CCONSTANT-SPEED RAMPS
OSCAR M. PERDOMO
Abstract.
It is easy to show that if the kinetic coefficient of friction between a block and a rampis µ k and this ramp is a straight line with slope − µ k then, this block will move along the ramp withconstant speed. A natural question to ask is the following: Besides straight lines, are there moreshapes of ramps such that a block will go down on the ramp with constant speed. Here we classifyall the possible shape of these ramps and, surprisingly Introduction
A basic computation shows that given a kinetic coefficient of friction µ k , the angle of a linear rampcan be adjusted so that a block on this ramp will move down with constant speed. In order tocancel the gravity force, the friction force and the normal force, it is enough to take a ramp given bya line with slope − µ k , this is, we must take the inclination of the ramp to be θ with tan( θ ) = µ k .See Figure 1.1 Μ k = coefficient of frictionIf tan H Θ L = Μ k , then the block will move down with constant speed Θ Figure 1.1.
A block sliding down under the effect of gravity and the friction force.This figure shows the three forces acting on the block. The dashed forces are just adecomposition of the gravity force.Let us assume that we want a block to move down with speed v and we know that the kineticcoefficient of friction is µ k . In this note we classify the shapes of all possible ramps on which thisblock will move down with speed v . Here is how you would built this constant-speed ramp . Given v and µ k = tan( δ ) , let us define a = gv sin δ , ( g = 9 . ms ) and Date : November 3, 2018. a r X i v : . [ m a t h . DG ] M a y ONSTANT-SPEED RAMPS 2 α ( s ) = (cid:18) s + 1 a ln(1 + e − as ) , a arccot ( e − as ) (cid:19) (1.1)This curve α has an U shape with two horizontal asymptotes, one at y = 0 and the second one at y = πa . See Figure 1.2. We have: If we clockwise rotate the curve α an angle δ (see Figure 1.3), then, the highest point divides thisrotated curve in two ramps on which a block can move with constant speed v under the assumptionthat the kinetic coefficient of friction is µ k = tan( δ ) The curve Α H s L Π a Figure 1.2.
This is the graph of the curve α , it has two horizontal asymptotes thatare separated πa . The parameter s in the parametrization provided in the definitionof α is arc-length. The curve Α H s L inclined an angle Θ - - - - Figure 1.3.
Graph of the two constant-speed ramps . At the highest point, the blockwith speed v must be placed on the top of the ramp to the right and it must beplaced upside down on the ramp to the left.The sequence of pictures at the end of the paper shows how the motion on the ramps takes placewhen the desire speed is ms and µ k = 0 . . The picture also displays the three forces acting on theblock under the assumption that the mass is Kg. On the highest part of ramp the normal is zero,the normal in the upper part of the ramp is pointing down, the bock does not fall down due to itsspeed.In section 2 we provide a first proof for the classification of two dimensional ramps. In section 3 weprovide a formal definition of ramp in order to state the result as a mathematical theorem. Thissection may be omitted. Section 4 deals with 3-dimensional ramps.The author would like to express his gratitude to Frank Gould, Roger Vogeler, Nidal Al-Masoudand Clifford Anderson for their valuable comments on this work.
ONSTANT-SPEED RAMPS 3 A first approach
Let us find all the possible curves with the property that a block, sliding down on it, will move withconstant speed v > under the assumption that the kinetic coefficient of friction is µ = tan( δ ) forsome constant δ between and π radians. Let us start by assuming that this curve is parametrizedby arc-length, this is, if α ( s ) = ( x ( s ) , y ( s )) denotes such a curve, then | α (cid:48) ( s ) | = 1 . Under thisassumption we can assume that for a smooth function θ ( s ) we have x (cid:48) ( s ) = cos( θ ( s )) and y (cid:48) ( s ) = sin( θ ( s )) The function θ ( s ) will help us to describe the curve α . It is clear that the vector n ( s ) = ( − sin( θ ( s )) , cos( θ ( s ))) is a unit vector perpendicular to α (cid:48) ( s ) and the chain rule give us that α (cid:48)(cid:48) ( s ) = θ (cid:48) ( s ) n ( s ) . Figure2.1 shows these two vectors Α ' H s L = H cos H Θ H s LL , sin H Θ H s LLL n H s L = H - sin H Θ H s LL , cos H Θ H s LLL Α H s L xy Figure 2.1.
