SSliding down over a horizontally moving semi-sphere
Roberto A. Lineros ∗ Departamento de F´ısica, Universidad Cat´olica del Norte,Avenida Angamos 0610, Casilla 1280, Antofagasta, Chile. (Dated: February 26, 2021)We studied the dynamics of an object sliding down on a semi-sphere with radius R . We considerthe physical situation where the semi-sphere is free to move over a horizontal surface. Also, weconsider that all surfaces in contact are friction-less. We analyze the values for the last contactangle θ (cid:63) , corresponding to the angle when the object and the semi-sphere detach one of each other,considering all possible scenarios with different values of m A and m B . We found that the last contactangle only depends on the ratio between the masses, and it is independent of the acceleration ofgravity and semi-sphere radius. In addition, we found that the largest possible value of θ (cid:63) occursfor the case when the semi-sphere does not move. On the opposite case, the minimum value of theangle occurs for m A (cid:29) m B occurring at the top of the semi-sphere. I. INTRODUCTION
In courses of Newtonian mechanics for engineers and physics students at university level, the concepts behindNewton’s laws are key for understanding the kinematics of objects under the effects of forces. It is a bit troublesomefor the students to understand the interplay between objects in contact due to the presence of reaction forces. Atthe university level, the focus on the study of vectorial mechanics is crucial for dealing with more difficult problemsinvolving many bodies and involving different vectorial basis like cartesian and cylindrical basis.The problem of a object sliding down on a circular path is an academic example to teach such concepts [1–3].Similar problems have been addressed with different approaches like the use of lagrangian mechanics [4], the case withfriction [5–7], or the experimental demonstration [8].In this manuscript, we consider the effect of a moving semi-sphere where its movement is the result of the reactionforce of the object on top of the semi-sphere.The manuscripts is organized as follow: In section II, We present the solution for the case where the semi-sphereremains still and use the results as benchmarks for the moving setup. In section III, We present the solution andanalysis for the moving semi-sphere. Finally, section IV are the conclusions.
II. SYSTEM WITH FIXED SEMI-SPHERE.
At a first stage, we consider the situation when semi-sphere B remains still (see figure 1). This is a known problemtaught in courses of Newtonian Mechanics at the university level. The equations of motion for the object A areconstructed using the second Newton’s Law and correspond to: (cid:88) (cid:126)F = (cid:126)N A − (cid:126)W A = m A (cid:126)a A , (1)where (cid:126)N A is the reaction force between objects and (cid:126)W A = m A g ˆ is the weight with g the gravity’s acceleration. Beforethe object A detaches from the semi-sphere, it moves along the surface following a circular path. The acceleration istherefore described as (cid:126)a A = (cid:126)α × (cid:126)r + (cid:126)ω × ( (cid:126)ω × (cid:126)r ) , (2)where (cid:126)α = α ˆ k and (cid:126)ω = ω ˆ k are the vectors of angular acceleration and angular velocity where ˆ ı × ˆ = ˆ k . The acceleration (cid:126)a A in the cylindrical basis where ˆ r × ˆ θ = ˆ k corresponds to (cid:126)a A = αR ˆ θ − ω R ˆ r . (3) ∗ [email protected] a r X i v : . [ phy s i c s . c l a ss - ph ] F e b Figure 1. (Left) Physical situation on the object A sliding down over a fixed semi-sphere B . The semi-sphere B does not presentfriction. Forces acting on A are shown in blue. (Right) Vector basis description. The equations of motions in that basis are: N A − m A g cos θ = − m A ω R , (4) − m A g sin θ = m A αR . (5)For the setup in 1, the angular acceleration and angular velocity are related to the angle θ by: α = − ¨ θ and ω = − ˙ θ .Using the latter expressions, the equations of motion are reduced to a couple of differential equations: gR cos θ − N A m A R = ˙ θ , (6) gR sin θ = d ˙ θdθ ˙ θ . (7)These equations are simplified via the substitution: f ( θ ) = ˙ θ , f (cid:48) ( θ ) = dfdθ , and κ = gR ; obtaining: κ cos θ − N A m A R = f ( θ ) , (8)2 κ sin θ = f (cid:48) ( θ ) . (9)Notice that the function f ( θ ) is related to the kinetic energy of the object A .These equations are simply solved by integrating