^2.} \]

2. Force diagrams and equations of motion

Choose the \(x\)-axis downhill along the road and the \(y\)-axis perpendicular to the road, pointing away from it.

We use the intended rigid-vehicle model: both cylinders remain in contact with the straight road, their centers have the same acceleration \(a\), and the body remains parallel to the road. Its angular acceleration is therefore zero. The body’s center of mass is midway between the axles and a perpendicular distance \[ b=h+\frac t2 \] above the axle line.

The force diagrams should show the following.

  • Body, including the massless undercarriage: weight \(5Mg\), vertically downward at its center of mass; at each axle position, a force with components \(-P_i\) along the road and \(+Q_i\) perpendicular to it.
  • Rear cylinder: weight \(Mg\), vertically downward through its center; axle-force components \(P_r\) downhill and \(Q_r\) toward the road; road reaction \(N_r\), perpendicular to the road at contact; friction \(f_r\), uphill at contact.
  • Front cylinder: the corresponding forces \(Mg,P_f,Q_f,N_f,f_f\).

The axle-force arrows are reference directions: an individual signed component may be negative. Friction points uphill in the downhill-sliding modes considered here and also supplies the torque for downhill rolling.

Newton’s second law for the body gives \[ 5Ma=5Mg\sin\theta-P_r-P_f \] and \[ Q_r+Q_f=5Mg\cos\theta. \]

Taking moments about the body’s center of mass, the two axle positions have road-parallel coordinates \(-l,+l\) and perpendicular coordinate \(-b\). Zero angular acceleration gives \[ l(Q_f-Q_r)-b(P_r+P_f)=0. \] Thus \[ \boxed{l(Q_f-Q_r)=b(P_r+P_f).} \]

For cylinder \(i\), translational motion gives \[ Ma=Mg\sin\theta+P_i-f_i, \] \[ N_i-Mg\cos\theta-Q_i=0. \] Only the road friction produces torque about the cylinder axis, because bearing friction is neglected: \[ \boxed{I\alpha_i=Rf_i.} \]

Adding the three downhill equations eliminates the internal axle forces: \[ \boxed{7Ma=7Mg\sin\theta-(f_r+f_f).} \]

The normal reactions satisfy \[ N_r+N_f=7Mg\cos\theta. \] Also, \[ N_f-N_r=Q_f-Q_r =\frac{b}{l}(P_r+P_f). \] Using the body’s downhill equation, \[ N_f-N_r=\frac{5Mb}{l}(g\sin\theta-a) =7Mq(g\sin\theta-a). \] Therefore \[ \boxed{ N_r(a)=\frac{7M}{2} \left[g\cos\theta-q(g\sin\theta-a)\right],} \] \[ \boxed{ N_f(a)=\frac{7M}{2} \left[g\cos\theta+q(g\sin\theta-a)\right].} \]

These equations include the transfer of load toward the front. Equivalently, \[ N_f-N_r=q(f_r+f_f). \]

The body length \(L\) does not enter: the body has no angular acceleration, and its center of mass is fixed by symmetry.

3. Possible motions and their accelerations

Assume the usual ordering \[ 0\le\mu_k\le\mu_s. \]

For a cylinder rolling without slipping, \[ a=R\alpha_i. \] Its rotational equation then requires \[ \boxed{f_i=\frac{I}{R^2}a=\kappa Ma.} \] This force must satisfy the static-friction bound \[ \kappa Ma\le\mu_sN_i. \]

For a cylinder whose contact point slides downhill, \[ f_i=\mu_kN_i,\qquad \boxed{\alpha_i=\frac{\mu_kN_i}{\kappa MR}.} \]

To check the direction of sliding, define the downhill velocity of the cylinder’s contact point relative to the road: \[ u_i=v-R\omega_i. \] Starting from rest, a constant-acceleration sliding solution has \[ u_i=(a-R\alpha_i)\tau. \] Thus uphill kinetic friction is consistent with downhill sliding only if \[ \boxed{a>R\alpha_i.} \]

All formulas below also require nonnegative normal reactions. A vanishing reaction marks possible loss of contact; beyond that boundary the assumed two-contact, non-pitching model no longer applies.