We are assuming that the parameter s is arc-length parameter. Thevectors α (cid:48) ( s ) and n ( s ) are perpendicular.Let us assume that β ( t ) = α ( vt ) describes the motion of the block. Since s = vt and s is arc-lengthparameter, then the speed of the block is constant, it is v . Since we have that β (cid:48)(cid:48) ( t ) = v α ( vt ) then,Figure 2.2 shows the free body diagram for the problem of making the block move along α withconstant speed v . - Μ Λ Α ' H s L N = Λ n H - mg L = m d Β dt H s L = m v Θ ' H s L n Free body diagram
Figure 2.2.
By Newton’s second law, the sum of the normal force N plus the weightplus the friction force must be mβ (cid:48)(cid:48) .As the free body shows the following equation must hold true. mv θ (cid:48) ( s ) n ( s ) = λ ( s ) n ( s ) − tan( δ ) λ ( s ) α (cid:48) ( s ) − (0 , mg ) (2.1)By doing the inner product of both sides of the equation 2.1 with the vector α (cid:48) ( s ) we obtain that ONSTANT-SPEED RAMPS 4 λ ( s ) = − mg cot( δ ) sin( θ ( s )) Likewise, by doing the inner product of both sides of the equation 2.1 with the vector n ( s ) , weobtain that mv θ (cid:48) = λ − mg cos( θ ) and therefore, if a = gv sin( δ ) , then θ (cid:48) ( s ) = − gv sin( δ ) sin( θ ( s ) + δ ) = − a sin( θ ( s ) + δ ) (2.2)Since the differential equation (2.2) does not have the variable s , then, all the solutions differ bya horizontal translation. This is, if θ ( s ) is a solution, then θ ( s + c ) is also a solution for everyreal number c . Due to the geometry of our problem we do not need to consider an integratingconstant, since just one solution will give us all the solutions α ( s ) . Recall that the equilibriumsolution of the differential equation (2.2) is θ ( s ) = − δ for all s . This solution corresponds to thecase α ( s ) = (cos( δ ) s, − sin( δ ) s ) which is the straight line ramp shown in Figure 1.1. When we solvethis differential equation by separation of variables we notice that we need to integrate the function csc( θ + δ ) . Instead of using the classical formula (cid:82) csc( u ) du = − ln(csc( u ) + tan( u )) we will use theformula (cid:82) csc( u ) du = ln(tan( u )) which led us to the formula θ ( s ) = − δ + 2 arctan( e − as ) It is clear that if γ ( s ) = ( z ( s ) , w ( s )) denotes a counterclockwise rotation of δ radians of the curve α ( s ) , then z (cid:48) ( s ) = cos(2 arctan( e − as )) and w (cid:48) ( s ) = cos(2 arctan( e − as )) Integrating the equations above we obtain that z ( s ) = s + ln(1 + e − as ) a and w ( s ) = 2 a arccot ( e − as ) The curve ( z ( s ) , w ( s )) is shown in Figure 1.2. As mentioned in the introduction, the solution thatwe are looking for is a clockwise rotation of the curve γ by an angle δ . In the next section we givea more detailed explanation of this solution.3. Understanding the solution of the ODE. A mathematical definition of ramp
Let us start this section with a definition.
Definition 1.
We will say that a curve γ : [ a , a ] → R is regular is γ (cid:48) ( t ) never vanishes. We say n : [ a , a ] → R is a normal of the curve γ if n ( t ) has length 1 and the inner product of n ( t ) and γ (cid:48) ( t ) is zero, this is, n ( t ) · γ ( t ) = 0 for all t . Definition 2.
