EN4: Dnamics and Vibrations Homework 3: Kinematics and Dnamics of Particles Due Frida Feb 15, 19 School of Engineering rown Universit Please submit our solutions to the MTL coding problems 4, 5, 6 b uploading a SINGLE file, with a.m etension, to Canvas. 1. Simple Newton s law problem. The European standard EN1195 1:1 regulates safe securing of loads for transportation. For a cargo secured b a single top-over lashing (see the figure) the give the following formula for the largest mass that can be transported without slipping on the truck-bed µ F sinα m = a µ g where µ is the coefficient of friction between the load and truckbed; a is the vehicle s acceleration (the formula assumes a > µ g ); F is the tension force in the lashing, α is the angle between the cable and the truck-bed (indicated as 75-9 degrees in the sketch), and g is the gravitational acceleration. The standard also states that for a load with COM at its center, the maimum mass that can be transported without tipping can be calculated from the formula F sinα m = a ( / ) H L g The goal of this problem is to derive these two formulas. j i F m H/ F k m j i F L/ L/ a 1.1 Draw a (D) free bod diagram showing the forces acting on the mass. ssume no slip occurs at the contact between the crate and truckbed.
j i F T mg m H/ T N N 1. Write down F= ma and M = COM for the mass [3 POINTS] Newtons law ( T + T ) i+ ( N + N mg F sin α) j= ma i Sum of moments about COM is zero (no rotation) ( T + T ) H / + ( N N ) L/ k = [ ] 1.3 Hence, find formulas for the reaction forces acting at and (ou will onl be able to calculate the sum of the horizontal forces acting at and see the people mover eample from class) We have 3 equations ( T + T ) = ma N + N mg F sinα = ( T + T) H / + ( N N) L/= We can solve these to see that N + N = mg + F sinα T + T = ma N = ( mg + F sin α ma ( H / L)) / N = ( mg + F sin α + ma ( H / L)) / [3 POINTS] 1.4 Use our solution to 1.3 to derive the two formulas in the European standard. t the onset of slip
T = µ N T = µ N T + T = µ ( N + N ) = µ ( mg+ Fsin α) µ ( mg + F sin α) = ma µ F sinα m = a µ g t the onset of tipping N = ( mg + F sin α ma ( H / L)) = F sinα m = a ( H / L ) g. Newtons laws in Normal-Tangential Coordinates. Suppose that the Virgin Hperloop vehicle (see HW) travels at constant speed 18 km/hr around a curve with constant radius R. n α k.1 If the magnitude of the vehicle s acceleration cannot eceed.g (g=9.81 m/s is gravitational acceleration) what is the smallest allowable value for the curve radius R? n t The vehicle is in constant speed circular motion around the track. The magnitude of the acceleration is therefore V / R. If V / R=.g it follows that (18 / 3.6) Rmin = = 45.9km (. 9.81) V [1 POINT]. Curved track will be banked at an angle α to minimize the lateral forces acting on the vehicle (and passengers inside the vehicle). Draw a free bod diagram showing the forces acting on the vehicle (ou can neglect air resistance). n k α n t T mg T N N [The friction forces can go upwards as well - since there is no slip either direction is fine.
.3 ssume that the vehicle travels at 18 km/hr around a curve with the minimum allowable radius calculated in 7.1. Find formulas for the reaction forces acting on the vehicle normal and tangent to the banked track, in terms of the mass m of the vehicle and the bank angle α. (You onl need to find the sum of the forces acting on the two wheels; there is no need to calculate the forces at and separatel) F=ma gives mv [( T + T)cos α + ( N + N)sin α] n+ [ ( T + T)sin α + ( N + N)cos α mg] k = n=.mgn R The n and k components give ( T + T)cos α + ( N + N)sinα =.mg ( T + T)sin α + ( N + N)cosα = mg N + N = mg(.sinα + cos α) T + T = mg(. cosα sin α) [GRDERS: it is not necessar to use n-k coordinates to solve this problem; coordinates normal and tangent to the track are also OK as long as the gravit and acceleration vectors are correctl epressed as components in the new basis vectors. People will get a sign in front of the tangential forces if the chose to make them act up the slope instead of down].4 Find the bank angle that will ensure that the force acting tangent to the track is zero. The force is zero if tanα =. α = 11.3 [1 POINT] 3. In this problem we will analze the motion of the first stage of the Falcon-9 launch vehicle To keep the calculation simple, we will assume that The mass of the rocket decreases with time as propellant is ejected, and can be calculated from the formula m= M µ t where M and µ are constants. ssume a vertical trajector and include gravit Neglect air drag ssume that the distance traveled b the vehicle (t) satisfies the differential equation d ( M µ t) = T ( M µ t) g dt (this looks like F=ma, but it is derived b considering the rate of change of momentum of the rocket and propellant appling Newton s law to an object that does not have constant mass is a bit trick).