Both cylinders roll without slipping.

Here \[ f_r=f_f=\kappa Ma. \] The combined downhill equation becomes \[ 7Ma=7Mg\sin\theta-2\kappa Ma. \] Hence \[ \boxed{ a_0=\frac{7g\sin\theta}{7+2\kappa} =\frac{2625}{3149}g\sin\theta \simeq0.8336g\sin\theta,} \] and \[ \boxed{\alpha_r=\alpha_f=\frac{a_0}{R}.} \]

The required friction at either contact is \[ f_0=\frac{7\kappa Mg\sin\theta}{7+2\kappa}. \]

Writing \[ \eta=\frac{2\kappa}{7+2\kappa}, \] the reactions are \[ N_{r,f}^{(0)} =\frac{7Mg}{2}\left(\cos\theta\mp q\eta\sin\theta\right). \] Since \(N_r^{(0)}\le N_f^{(0)}\), the rear contact imposes the stricter static-friction condition: \[ f_0\le\mu_sN_r^{(0)}. \] Rearrangement gives \[ \eta(1+q\mu_s)\tan\theta\le\mu_s, \] or \[ \boxed{ \tan\theta\le \frac{(7+2\kappa)\mu_s}{2\kappa(1+q\mu_s)}.} \]

As an independent check, pure rolling has kinetic energy \[ K=\frac12(7M)v^2 +2\left(\frac12I\frac{v^2}{R^2}\right) =\frac12(7+2\kappa)Mv^2. \] Equating its increase to the loss of gravitational potential energy gives the same \(a_0\).

Rear cylinder slides; front cylinder rolls.

Now \[ f_r=\mu_kN_r(a),\qquad f_f=\kappa Ma. \] Therefore \[ (7+\kappa)Ma =7Mg\sin\theta-\mu_kN_r(a). \] Substituting the normal reaction, \[ (7+\kappa)a =7g\sin\theta -\frac{7\mu_k}{2} \left[g\cos\theta-q(g\sin\theta-a)\right]. \] Collecting the terms containing \(a\), \[ \boxed{ a_{\rm SR} =\frac{7g[(2+q\mu_k)\sin\theta-\mu_k\cos\theta]} {14+2\kappa+7q\mu_k}.} \] The angular accelerations are \[ \boxed{ \alpha_r=\frac{\mu_kN_r(a_{\rm SR})}{\kappa MR}, \qquad \alpha_f=\frac{a_{\rm SR}}R.} \]

The consistency conditions are \[ \boxed{ a_{\rm SR}> \frac{\mu_kN_r(a_{\rm SR})}{\kappa M},\qquad \kappa Ma_{\rm SR}\le\mu_sN_f(a_{\rm SR}).} \] The first makes the rear contact slide downhill; the second allows the front to remain rolling.

This is the mixed branch to test when the rear loses static grip. If the front cannot supply its required static friction, both cylinders must slide.

Rear cylinder rolls; front cylinder slides.

For completeness, the other mixed contact combination has \[ f_r=\kappa Ma,\qquad f_f=\mu_kN_f(a). \] Thus \[ (7+\kappa)Ma =7Mg\sin\theta-\mu_kN_f(a), \] which yields \[ \boxed{ a_{\rm RS} =\frac{7g[(2-q\mu_k)\sin\theta-\mu_k\cos\theta]} {14+2\kappa-7q\mu_k}.} \] Its angular accelerations are \[ \boxed{ \alpha_r=\frac{a_{\rm RS}}R,\qquad \alpha_f=\frac{\mu_kN_f(a_{\rm RS})}{\kappa MR}.} \] Its consistency conditions are \[ \boxed{ \kappa Ma_{\rm RS}\le\mu_sN_r(a_{\rm RS}),\qquad a_{\rm RS}> \frac{\mu_kN_f(a_{\rm RS})}{\kappa M}.} \]

This is a conditional sliding-state solution. It does not describe the front being the first cylinder to lose grip from the both-rolling branch, because that branch requires equal friction at the two contacts while the front has the larger normal reaction.