A ramp in the plane R is an ordered pair ( γ, n ) where γ : [ a , a ] → R is a regularcurve and n : [ a , a ] → R is a normal. We will interpret the ramp ( γ, n ) as a portion of the planewhose boundary contains γ and its outer normal vector along is n . Example 1. ( γ , n ) where γ , n : [0 , π ] → R are given by γ ( t ) = 5(cos( t ) , sin( t )) and n ( t ) =(cos( t ) , sin( t )) is an example of a ramp. This ramp is shown in the first two images in Figure 3.1. ( γ , n ) where γ , n : [0 , π ] → R are given by γ ( t ) = 5(cos( t ) , sin( t )) and n ( t ) = − (cos( t ) , sin( t )) is an example of a ramp. This ramp is shown in the last two images in Figure 3.1. The following definition is based on Newton’s second law.
ONSTANT-SPEED RAMPS 5 Γ H t L n H t L - - - - - - Γ H t L n H t L - - - - - - Figure 3.1.
A normal vector of a curve helps us to define the part of a curve wherewe want a object to slide on a ramp
Definition 3. A ramp with external force is a triple ( γ, n, F ) where F : [ a , a ] → R is asmooth function and ( γ, n ) is a ramp. Given a positive number m , a solution of the ramp withexternal force ( γ, n, F ) is given by a curve β ( t ) = γ ( h ( t )) (Notice that β is just a re-pametrizationof the curve γ and it is completely determined by the function h : [ a , a ] → R ) and a nonnegativefunction λ : [ a , a ] → R such that F ( h ( t )) + λ ( t ) n ( h ( t )) = mβ (cid:48)(cid:48) ( t ) Remark . In Definition 3, the term λ ( t ) n ( t ) represents the force that the surface of the ramp isdoing on the object that moves on the ramp under the action of the external force F . By Newton’sthird law, − λ ( t ) n ( t ) is the force that the object is doing to the ramp. The condition λ ≥ is neededso that the object stays on the ramp. Example 2. If ( γ, n ) is a ramp, m > and F ( t ) = (0 , − . m ) , then the triple ( γ, n, F ) representsthe action of the gravity acting on a particle with mass m that moves on the ramp without friction. Example 3.
Let us consider the ramp ( γ, n ) where γ, n : [ − . , . → R are given by γ ( t ) =( t, t ) and n ( t ) = ( t √ t +1 , − √ t +1 ) . If F ( t ) = (0 , − . then, the ramp with external force ( γ, n, F ) has no solution. Figure 3.2 shows this ramp. There is not solution for thisramp if F H t L = H - L - - - - Figure 3.2.
Some ramps with external force do not have a solution.Definition 3 provides the definition of the solution of a particle moving on the ramp under the actionof the force F . In the case that F denotes all the forces acting on the particle that moves on theramp but the friction force, the definition of solution needs to be changed to: ONSTANT-SPEED RAMPS 6
Definition 4.
Given a ramp with external force ( γ, n, F ) defined on the interval [ a , a ] and twopositive numbers µ < and m , a solution of the ramp with external force ( γ, n, F ) andkinetic coefficient of friction µ is given by a curve β ( t ) = γ ( h ( t )) and a nonnegative function λ : [ a , a ] → R such that F ( h ( t )) + λ ( t ) n ( h ( t )) − µλ ( t ) β (cid:48) ( t ) | β (cid:48) ( t ) | = mβ (cid:48)(cid:48) ( t ) Remark . Let us define an elementary function to be a function of one variable with real valuesbuilt from a finite number of trigonometric, exponential, constant, n th functions and their inverses,through composition and using the four basic elementary operation (+, -, × , ÷ ). When we wantto define basic examples of curves defined by arc-length parameter, that is, if we want to define γ ( s ) = ( x ( s ) , y ( s )) such that x (cid:48) ( s ) + y (cid:48) ( s ) = 1 for all t , then, the first thing that comes to ourmind if to find easy possibilities for x (cid:48) ( t ) and y (cid:48) ( t ) . The easiest one would be a pair of numbers c and c such that c + c = 1 . If we make x (cid:48) ( s ) = c and y (cid:48) ( s ) = c then after and easy integrationwe obtain that the curve γ is a straight line. If we want to use the fact that cos( as ) + sin( as ) = 1 and we decide to make x (cid:48) ( s ) = cos( as ) and y (cid:48) ( s ) = sin( as ) , then, after an easy integration we obtainthat γ should be a circle. The most important curves in this paper are those that are obtained byusing the identity tanh ( t ) + sech ( t ) = 1 for all t That is, we will be using curves that satisfy x (cid:48) ( s ) = tanh ( as ) and y (cid:48) ( s ) = sech ( as ) .The following lemma is a direct computation and provides a definition of the curve α whose graphis shown in Figure 1.2 Lemma 3.3.