3.1 Write a MTL Live Script to solve the differential equation for, (using the dsolve function) d with initial conditions = = at time t=. Please upload our script to CNVS. dt The live script gives This can be simplified to { } = Tt/ µ gt / + ( T / µ )( M µ t)log ( M µ t)/ M [3 POINTS] 3. Hence (using our Live Script ) plot the speed as a function of time. Use the following values for parameters: Total initial mass M=575kg Propellant mass flow rate µ = 5 kg / s 6 Thrust T = 7.6 1 N Time interval <t<16 sec. You can compare our prediction with the data from HW, if ou are curious. For eample, the velocit at the end of the first stage burn was measured to be 1848 m/s. Falcon 9 velocit 15 Speed (m/s) 1 5 4 6 8 1 1 14 16 Time (s)
4. We can set up a more sophisticated model of the Falcon-9 b using the MTL ode solver to integrate the equations of motion. For the complete calculation, assume that The mass of the rocket can be calculated from the formula 55 5t t < 165s m= 1375 63( t 165) 165 < t < 584s 7 t > 584s The mass is in kg, and the formula includes both stages of the rocket The thrust varies with time as 6 7.6 1 t < 165s 6 T = 1.6 1 165 < t < 584s t > 584s drag force F = 5ρv D (in Newtons) acts on the rocket, where v is the rocket s speed in m/s, and ρ is the air densit, 3 which varies with altitude as ρ = 1.ep( / 8.5 1 ) kg m -3, where is in m. ssume as an approimation that the trajector is vertical, and neglect variations in gravitational force. 4.1 Show that the equation of motion d ( M µ t) = T ( M µ t) g 5ρv dt can be written in the form required b the MTL ode solver as follows d v dt v = ( T 5 ρv ) / m g dv d Using a = and v = and rearranging the result gives the equations stated. [1 POINT] dt dt 4. Write a MTL script that will calculate and v for a time interval <t<7s. Plot the variation of speed v with time, and compare the prediction with the eperimental data (use the data file from HW, and plot the prediction and eperiment on the same graph) 8 Falcon 9 velocit 7 Prediction Measurement 6 5 Speed (m/s) 4 3 1 1 3 4 5 6 7 8 Time (s) [4 POINTS]
5. Dnamic Kingdon trap is a device for storing charged particles (see e.g. this patent). It consists of a wire that runs down the ais of a clinder. time varing electric field is applied to the wire, which subjects the particle to a radial force 1 F= qv ( + VcosΩt) e r πεr Here, V, V are a static and fluctuating voltage applied between the wire and clinder; Ω is its frequenc, q is the charge of the trapped ion, and ε is the permittivit of free space. The goal of this problem is to analze the motion of an ion in the trap. We will assume that the particle moves in the r, θ plane, and so must solve for [ r, θ ] as a function of time e θ ion j Wire r Clinder θ i e r 5.1Write down Newton s laws for the particle (use polar coordinates) and show that the er, e components of the equation can be re-arranged into the form required b matlab b re-writing them as r v d θ ω = dt v rω + ( λ + µ cos Ωt) / r ω vω / r Find formulas for λµ, in terms of V, V, ε, qm,. Newton s law in polar coordinates gives 1 d r dθ d θ dr dθ qv ( + VcosΩ t) er = m r er + m r + e θ πεr dt dt dt dt dt d r dv qv q V 1 d d v = = cos t r + Ω + ω = = dt m dt π ε πmε r dt dt r θ ω ω So if we define λ = qv / ( πmε ) µ = q V / ( πmε) we obtain the formulas stated. 5. Write a MTL code that will calculate [ r, θ,, v ω ]. For the parameter values listed below, plot the path of the particle, for a time period < t < 5 time units. You can use the following plot function to do the polar plot for ou polarplot(theta,r,'displaname','(problem a)','color',[1,,]); where theta is a vector of values of θ and r is a vector of values of r at successive time intervals (these can be etracted from the matri that stores values of [ r, θ,, v ω ] returned b ode45). θ
