# 20090916

October 13, 2017 | Author: Elvio Junges | Category: Acceleration, Velocity, Coordinate System, Physical Phenomena, Theoretical Physics

dynamics...

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Dynamics 4600:203

Homework 02 Due: September 16, 2009 Name:

Please denote your answers clearly, i.e., box in, star, etc., and write neatly. There are no points for small, messy, unreadable work. . . please use lots of paper. Problem 1: Hibbeler, 12–20 A stone A is dropped from rest down a well of depth 80 ft, and in 1 s another stone B is dropped from rest. Determine the time interval between the instant A strikes the water and the instant B strikes the water. Also, at what speed do they strike the water?

B y A

ℓ = 80 ft

ˆ

ˆı

Solution: The time interval required for particles a and B to fall the depth of the well is identical. Therefore, if B is released 1 s after A, then B will strike the water 1 s after A, regardless of how long it actually takes to travel that distance. We describe the position of the particle A with respect to the ground with the coordinate y, so that its acceleration is defined as aA = −¨ y ˆ. A free-body diagram for the particle is shown to the right, and the gravitational force is the only force acting on the particle as it falls. Therefore, the equation of motion reduces to X F = −m g ˆ = −m y¨ ˆ = m aA ,

−m g ˆ

which yields, in the ˆ direction, y¨ = g. Notice that the gravitational force produces a positive second-derivative. Can you explain why? From the equation of motion, with y(0) ˙ = 0 and y(0) = 0, the velocity and position of the particle become g t2 y(t) ˙ = g t, y(t) = . 2 The particle strikes the water when y(tf ) = ℓ = 80 ft, which occurs at time s 2ℓ tf = = 2.23 s, g √ 2 g ℓ = 71.8 ft/s. The velocity at this instant is   vA (tf ) = −y(t ˙ f ) ˆ = − 71.8 ft/s ˆ .

and at this instant y(t ˙ f) =

1

Problem 2: Hibbeler, 12–23 The acceleration of a rocket traveling upward is given by a = (6 + 0.02 s) m/s2 , where s is the distance traveled in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0. Solution: Note that we have the acceleration as a function of position, so that: Z s Z v (6 + 0.02 s) ds, v dv = 0

0

v2 2

=

6 s + 0.01 s2 ,

and so that solving for the speed v=

p

12 s + 0.02 s2 .

To determine the position as a function of time, we integrate again: Z s Z t ds , dτ = v(s) 0 0 Z s √ Z s ds ds √ √ = 50 t = 2 12 s + 0.02 s 600 s + s2 0 0 To solve this integral we use the following Z hp 1 ai √ + C. du = ln a u + u2 + u + 2 a u + u2 With this, the above integral reduces to i  √  hp t = 50 ln 600 s + s2 + s + 300 − ln(300) . Therefore, the time required to travel an altitude of s = 100 m is t = 5.62 s . Problem 3: Hibbeler, 12–47 The v − t graph for the motion of a train as it moves from station A to station B is shown. Draw the a − t graph and determine the average speed and the distance between the stations.

x˙ (ft/s)

40

30

Solution: 2

90

120

t (s)

The displacement of the train along its direction of travel can be described with the coordinate x, so that its velocity and acceleration are given as x˙ and x ¨ respectively. For the given x−t ˙ graph, the acceleration is constant from 0 ≤ t < τ1 = 30 s as the train moves from rest to velocity v⋆ = 40 ft/s, with a value x ¨ = a0 ≡

x ¨ a0

4 v⋆ = ft/s2 . τ1 3

τ2

T

τ1

For the interval of constant velocity from τ1 = 30 s ≤ t < τ2 = 90 s, the acceleration is zero, and for the final interval the acceleration is x ¨ = −a0 . The x ¨ − t graph is shown to the right.

t (s)