For any non zero real number a , the curve α ( s ) = ( x ( s ) , y ( s ))) = (cid:18) s + 1 a ln(1 + e − as ) , a arccot ( e − as ) (cid:19) (3.1) is an arc-length parametrized curve. Moreover, x (cid:48) ( s ) = tanh(as) and y (cid:48) ( s ) = sech(as) Lemma 3.4.
For any positive number δ smaller than π , if α δ represents the curve obtained byrotating an angle of δ radians the curve α = ( x ( s ) , y ( s )) defined in Lemma 3.3, this is, if α δ ( s ) = ( x δ ( s ) , y δ ( s )) = (cos( δ ) x ( s ) + sin( δ ) y ( s ) , − sin( δ ) x ( s ) + cos( δ ) y ( s )) (3.2) then, the maximum value for the second entry of α δ is achieved when s = arcsinh( cot( δ ) ) a . Also, wehave that α (cid:48)(cid:48) δ ( s ) · ( y (cid:48) δ ( s ) , − x (cid:48) δ ( s )) = a sech ( as ) (3.3) The graph of the curve α δ is shown in Figure 1.3. ONSTANT-SPEED RAMPS 7
Proof.
A direct computation using Lemma 3.3 shows that y (cid:48) δ ( s ) = − sin( δ ) tanh ( as )+cos( δ ) sech ( as ) ,then we can see that the only solution of the equation y (cid:48) δ ( s ) = 0 is s = arcsinh( cot( δ ) ) a . In order toprove the identity 3.3 we point out that since the inner product is invariant under rotations, thisidentity is equivalent to show that α (cid:48)(cid:48) ( s ) · ( y (cid:48) ( s ) , − x (cid:48) ( s )) = a sech ( as ) which follows because, x (cid:48)(cid:48) ( s ) y (cid:48) ( s ) − y (cid:48)(cid:48) ( s ) x (cid:48) ( s ) = a sech ( as )+ a tanh ( as ) sech ( as ) = a sech ( as ) ( sech ( as )+ tanh ( as )) = a sech ( as ) (cid:3) Remark . In the previous proof we got that y (cid:48) δ ( s ) = − sin( δ ) sech ( as )( sinh ( as ) − cot( δ )) . It followsthat y (cid:48) δ ( s ) > if s < s and y (cid:48) δ ( s ) < if s > s . Definition 5.
For any positive number δ smaller than π and any a > we define the ramps ( γ δ , n δ ) and (˜ γ δ , ˜ n δ ) by γ δ ( s ) = α δ ( s − s ) and n δ ( s ) = ( y (cid:48) δ ( s − s ) , − x (cid:48) δ ( s − s )) and ˜ γ δ ( s ) = α δ ( s + s ) and ˜ n δ ( s ) = ( − y (cid:48) δ ( s + s ) , x (cid:48) δ ( s + s )) where s = arcsinh( cot( δ ) ) a and the maps γ δ and n δ are defined in Lemma 3.4 and all the function α δ , n δ , ˜ α δ , ˜ n δ are defined on the interval [0 , ∞ ) . These two ramps are shown in Figure 3.3. Ramp H Γ ∆ , n ∆ L with ∆ = - - - H Γ ~∆ , n ~∆ L with ∆ = - - - Figure 3.3.