over the θ angle in equation 9 and after by replacing in equation8. When considering the initial conditions: θ ( t = 0) = 0, and ˙ θ ( t = 0) = 0 + , then the reaction force and the angularvelocity squared are: N A = m A g (3 cos θ − , (10) f ( θ ) = ˙ θ = 2 κ (1 − cos θ ) . (11)It is important to remark that the initial position, θ ( t = 0) = 0, is a unstable equilibrium point. In order to break thesymmetrical evolution of sliding down to any slide of the semi-sphere, it is important to indicate an initial directionof movement.The equations of motion are valid only for the regimen when N A ≥ A moving over thesemi-sphere. The case N A = 0 sets the last contact angle θ (cid:63) that in this case corresponds to:cos θ (cid:63) = 23 → θ (cid:63) (cid:39) . ◦ . (12)This value represents a benchmark to compare with the case of a moving semi-sphere. III. SYSTEM WITH A MOVING SEMI-SPHERE
In this part, we allow the semi-sphere to freely move over a surface without friction. The aim of this freedom isknowing the impact on the value of the last contact angle θ (cid:63) . The physical setup is presented in figure 2. Figure 2. (Left) Physical situation on the object A sliding down over a moving semi-sphere B . The semi-sphere B does notpresent friction with both object A neither the flat surface. Forces acting on A are shown in blue. Forces acting on B are shownin dark pink. Accelerations are shown in orange. (Right) Unit vector basis description. A. Equations of motions
Similarly to the case with fixed semi-sphere, the object A is affected by the reaction force with the semi-sphere andits weight. Then the equation of motion for A is: (cid:126)N A − (cid:126)W A = m a (cid:126)a A . (13)On the other hand, the semi-sphere B is now free to move and therefore its equation of motion is: (cid:126)N B − (cid:126)N A − (cid:126)W B = m B (cid:126)a B , (14)where (cid:126)N B is the reaction force between the flat surface and the semi-sphere and W B = m B g ˆ is its weight. Due tothe semi-sphere is constrained to move horizontally then (cid:126)a B = a Bx ˆ ı . Besides, we need to include the relative motionof the object A over the surface of B : (cid:126)a A = (cid:126)a B + (cid:126)α × (cid:126)r + (cid:126)ω × ( (cid:126)ω × (cid:126)r ) , (15) (cid:126)a A = a Bx ˆ ı + αR ˆ θ − ω R ˆ r , (16)which is a circular motion and it is valid while A is in contact with B . Notice that the origin of the polar basis(ˆ r, ˆ θ ) moves with B and it is in O . Also, it is important to remark that (cid:126)a B = (cid:126)a O because the semi-sphere B is in ahorizontal motion with no rotation.Up this point, the acceleration of A in the polar basis is: (cid:126)a A = (cid:0) a Bx sin θ − ω R (cid:1) ˆ r + ( − a Bx cos θ − αR ) ˆ θ , (17)which allow us to get the full set of equations of motion: N A − m A g cos θ = m A (cid:0) a Bx sin θ − ω R (cid:1) , (18) − m A g sin θ = m A ( − a Bx cos θ − αR ) , (19) − N A sin θ = m B a Bx , (20) N B − N A cos θ − m B g = 0 . (21)After inspection, the last 2 equations lead to: a Bx = − N A sin θm B , (22) N B = m B g + N A cos θ , (23)and those can be used to reduce the full set of equations of motion to two equations: gR cos θ − N A m A R (cid:18) m A sin θm B (cid:19) = f ( θ ) , (24)2 gR sin θ + 2 N A m B R sin θ cos θ = f (cid:48) ( θ ) , (25)where f ( θ ) = ˙ θ and f (cid:48) ( θ ) = 2¨ θ . Notice, we use same relations for the angular acceleration and velocity with respectto the angle θ : α = − ¨ θ , and ω = − ˙ θ . B. Solving the equations of motion
The solution of the system of equation cannot be performed as in the fixed semi-sphere case because the reactionforce N A is present in both equations and it depends on the angle. Nevertheless, the reaction force N A can be isolatedfrom the equations and be written in terms of the dynamical variables: N A ( θ ) = m A R κ cos θ − f ( θ )1 + β sin θ , (26)where β = m A /m B is the ratio between the masses and κ = g/R is the ratio between the acceleration of gravity andthe radius of the semi-sphere. Here the reaction force depends on the angular velocity encoded in f ( θ ).After removing the explicit dependence of N A , we get the differential equation for f ( θ ):1 + β sin θ θ f (cid:48) ( θ ) + β cos θ f ( θ ) − κ (1 + β ) = 0 . (27)This differential equation is analytically solvable and when including the initial condition f (0) = 0, the solution is: f ( θ ) = 2 κ (1 + β )(1 − cos θ )1 + β sin θ . (28)This solution is valid for N A ( θ ) ≥ ≤ θ ≤ θ (cid:63) where θ (cid:63) corresponds to the last contactangle. C. Finding the last contact angle
