ex mathcad 5th dynmeriam

Sample Problems fromSolving Dynamics Problems in Mathcad

by Brian D. HarperOhio State University

Solving Dynamics Problems in Mathcad is a supplement tothe textbook Engineering Mechanics: Dynamics (5thEdition) by J.L. Meriam and L.G. Kraige, Wiley, 2001.

2 Sample Problems Mathcad

2.2 Problem 2/94 (Rectangular Coordinates)

A projectile is fired into an experimental fluid attime t = 0. The initial speed is vo and the angle tothe horizontal is θ. The drag on the projectileresults in an acceleration term aD = -kv, where kis a constant and v is the velocity of theprojectile. Determine the x- and y-components ofboth the velocity and the displacement asfunctions of time. What is the terminal velocity?For the case vo = 40 m/s and θ = 60°, plot (on asingle graph) the trajectory (y versus x) of theprojectile for k = 0.2, 0.4 and 1 sec-1. Include theeffects of gravitational acceleration.

Problem Formulation

The x and y components of the acceleration are:

= ax −k vx and = ay − − g k vy

These equations can be integrated to yield the veloc

= ay

dvy

dt = − − g k vy .

Rearranging terms yields = dt −dvy

+ g k vy. The righ

from = vy0 v0 ( )sin θ to vy , thus

= t − d⌠

⌡

v0 ( )sin θ

vy

1 + g k s s .

Mathcad will be used to solve this equation for vy a

kge

kgvv kt

y −

+= −θsin0

vy will then be integrated over time to yield y.

ities. To illustrate, consider

t hand side is to be integrated

s a function of time, yielding


( ) tkge

kgv

ky kt −−

+= −1sin1

0 θ

An analogous approach is used for the x components yielding,

ktx evv −= )cos( 0 θ ( )kte

kv

x −−= 1cos0 θ

Before proceeding, let's observe that the terminal velocity, which by definition isconstant, can be determined by setting the acceleration equal to zero. Thus, the

components of the terminal velocity are = vx 0 and = vy −gk . In other words, at

long times the projectile will be moving down at a constant velocity of gk .

Mathcad Worksheet

Symbolic calculations

t

v0 sin θ( )⋅

vy

s1−

g k s⋅+

⌠⌡

d solve vy,exp t− k⋅( ) g⋅ exp t− k⋅( ) v0⋅ sin θ( )⋅ k⋅ g−+( )

k→

Now we copy and paste the result for vy into the following integral, changing tback to the dummy integration variable s.

y

0

t

sexp s− k⋅( ) g⋅ exp s− k⋅( ) v0⋅ sin θ( )⋅ k⋅ g−+( )

k

⌠⌡

d:=

t

yg− v0 sin θ( )⋅ k⋅− g t⋅ k⋅ exp t k⋅( )⋅−( )

k2exp t− k⋅( )⋅

g v0 sin θ( )⋅ k⋅+( )k2

+→

t

v0 cos θ( )⋅

vx

s1−

k s⋅⌠⌡

d solve vx, exp t− k⋅( ) v0⋅ cos θ( )⋅→

x0

tsexp s− k⋅( ) v0⋅ cos θ( )⋅

⌠⌡

d:=t


x1−

kexp t− k⋅( )⋅ v0⋅ cos θ( )⋅

1k

v0⋅ cos θ( )⋅+→

Numerical results

Results for x and y are copied and pasted from the symbolic results above.

g 9.81:= v0 40:=

θ 60π

180⋅:=

y k t,( )g− v0 sin θ( )⋅ k⋅− g t⋅ k⋅ exp t k⋅( )⋅−( )

k2exp t− k⋅( )⋅

g v0 sin θ( )⋅ k⋅+( )k2

+:=

x k t,( )1−

kexp t− k⋅( )⋅ v0⋅ cos θ( )⋅

1k

v0⋅ cos θ( )⋅+:=

The trajectory of the projectile can be obtained by plotting y as a function of x. Since we do nothave y explicitly as a function of x, we must use a parametric plot. Note in the following thatthree different functions are entered on both the horizontal and vertical axes.

t 0 0.05, 6..:=

0 10 20 30 40 50 60 70 800

10

20

30

40

50trajectory of projectile

y .2 t,( )

y .4 t,( )

y 1 t,( )

x .2 t,( ) x .4 t,( ), x 1 t,( ),


3.6 Problem 3/358 (Curvilinear Motion)

The 26-in. drum rotates about a horizontal axiswith a constant angular velocity Ω = 7.5 rad/sec.The small block A has no motion relative to thedrum surface as it passes the bottom position θ =0. Determine the coefficient of static friction µsthat would result in block slippage at an angularposition θ; plot your expression for 0 ≤ θ ≤ 180°.Determine the minimum required coefficientvalue µmin that would allow the block to remainfixed relative to the drum throughout a fullrevolution. For a friction coefficient slightly lessthan µmin, at what angular position θ wouldslippage occur?