The two ramps define in Definition 5Let us state and proof the main theorem in this section.
Theorem 3.6.
Given an angle δ between and π radians and a positive speed v > . Let usconsider µ = tan( δ ) , a = gv sin δ and F ( t ) = (0 , − mg ) with g = 9 . . We have a.) For the ramp ( γ δ , n δ ) given in Definition 5, β ( t ) = γ δ ( vt ) and λ ( t ) = mg cot( δ ) y (cid:48) δ ( s − vt ) is a solution for the ramp with external force ( γ δ , n δ , F ) and kinetic coefficient of friction tan( δ ) . b.) For the ramp (˜ γ δ , ˜ n δ ) given in Definition 5, β ( t ) = ˜ γ δ ( vt ) and λ ( t ) = − mg cot( δ ) y (cid:48) δ ( s + vt ) ONSTANT-SPEED RAMPS 8 is a solution for the ramp with external force (˜ γ δ , ˜ n δ , F ) and kinetic coefficient of friction tan( δ ) . c.) Since in both cases | β (cid:48) ( t ) | = v , then these motions have constant speed.Proof. Let us prove part a.) . First of all, notice that as it is required by Definition 4, λ > byremark 3.5. From the definition of β we obtain that β (cid:48) ( t ) = − vα (cid:48) δ ( s − vt ) and β (cid:48)(cid:48) ( t ) = v α (cid:48)(cid:48) δ ( s − vt ) therefore the equation F ( h ( t )) + λ ( t ) n ( h ( t )) − µλ ( t ) β (cid:48) ( t ) | β (cid:48) ( t ) | = mβ (cid:48)(cid:48) ( t ) is equivalent to (0 , − mg ) + λ ( t ) n δ ( vt ) + tan( δ ) λ ( t ) α (cid:48) δ ( s − vt ) = mv α (cid:48)(cid:48) δ ( t ) (3.4)Notice that the vector u = n δ ( vt ) and u = α (cid:48) δ ( s − tv ) form an orthonormal basis. Therefore inorder to prove equation (3.4) it is enough to prove that the dot product with u and u of the lefthand side (LHS) and right hand side (RHS) of the equation is the same. The dot product of theLHS of equation (3.4) with u is equal to mg x (cid:48) δ ( s − vt ) + λ ( t ) = mg x (cid:48) δ ( s − vt ) + cot( δ ) y (cid:48) δ ( s − vt )= mg (cos( δ ) tanh ( a ( s − vt )) + sin( δ ) sech ( a ( s − vt ))) + mg cot( δ ) ( − sin( δ ) tanh ( a ( s − vt )) + cos( δ ) sech ( a ( s − vt )))= mg sin( δ ) sech ( a ( s − vt )) Using Equation (3.3) we get that the dot product of the RHS of equation (3.4) with u is equal to mv a sech ( a ( s − vt )) = mg sin( δ ) sech ( a ( s − vt )) Therefore LHS · u =RHS · u . It is easy to check that the dot product of the RHS of equation(3.4) with u vanishes. On the other hand, the dot product of the LHS of the equation (3.4) with u equals to − mg y (cid:48) δ ( s − vt ) + v tan( δ ) λ = − mg y (cid:48) δ ( s − vt ) + tan( δ ) mg cot( δ ) y (cid:48) δ ( s − vt ) = 0 Therefore part a.) follows. Part b.) is similar. Part c.) follows because | α (cid:48) | = 1 and since α δ is arotation of α then | α (cid:48) δ | = 1 , then, β (cid:48) ( t ) = − vα (cid:48) ( s − vt ) and | β (cid:48) ( t ) | = v (cid:3) ONSTANT-SPEED RAMPS 9 In this section we describe ramps in the space on which an object can move with constant speed v under the assumption that the kinetic coefficient of friction is µ . We will prove that fixing v and µ ,there are as many different ramp as continuous unit tangent vector fields in the south hemisphere.In the correspondence that we will establish, the ramp that we defined in section 2 corresponds toa particular choice of the a tangent vector field.For curves that represent planar ramps, we got a description of them by studying first their velocityvector. Recall that the function θ ( s ) was used to described a point in the lower part of the semicirclethat represented a possible velocity vector of the curve that describe the ramp. See Figure 4.1. Whenthis curve is free to move in the whole 3-dimensional space, the tangent unit vector to the curvelies in the unit sphere and since the block must go down the ramp we can assume that the tangentunit vector of the ramp lies in the shouth hemisphere. See Figure 4.1. H y , y L ΘΑ ' H s L = H y , y L = H cos Θ , sin Θ L - - - - - - - - - - - - Α ' H s L = H y , y , y L - - - - - - Figure 4.1.