The angle θ (cid:63) is obtained by solving the following equation: N A ( θ (cid:63) ) = m A R κ cos θ (cid:63) − f ( θ (cid:63) )1 + β sin θ (cid:63) = 0 , (29)where f ( θ (cid:63) ) is the solution of the differential equation evaluated at θ (cid:63) . Due to 0 ≤ θ (cid:63) ≤ π/
2, equation 29 correspondsto a depressed cubic equation with the form: H ( ξ ) = sin (cid:16) π τ (cid:17) ξ − ξ + 2 = 0 (30)where ξ = cos θ (cid:63) and sin (cid:16) π τ (cid:17) = β β . We introduce the τ -parameter ranging 0 ≤ τ ≤ m A /m B such as: β = tan (cid:16) π τ (cid:17) , (31)where at the extremes: τ → β → τ → β → ∞ . In figure 3, we present the function H ( ξ ) forvarious values of τ .The limit of fixed semi-sphere is reached when τ → m A → m B → ∞ .In this limit, equation 30 corresponds to: − ξ + 2 = 0 , (32)with solution ξ = cos θ (cid:63) = 23 . This case agrees with the solution obtained in section II.The other limit, τ →
1, gives the equation: ξ − ξ + 2 = ( ξ + 2)( ξ − = 0 , (33) - - Figure 3. H ( ξ ) versus ξ = cos θ for τ = 0 , ,
1. The intersection of each curve H ( ξ ) with the x -axis produces the solutionwhere ξ = cos θ (cid:63) . with 3 real solutions: ξ = 1, ξ = 1 and ξ = −
2. The only physical solutions are ξ , corresponding to an angle θ (cid:63) = 0. This means when m A → ∞ the angle of last contact between A and B is at the top of the semi-sphere and hap-pens at begging of the movement. In this case, the semi-sphere moves fast enough to not be in contact with the object A .In the intermediate cases, 0 < τ <
1, the solutions for equation 30 corresponds to: ξ = 21 + 2 cos (cid:16) π τ (cid:17) , (34) ξ = √ (cid:16) π τ (cid:17) − sin (cid:16) π τ (cid:17) sin (cid:16) π τ (cid:17) , (35) ξ = − √ (cid:16) π τ (cid:17) + sin (cid:16) π τ (cid:17) sin (cid:16) π τ (cid:17) , (36)(37)which are obtained by solving the depressed cubic equation. However, in order to get the real values the solutions ofequation 30 need to be rephased the roots by factors of e iπ/ when the equation is solved using Vieta’s substitution [9].From the 3 roots, ξ has a physical meaning: ξ = cos θ (cid:63) . The dependence of the last contact angle θ (cid:63) in terms of τ is shown in figure 4. We observe that the solution of the cubic equation include the extreme limits. In addition, thevalue of θ (cid:63) indicates that the largest value corresponds to the fixed semi-sphere case and the lowest corresponds when m A → ∞ or m B → B and the object A get immediatelydetached upon the first contact. IV. CONCLUSIONS
We present the analytical solution for the problem of an object sliding down on a semi-sphere of radius R . Besides,the semi-sphere is on a friction-less surface. The key effect to consider is the reaction force between the object andthe semi-sphere. This force provokes the semi-sphere to move horizontally and the object A to descend down keepingcontact with the semi-sphere. If the velocity of the object A is larger enough that the reaction force between theobject and the surface is null, then both objects detach one of each other. The angle in which that occurs is the lastcontact angle and it was calculated analytically. Figure 4. Last contact angle θ (cid:63) versus τ . Red line shown the functional dependence in τ . Grey line show the fixed-sphere case θ (cid:63) = 48 . ◦ We found that the case of a fixed semi-sphere (equivalent to m B (cid:29) m A ) gives the maximum possible last contactangle among any physical configuration of masses. In addition, we found that this angle depends only on the ratio ofthe masses m A and m B and it is independent of the value of the acceleration of gravity or the radius of the semi-sphere.The problem discussed in the manuscript present a general physical scenario that might be worth to be used as anadvance example in courses of Newtonian mechanics at university level. ACKNOWLEDGMENTS
We thank Nicolas Rojas for useful discussion and comments. RL was supported by Universidad Cat´olica del Nortethrough the Publication Incentive program No. CPIP20180343 and CPIP20200063. [1] F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, and B. P. Self,
Vector Mechanics for Engineers (2021).[2] R. C. Hibbeler,
Engineering mechanics (Pearson Education, Harlow, 2017).[3] W. Riley,