Problem Formulation

From the free body and mass acceleration diagrams wehave,

[ ]nn maF =Σ 2cos Ω=− mrmgN θ

[ ]tt maF =Σ 0sin =− θmgF

For impending slip we have NF sµ= . Substituting Finto the above and solving gives,

θθ

θθ

µcos8925.1

sincos

sin2 +

=Ω+

=rg

gs

The last two questions can be answered only after plotting µ
s as a function of θ.


Mathcad Worksheet

Given

N m g⋅ cos θ( )⋅− m r⋅ Ω2

⋅

µs N⋅ m g⋅ sin θ( )⋅− 0

Find µs N,( )g

sin θ( )

g cos θ( )⋅ r Ω2

⋅+( )⋅

m g⋅ cos θ( )⋅ m r⋅ Ω2

⋅+

→

g 32.2:= r1312

:= Ω 7.5:=

µs θ( ) gsin θ( )

g cos θ( )⋅ r Ω2

⋅+( )⋅:=

θ 0 0.02, π..:=

0 30 60 90 120 150 1800

0.2

0.4

0.6

0.8coefficient of static friction

(degrees)

µs θ( )

θ180π

⋅


If the block is not to slip at any angle θ, the coefficient of friction must be greaterthan or equal to any value shown on the plot above. Thus, the minimum requiredcoefficient value µmin that would allow the block to remain fixed relative to thedrum throughout a full revolution is equal to the maximum value in the plotabove. The location where this maximum occurs can be found by solving theequation 0/ =θµ dd s for θ. This θ can then be substituted into µs to yield therequired value for µmin. Here’s how we do this with Mathcad.

xµs x( )d

d0 solve x,

2.1275232852017454951−

2.1275232852017454951

→

µs 2.1275( ) 0.622=

From the above we see that µmin = 0.622. If µs is slightly less than this value, theblock will slip when θ = 2.128 rads (121.9°).

6.6 Problem 6/204 (Impulse/Momentum)

Determine the minimum velocity v that thewheel may have and just roll over theobstruction. The centroidal radius of gyration ofthe wheel is k, and it is assumed that the wheeldoes not slip. Plot v versus h for three cases: k =½, ¾, and 1 m. For each case take r = 1 m.

Problem Formulation

During Impact: Conservation of Angular Momentum

As usual, we neglect the angular impulse of theweight during the short interval of impact. With thisassumption we have conservation of angularmomentum about point A. Immediately beforeimpact, the center of the wheel is not moving in acircular path about A and we need to use the formulafor general plane motion. Note that 2mkI = .

( ) ( )hrmvrvmkhrmvmkdvmIH A −+=−+=+= 22ωω


We will use primes to denote the state immediately after impact. Since the wheelnow rotates about A we can use the simpler formula ωAA IH = . Note that, by theparallel axis theorem, 2mrII A += ( )22 rkm += .

( )rvrkIH AA′

+=′=′ 22ω

Setting ′= AA HH and solving yields,

+−=′

221rk

rhvv

After Impact: Work-Energy

( ) mghIVT Ag +′−==∆+∆ 220210 ω

( ) mghrvrkm =

′

+2

22

21

Substituting the result for v′ into the aboveequation followed by simplification yields,

( )rhrkrkghr

v−+

+= 22

222

Mathcad Worksheet

Ibar m k2⋅:= k IA m k2 r2+( )⋅:= r

HA Ibarvr

⋅ m v⋅ r h−( )⋅+:= h HAp IAvpr

⋅:=r

HA HAp solve vp, vk2 r2 r h⋅−+( )

k2 r2+( )⋅→

vp vk2 r2 r h⋅−+( )

k2 r2+( )⋅:=h


eqn12

IA⋅vpr

2

⋅ m g⋅ h⋅:= h

eqn solve v,

2 r2⋅ g⋅ h⋅ 2 g⋅ h⋅ k2⋅+( )12 r

k2 r2 r h⋅−+( )⋅

2 r2⋅ g⋅ h⋅ 2 g⋅ h⋅ k2⋅+( )12

−r

k2 r2 r h⋅−+( )⋅

→

g 9.81:= r 1:=

Now we copy and paste the first solution above.

v k h,( ) 2 g⋅ h⋅ r2⋅ 2 g⋅ h⋅ k2⋅+( )12

r

k2 r2 r h⋅−+( )⋅:=

h 0 0.01, 1..:=

0 0.5 10

5

10

15

20velocity (m/s)

(meters)

v12

h,

v34

h,

v 1 h,( )

h

ex mathcad 5th dynmeriam

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