(a) λ = 1, µ =, Ω= (this is a static trap the electric field does not var with time), with initial conditions r = 1, θ =, v=, ω = 1. For this case ou will need to specif the time intervals ou would like in the solution, otherwise ode45 will take huge time steps and the trajector will look funn. You can do this, eg, with [time,sols] = ode45(@(t,w) trap(t,w,lambda,mu,omega),[:.1:5],initial_w); Here, the :.1:5 tells MTL ou would like solution values ever.1 time intervals. (b) λ = 1, µ =, Ω= with initial conditions r =.8 θ =, v=, ω =.4. (use hold on to put this on the same plot as (a). The static trap is stable a particle remains trapped, as long as nothing slows it down. (c) λ = 1, µ = 8, Ω= (this is a dnamic trap the electric field varies with time) with initial conditions r =.5 θ =, v=, ω =.5. Plot this case on the same graph as (a) and (b). Particles with the right radius and velocit are trapped. (d) λ = 1, µ = 8, Ω= with initial conditions = 1.5, =, v =, v =.5. Run this case for 1 time units, and plot it on a new figure. dnamic trap will epel particles that are too fast or slow, and are too close or too far from the wire 9 9 1 4 6 1 6 15 3 3 15 15 1 3 1 5 (Case a) 18 (Case b) (Case c) 18 1 33 1 33 4 7 3 4 7 3 [4 POINTS] j i m D H
6. The figure shows an idealization of a person with mass m on a zip-line. The person starts at end of the line and travels to end, a horizontal distance D and vertical distance H below. The cable is idealized as a spring with stiffness k and un-stretched length L, with L > D + H. The tension in the cable is related to its stretched length b T= kl ( L ). 6.1 Show that the length of the cable is L D H = + + ( ) + ( ) (this is just geometr) E D- H C -H ppl Pthagoras to triangles CE and CE to get the two answers. 6. Draw a free bod diagram showing the forces acting on the mass (include the cable tension, which is 1 the same on both sides of the mass). ssume that an air drag force with magnitude FD = ρ CDV and direction opposite to the velocit vector acts on the rider. Here, ρ is the air densit, is the frontal area of the rider; The FD is shown below C D is the drag coefficient and V = v + v is the rider s speed. θ 1 T T θ m mg FD 6.3 Use Newton s law to derive equations of motion for the mass, and show that the can be re-arranged in the following form (suitable for the MTL ODE solver) v d1 = + v d = d = ( D ) + ( H) dt v ( T / m)( D )/ d ( T / m) / d1 cvv 1 v g ( T / m)( H)/ d ( T / m) / d c C 1 cvv = D m ρ 1 (i) Note that the drag force can be epressed as a vector as FD = ρ CDV ( v + v )/ V i j
(ii) ppl Newton s law dv dv ( T cosθ1+ T cos θ) i+ ( mg T sinθ1 T sin θ) j+ FD = m i+ j dt dt (iii) Trig shows that cos θ = / + cos θ = ( D ) / ( D ) + ( H) 1 sin θ = / + sin θ = ( H) / ( D ) + ( H) 1 Collecting (1,,3) and using v = d / dt v = d / dt gives the solution Use ode45 to solve the equation. Use the following values for parameters Rider mass m 8kg Dimensions D=4m, H=3m Drag parameters: =1m, ρ = 1. kg m -3 C =. Line stiffness k=1 6 Nm -1 and unstretched length Initial position D L = D + H + m =, = ( L D H ) / ( L H), v = v = [3 POINTS] dd an event function to our script that will stop the calculation when the rider reaches =D. Hence, plot (i) The trajector of the rider; and (ii) a graph of the riders speed as a function of time. You can compare our predictions to data for a number of ziplines here Trajector 5 Speed 1 5 15 (m) -5 Speed (m/s) 1-1 5-15 5 1 15 5 3 35 4 (m) 5 1 15 5 Time (s) [4 POINTS] 6.4 The rider arrives at =D at over 1 m/s. How would ou change the design to reduce the speed to below 7 m/s (but without changing H or D)? Eperimenting with the MTL (that s what this sort of design simulation is useful for ou don t have to design our ride b trial and error) shows that the rider slows down if the unstretched length of the cable is increased. n unstretched length L = D + H + 5.5 m would work. Making the cable too slack is risk, because riders might not make it to the end.
5 Speed 15 Speed (m/s) 1 5 5 1 15 5 Time (s) Other slightl less straightforward solutions include reducing the stiffness of the cable k (but it has to be reduced a lot, and the resulting cable stretch is a problem); restricting the ride to people with ver low mass or large frontal area to increase air drag. You can of course also add a brake to the sstem but that s a lot more epensive than just adding a couple of meters of cable.