978-0073380308 Chapter 5 Solution Manual Part 28

Problem 5.179

A Pelton impulse wheel, as shown in Fig. P5.136(a), is typically found in hydroelectric power plants and

consists of a wheel with a series of buckets attached at the periphery. As shown in Fig. P5.136(b), water

jets impinge on the buckets and cause the wheel to spin about its axis (labeled

O

). Let

vw

and

.Pm

f/nz

be

the speed and the mass ﬂow rate of the water jets at the nozzles (the nozzles are stationary), respectively.

As the wheel spins, a given water jet will impinge on a given bucket only for a very small portion of the

bucket’s trajectory. This fact allows us to model the motion of a bucket relative to a given jet (during the

time the bucket interacts with that jet) as essentially rectilinear and with constant relative speed, as was

done in Example 5.17. Although each bucket moves away from the jet, the fact that they are arranged in

a wheel results in an effective mass ﬂow rate experienced by the vanes is

.Pm

f/nz

instead of the reduced

mass ﬂow rate computed in Eq. (6) of Example 5.17. With this in mind, consider a bucket, as shown in Fig.

P5.136(c), that is moving with a speed

v0

horizontally away from a fixed nozzle, but subject to a mass

ﬂow rate

.Pm

f/nz

. The inside of the bucket is shaped so as to redirect the water jet laterally out (away from

the plane of the wheel). The angle

✓

describes the orientation of the velocity of the ﬂuid relative to the

(moving) bucket at

B

, the point at which the water leaves the bucket. Determine

✓

and

v0

such that the

power transmitted by the water to the wheel is maximum. Express v0in terms of vw.

Photo credit: Courtesy of Andritz Hydro, Austria

Solution

Due to the symmetry of the shape of a bucket, we can study the ﬂow over half of a bucket. Using the

arguments presented in Example 5.17, under the assumption that the bucket is moving at constant velocity,

we can choose a control volume moving with the bucket. It is sufficient to study the motion only in the

horizontal direction. As explained in Example 5.17, the velocity of the water ﬂow over the vanes must be

understood as relative velocity of the water with respect to the vanes.

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Dynamics 2e 1121

Note that we have ignored the forces in the

y

direction. This is due to the fact that the forces in the

y

direction

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Problem 5.180

In Major League Baseball, a pitched ball has been known to hit the head of the batter (sometimes

unintentionally and sometimes not). Let the pitcher be, for example, Nolan Ryan who can throw a

51

8oz

baseball that crosses the plate at

100 mph

. Studies have shown that the impact of a baseball with a person’s

head has a duration of about

1ms

. So using Eq. (5.9) on p. 315 and assuming that the rebound speed of the

ball after the collision is negligible, determine the magnitude of the average force exerted on the person’s

head during the impact.

Solution

By a straightforward application of Eq. (5.9) on p. 315, we have

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Dynamics 2e 1123

Problem 5.181

A

0:6 kg

ball that is initially at rest is dropped on the ﬂoor from a height of

1:8

m and has a rebound height of

1:25

m. If the ball spends a total of

0:01

s

in contact with the ground, determine the average force applied to the ball by

the ground during the rebound. In addition, determine the ratio between the

magnitude of the impulse provided to the ball by the ground and the magnitude

of the impulse provided to the ball by gravity during the time interval that the

ball is in contact with the ground. Neglect air resistance.

Solution

We assume that the ball is only subject to (constant) gravity and, when in contact with the ground,

to a reaction force normal to the ground. Hence, the preimpact velocity can be determined using

constant acceleration equations, as follows:

p2ghi;(1)

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impulse provided by gravity to the ball during the impact is

t.mg O|/

. Hence, referring to the expression

larger than the corresponding impulse provided by gravity.

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Dynamics 2e 1125

Problem 5.182

A person

P

is initially standing on a cart on rails, which is moving

to the right with a speed

v0D2m=s

. The cart is not being propelled

by any motor. The combined mass of person

P

, the cart, and all that

is being carried on the cart is

270 kg

. At some point a person

PA

standing on a stationary platform throws to person

P

a package

A

to

the right with a mass

mAD50 kg

. Package

A

is received by

P

with a

horizontal speed

vA=P D1:5 m=s

. After receiving the package from

A

, person

P

throws a package

B

with a mass

mBD45 kg

toward

a second person

PB

. The package intended for

PB

is thrown to the

right, i.e., in the direction of the motion of

P

, and with a horizontal

speed

vB=P D4m=s

relative to

P

. Determine the final velocity of

the person

P

. Neglect any friction or air resistance acting on

P

and

the cart.

Solution

Balance Principles.

Letting the subscripts

3

and

4

denote the time instants right before and right after the

person

P

throws the package, and applying the impulse-momentum principle as conservation of momentum

in th xdirection, we have

Force Laws. Again, all forces are accounted for on the FBD.

Kinematic Equations. We note that

Computation.

Substituting Eqs. (5) into Eq. (4), using the result in Eq. (3), and solving for

.vPx/4

, we

have

.vPx/4D.vPx/1CmA

vA=P mB

vB=P :(6)

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Dynamics 2e 1127

Problem 5.183

A Ford Excursion

A

, with a mass

mAD3900 kg

, traveling with a speed

vAD85 km=h

, collides head-on

with a Mini Cooper

B

, with a mass

mBD1200 kg

, traveling in the opposite direction with a speed

vBD40 km=h

. