Chapter 1 HW Solution

ME 516 Spring 2005 Chapter 1 HW Solution

Problem 4. Using ﬁxed Cartesian coordinates, the acceleration of the particle is given by

a =x ¨i +y ¨j (1)

The approach is to integratex ¨ to getx ˙ then x; then gety ˙,y ¨. Use (1.3.15) to get radius of curvature ρ. Sincex ¨(t) = 1 + 0.2t, integration plus initial conditions yields

x˙(t) = t + 0.1t2 (2) t2 0.1t3 x(t) = + − 0.6 (3) 2 3

Then since the constraint equation is y = −x2 + x, we have (using the chain rule)

dy dx y˙(t) = = (1 − 2x)x ˙ (4) dx dt d d dx d dx˙ y¨(t) = [(1 − 2x)x ˙] = [·] + [·] = −2x ˙ 2 + (1 − 2x)¨x (5) dt dx dt dx˙ dt

Evaluate bothx ¨(t) andy ¨(t) at t = 0.5, and you get

a =x ¨i +y ¨j = 1.1i + 1.5845j (6)

Radius of curvature is given by (note y0 = dy/dx and y00 = d2y/dx2)

3/2 1 + y02 ρ = = 5.209 (7) |y00|

Problem 8. (a) Here we have

hθ z = (8) 10π which gives position vector hθ r = Re + k (9) R 10π Note that R is constant, and one can either diﬀerentiate r with respect to time or use text equations [1.3.47] and [1.3.48] to ﬁnd velocity and acceleration. The results are

hθ˙ v = Rθ˙e + k (10) θ 10π

hθ¨ a = −Rθ˙2e + Rθ¨e + k (11) r θ 10π

(b) The normal unit vector is opposite the radial unit vector,while the tangent unit vector is upward along the vehicle path, so

en = −er (12)

et = cos φeθ + sin φk (13)

1 ME 516 Spring 2005 where

h φ = tan−1 (14) 10πR

Radius of curvature ρ is found using the fact that along a curved path the normal acceleration is related to radius of curvature and velocity.

v2 v · v h2 ρ = = = R 1 + 2 2 (15) an Rθ˙2 100π R

Problem 14. Apply F = ma using acceleration in spherical coordinates. A sketch of the system and the free-body diagram is shown below.

Figure 1: Sketch of spherical pendulum and free-body diagrams.

The resultant force F is

F = mg cos βer + mg sin βeφ − k(L − Lo)er = ma (16) where a is expressed in spherical coordinates using [1.3.62]. Probably the trickiest issue here is changing from φ to β. The resulting equations of motion are:

˙2 2 ˙2 er : m(L¨ − Lθ sin β − Lβ ) + k(L − Lo) − mg cos β = 0 (17) ¨ ˙ ˙ ˙ eθ : m(Lθ sin β + 2L˙ θ sin β + 2Lθβ cos β) = 0 (18)

¨ ˙ ˙2 eφ : m(Lβ − 2L˙ β + Lθ sin β cos β) + mg sin β = 0 (19)

If one were given numerical values for the parameters and initial conditions, one could numerically integrate the equations of motion to simulate the motion of this spherical pendulum.

2 ME 516 Spring 2005

Problem 18. The path of the bead is constrained by z = x2/4. Since the constraint equation is expressed in ﬁxed Cartesian coordinates, that’s the coordinate system I’ll use in the problem. My free-body diagram is shown in Figure 2 below.

Figure 2: Free-body diagram of the bead sliding on the wire.

Force balances in the x and z directions yield (As shown in the ﬁgure, I drew my free-body diagram with normal force N pointing toward the center of curvature, and θ is the angle of the tangent)

− N sin θ = mx¨ (20) N cos θ − mg = mz¨ (21)

Eliminating N yields equation of motion mx¨ cos θ mz¨ + + mg = 0 (22) sin θ The slope of the path θ is given by dz x tan θ = = (23) dx 2 We need to eliminate both z and θ. Diﬀerentiating z we get 1 z˙ = xx˙ (24) 2 1 z¨ = x˙ 2 + xx¨ (25) 2

Substitutingz ¨ and tan θ in the equation of motion, we get

x¨ x2 + 4 + xx˙ 2 + 2gx = 0 (26)

This is a surprisingly complex equation of motion for such an apparently simple problem (in my opinion).

Problem 29. I used curvilinear coordinates for this problem (one could also use cylindrical, but since the problem is planar z is not needed). My coordinates and a free-body diagram are shown below.

Figure 3: Coordinates and free-body diagram for Problem 29.

3 ME 516 Spring 2005

A force balance in the normal direction yields

mv2 mg cos θ − N = (27) R Contact is lost when N = 0, thus

v2 cos θ = (28) gR

The normal force N does no work (it’s always perpendicular to path) and there is no friction, hence this is a conservative system. From conservation of energy we have 1 1 T + V = T + V =⇒ mv2 + mgR = mv2 + mgy (29) 1 1 2 2 2 o 2 where the xy coordinate system has its origin at the center of the cylinder, so

y = R cos θ (30)

Solving for v from (29) yields

2 2 v = vo + 2gR − 2gR cos θ (31)

From (28) we have

2 2 gR cos θ = v = vo + 2gR − 2gR cos θ, (32) hence

v2 + 2gR θ = cos−1 o (33) 3gR