−a0

The distance d1 traveled by the train during the initial acceleration and final deceleration are identical, and determined to be v⋆ τ 1 = 600 ft. d1 = 2 During the middle interval of duration ∆t = τ2 − τ1 ≡ 60 s and constant velocity v⋆ , the train covers a distance d2 = v⋆ ∆t = 2400 ft. Therefore the total distance s between stations is s = 2 d1 + d2 = 3600 ft . As a result, the average velocity is x˙ average = 30 ft/s . Problem 4: Hibbeler, 12–81 Show that if a projectile is fired at an angle θ from the horizontal with an initial velocity v0 , the maximum range the projectile can travel is given by Rmax = v02 /g, where g is the acceleration of gravity. What is the angle θ for this condition? Solution: The position of the ball is described by rP O = x(t) ˆı +y(t) ˆ. When subject to gravity, and initial velocity,

v0

(x(t), y(t))

ˆı + y(0) vP = x(0) ˙ ˙ ˆ = v0 (cos θ ˆı + sin θ ˆ) , the equations for the motion of the particle are written as x(t) = (v0 cos θ) t,

y(t) = −g

t2 + (v0 sin θ) t. 2

Therefore, when y is written as a function of x, this reduces to   v02 sin(2 θ) sin θ gx g 2 x − . x + x = y(x) = − 2 2 v0 cos2 θ cos θ v02 (1 + cos(2 θ)) g 3

The range occurs when y = 0, or solving for x R=

v02 sin(2 θ) . g

The maximum range thus occurs for 2 θ = 90◦ , or θ = 45◦ ,

−→

Rmax =

Problem 5: Hibbeler, 12–85 The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity vA at which it was launched, the angle of release θ, and the height h.

v02 . g

See text for figure.

Solution: We measure the position of the ball as rAO = x(t) ˆı + y(t) ˆ with respect to the base of the catapult. Because the ball is launched at a height a = 3.5 ft above the ground, the initial position is given as x(0) = 0 ft,

y(0) = a = 3.50 ft.

In addition, the initial velocity is given as vA (0) = v0 (cos θ ˆı +sin θ ˆ), so that the initial vlaues of x˙ and y˙ are x(0) ˙ = v0 cos θ,

y(0) ˙ = v0 sin θ,

where v0 is the initial speed and θ is the initial inclination of the trajectory. With these the velocity and position of the ball can be written as x(t) ˙ = v0 cos θ,

y(t) ˙ = −g t + v0 sin θ, g y(t) = − t2 + (v0 sin θ) t + a. 2

x(t) = (v0 cos θ) t,

When the ball reaches its maximum height at t = tf = 1.5 s, the following conditions hold x(tf ) = b = 18 ft,

y(t ˙ f ) = 0 ft/s,

and y(tf ) = h. We can use these to find that v0 cos θ =

b = 12 ft/s, tf

v0 sin θ = g tf = 48.3 ft/s.

From these, the initial speed v0 and inclination θ are v0 =

s

b tf

2

2

+ (g tf ) = 49.8 ft/s ,

4

tan θ =

g t2f = 4.02 −→ θ = 76.0◦ . b

Finally, we can evaluate y(t) at the final time to find that y(tf ) = h =

g 2 t + a = 39.7 ft . 2 f

Problem 6: Hibbeler, 12–110 The Ferris wheel turns such that the speed of the passengers is increased by v˙ = (4 t) ft/s2 , where t is in seconds. If the wheel starts from rest when θ = 0◦ , determine the magnitude of the velocity and acceleration of the passengers when the wheel turns θ = 30◦ .

r = 40 ft

Solution: The velocity and acceleration of the passengers can be described in terms of normal and tangential coordinates as ˆt , vP = v e