Both cylinders slide.

In this case, \[ f_r+f_f=\mu_k(N_r+N_f)=7\mu_kMg\cos\theta. \] Hence \[ \boxed{a_{\rm SS}=g(\sin\theta-\mu_k\cos\theta).} \]

Since \[ g\sin\theta-a_{\rm SS}=\mu_kg\cos\theta, \] the normal reactions reduce to \[ \boxed{ N_{r,f}^{\rm SS} =\frac{7Mg\cos\theta}{2}(1\mp q\mu_k).} \] Therefore \[ \boxed{ \alpha_{r,f} =\frac{7\mu_kg\cos\theta}{2\kappa R} (1\mp q\mu_k).} \]

The front has the larger angular acceleration. Consequently, the stricter condition for both contact points to slide downhill from rest is \[ a_{\rm SS}>R\alpha_f. \] For \(0\le\theta<\pi/2\), this is \[ \boxed{ \tan\theta> \mu_k\left[1+\frac{7}{2\kappa}(1+q\mu_k)\right].} \] Maintaining both contacts also requires \[ q\mu_k\le1. \]

If \(\mu_k=0\), the cylinders undergo pure translational sliding from rest: \[ a=g\sin\theta,\qquad \alpha_r=\alpha_f=0. \]

The distinction between the onset of sliding and its continuation matters: static friction determines whether a contact loses grip, whereas the kinetic conditions above determine whether an assumed sliding state is consistent. A kinetic solution should not replace an available no-slip solution merely because \(\mu_k<\mu_s\).

4. Motion on the lower-friction section

During the initial pure-rolling distance \(d\), the acceleration is \(a_0\). Starting from rest, \[ \boxed{v_d=\sqrt{2a_0d},\qquad \omega_{r,d}=\omega_{f,d}=\frac{v_d}{R}.} \]

We neglect the short transition during which only one cylinder has reached the new surface, using the stated large-distance approximation. There is no impulsive change in either linear or angular velocity.

Once both cylinders slide on the new surface, replace \(\mu_k\) by \(\mu_k'\) in the preceding both-sliding results: \[ a_1=g(\sin\theta-\mu_k'\cos\theta), \] \[ N_{r,f}'=\frac{7Mg\cos\theta}{2}(1\mp q\mu_k'), \] \[ \alpha_{r,f}'= \frac{7\mu_k'g\cos\theta}{2\kappa R}(1\mp q\mu_k'). \] These quantities are constant.

Over the remaining distance \(s-d\), \[ v^2=v_d^2+2a_1(s-d), \] so both cylinder centers have velocity \[ \boxed{ v_r=v_f= \sqrt{ \frac{14gd\sin\theta}{7+2\kappa} +2g(\sin\theta-\mu_k'\cos\theta)(s-d) }.} \]

The elapsed time on this section follows from the average velocity: \[ s-d=\frac{v+v_d}{2}\Delta\tau, \] hence \[ \boxed{\Delta\tau=\frac{2(s-d)}{v+v_d}.} \]

Integrating the constant angular accelerations gives \[ \boxed{ \omega_r= \frac{v_d}{R} +\frac{7\mu_k'g\cos\theta}{2\kappa R} (1-q\mu_k')\frac{2(s-d)}{v+v_d},} \] \[ \boxed{ \omega_f= \frac{v_d}{R} +\frac{7\mu_k'g\cos\theta}{2\kappa R} (1+q\mu_k')\frac{2(s-d)}{v+v_d}.} \]

The premise that both cylinders begin sliding requires a consistent downhill-sliding regime. Indeed, immediately after entry, \[ v-R\omega_i=(a_1-R\alpha_i')\Delta\tau, \] so the condition is \(a_1>R\alpha_f'\), together with nonnegative reactions. The reduced static coefficient \(\mu_s'\) determines loss of grip, but does not enter the subsequent velocities once both contacts are sliding.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.