The unit tangent vector of a possible ramp in the plane lies in thelower part of the unit semicircle, if the ramp is free to move in the whole space, thenthe unit tangent vector lies in the lower hemisphere of the unit spherein order to completely determine the ramp we need to specify the desire direction of the normal tothe ramp, see Figure 4.2. We can do this by choosing a unit normal vector field H ( y ) defined in thelower hemisphere and requesting the normal vector of the ramp to be equal to N ( y ) anytime thetangent unit vector to the ramp is the vector y . See Figure 4.3. - - - - - - - - - - - - - - - Figure 4.2.
To built a ramp we need to decide about the normal direction at everypoint in the ramp.We will prove that for any continuous choice of a normal field N ( s ) in the lower hemisphere, thereexist a family of ramp on which a motion with constant speed is possible under the effect of thegravity force and the force of friction. More precisely we have. ONSTANT-SPEED RAMPS 10 Α ' H s L = H y , y , y L - - - - - - Figure 4.3.
We will make the choice of the normal direction on the ramp to dependon the unit tangent vector.
Theorem 4.1.
Let
Σ = { ( y , y , y ) ∈ R : y + y + y = 1 and y ≤ } denote the southhemisphere and let N : Σ → R be any unit tangent vector field. This is, for any y ∈ Σ , N ( y ) hasnorm 1 and N ( y ) is perpendicular to y . For any positive numbers v , µ < and any y ∈ Σ thereexists a unique curve α ( s ) parametrize by arc-length, such that α (0) = (0 , , , α (cid:48) (0) = y and themotion along β ( t ) = α ( vt ) in the ramp given by R ( s, r ) = α ( s ) + r α (cid:48) ( s ) × N ( α (cid:48) ( s )) represents a motion that is a solution of Newton’s second law under the assumption that the onlyforces action on this moving particle are gravity and a friction force with kinetic constant coefficient µ . Recall that since the parameter s is arc-length parameter, then | β (cid:48) ( t ) | = v for all t ; this is, themotion has constant speed v .Proof. Recall that we are denoting by u · w the dot (or inner) product of the two vectors u and w .A direct computation shows that if e = (0 , , , then e T : Σ → R given by e T ( y ) = e − ( e · y ) y is a tangent vector field (notice that e T ( y ) must be in the tangent space T y Σ because e T ( y ) · y = 0 )Let λ : Σ → R be the function given by λ ( y ) = − gµ e · y = − gµ y Notice that since y ≤ then λ ≥ and λ only vanishes on the boundary of Σ , (recall that theboundary of Σ is given by the equation y = 0 ). Let us consider the tangent vector field X = − v (cid:0) ge T − λN ( y ) (cid:1) = − gv (cid:18) e T + y µ N ( y ) (cid:19) (4.1)Notice that along points in the boundary of Σ , X ( y ) = − gv e points toward Σ . It follows that anyintegral curve of the tangent vector field X remain in Σ . Let γ ( s ) be the integral curve of the vectorfield X that satisfies γ (0) = y . We will prove the Theorem by showing the curve α ( s ) given by α ( s ) = (cid:90) s γ ( u ) du satisfies all the conditions. Clearly, α (0) = (0 , , and α (cid:48) (0) = γ (0) = y . We also have that s is arc-length parameter because | α (cid:48) ( s ) | = | γ ( s ) | = 1 . A direct computation show that the normalvector of the surface (or ramp) R along the curve α (along points in the surface with r = 0 ) is givenby N ( α ( s )) . We