ˆt + aP = v˙ e

v2 ˆn , e ρ

where v is the given speed and the radius of curvature is simply the radius of the wheel, i.e., ρ = r = 40 ft. When the wheel starts from rest the speed can be determined to be Z t v(τ ˙ ) dτ = (2 t2 ) ft/s, v(t) = v(0) + 0

while the distance s that the passengers travel is given by s˙ = v, so that   Z t 2 3 s(τ ˙ ) dτ = s(t) = s(0) + ft. t 3 0 In terms of the angle θ this distance can also be expressed as s(t) = r θ(t), so if during a time t = tf the wheel turns by an angle θ(tf ) = π/6 rad = 30◦ , the time tf required can be determined as 1/3    3 r θ(tf ) 2 3 ft, −→ tf = t = 3.16 s. r θ(tf ) = s(tf ) = 3 2 Finally, at this time the speed becomes v(tf ) = 19.9 ft/s, while v(t) ˙ = 12.6 ft/s2 , so that the magnitude of the velocity and acceleration are

kvP (tf )k = v(tf ) = 19.9 ft/s ,

kaP (tf )k =

5

s

v˙ 2 (tf ) +



v 2 (tf ) r

2

= 16.0 ft/s ,

Problem 7: Hibbeler, 12–145 A truck is traveling along the horizontal circular curve of radius r = 60 m with a speed of 20 m/s which is increasing at 3 m/s2 . Determine the truck’s radial and transverse components of acceleration. P rP O

O

Solution: Given the path of the truck, it is natural to describe its position in terms of polar coordinates, so that rP O vP aP

ˆθ e

ˆr , = re ˆ + r θ˙ e ˆ , = r˙ e θ  r   2 ˙ ˆr + r θ¨ + 2 r˙ θ˙ e ˆθ . = r¨ − r θ e

ˆ ˆr e θ ˆı

P rP O

With constant radius of the curve, r˙ = 0 and r¨ = 0, so that the kinematics reduce to rP O vP aP

ˆ , = re  r ˆθ , = r θ˙ e     ˆr + r θ¨ e ˆθ . = −r θ˙2 e

O

The speed of the truck, 20 m/s, is given in terms of the coordinates as kvP k = r θ˙ = 20 m/s, while its rate of change is

d (kvP k) = r θ¨ = 3 m/s2 . dt

Notice that the rate of change of the speed is different from the magnitude of the acceleration. From these, we can determine θ˙ and θ¨ as 1 θ˙ = rad/s, 3

1 θ¨ = rad/s2 . 20

Finally, with these values the acceleration of the truck can be written as        20 ˆθ = − m/s2 e ˆr + 3 m/s2 e ˆθ . ˆr + r θ¨ e aP = −r θ˙2 e 3 6

Problem 8: Hibbeler, 12–149 The slotted link is pinned at O, and as a result of the constant angular velocity θ˙ = 3 rad/s it drives the peg P for a short distance along the spiral guide r = (0.4 θ) m, where θ is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m.

See text for figure.

Solution: a) In terms of the polar coordiantes r and θ, velocity and acceleration of the peg are       ˆr + r θ˙ e ˆr , ˆr + r θ¨ + 2 r˙ θ˙ e ˆr . vP = (r) ˙ e aP = r¨ − r θ˙2 e While behavior of the angular coordinate is given directly as θ˙ = 3 rad/s,

the radial coordinate r is given in terms of θ. However, r˙ can be determined through the chain rule as r˙ =

dr dt

= =

dr dθ , dθ dt (0.4 m) (3 rad/s) = 1.2 m/s.

Likewise, r¨ = 0 m/s2 , so that when the particle leaves the slot at r = 0.5 m, the velocity and acceleration of the particle are   ˆr + r θ˙ e ˆr = (1.2 m/s) e ˆr + (1.5 m/s) e ˆr , vP = (r) ˙ e

      ˆr + r θ¨ + 2 r˙ θ˙ e ˆr , ˆr + 7.2 m/s2 e ˆr = −4.5 m/s2 e aP = r¨ − r θ˙2 e At the instant the particle leaves the slot, its speed is v = 1.92 m/s and the magnitude of its acceleration is a = 8.49 m/s2 .

7