will now proof that Newton’s second law holds true for β ( t ) = α ( vt ) . Notice that ONSTANT-SPEED RAMPS 11 since v is constant, then β (cid:48) ( t ) = vα (cid:48) ( vt ) = vγ ( s ) and β (cid:48)(cid:48) ( vt ) = v γ (cid:48) ( vt ) = v γ (cid:48) ( s ) . If m denotes themass of the particle and λ is defined as in the beginning of the proof, we have that mβ (cid:48)(cid:48) ( t ) = mv γ (cid:48) ( s ) = − mv v (cid:0) ge T ( γ ) − λN ( γ ) (cid:1) = m ( − ge + g ( e · γ ) γ − gµ ( e · γ ) N ( γ ( t )))= − gme + mλN ( γ ( s )) − mλµγ ( s ) From the equation above we conclude that then normal force made from the ramp to the block hasmagnitude mλ . Notice that the last equation above is Newton’s second law.It is clear the the form of the ramp depends on the desire constant speed that we want on the ramp.For example if the block is to travel outside down on some portion of the ramp and the speed v isnot much then the curvature of that portion of the ramp must be big so that the block does not falloff the ramp. The following corollary explains how to change the ramp if we want to change theconstant speed or if we want to change the gravity force. (cid:3) Corollary 4.2. a.)
If a ramp R ⊂ R does the work of allowing a block to move down with constantspeed v and gravity g , then, for any positive κ the ramp κR = { κ ( x, y, z ) : ( x, y, z ) ∈ R } does thework of allowing a block to move down with constant speed √ κv and gravity g . b.) If a ramp R ⊂ R does the work of allowing a block to move down with constant speed v andgravity g , then, for any positive κ the ramp κR = { κ ( x, y, z ) : ( x, y, z ) ∈ R } does the work ofallowing a block to move down with constant speed v and gravity gκ .Proof. This corollary is a consequence of Equation (4.1). We can check that if γ ( t ) is an integralcurve of the vector field X , and α ( s ) = (cid:82) s γ ( u ) du , then ˜ γ ( τ ) = γ ( τκ ) is a solution of the vector field κ X , which can be interpreted as either the vector field coming from the tangent unit vector field N ( y ) with velocity √ κ v and gravity g or it can be interpreted as the vector field coming from thetangent unit vector field N ( y ) , with velocity v and gravity gκ . The result follows by noticing that ˜ α ( s ) = (cid:90) s ˜ γ ( u ) du = (cid:90) s γ ( uκ ) du = κ (cid:90) sκ γ ( v ) dv = κ α ( sκ ) (cid:3) Remark . If a ramp R on Earth has the property than an object will fall down with constantspeed v , then, if we dilate this ramp by a factor of , the same block will move down with the sameconstant speed v on this dilated ramp when it is placed on the Moon. ONSTANT-SPEED RAMPS 12
Normal = - < acceleration = - < friction = < - - - - Normal = - < acceleration = - < friction = < - - - - Normal = - < acceleration = - < friction = < - - - - Normal = - < acceleration = - < friction = < - - - - Normal = - < acceleration = - < friction = < - - - - Normal = - < acceleration = - < friction = < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Current address : Department of Mathematics, Central Connecticut State University, New Britain, CT 06050,
E-mail address : [email protected] ONSTANT-SPEED RAMPS 13
Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Normal = < acceleration = < friction = - < - - - - Figure 4.4.
Motion on the longest parts of the ramps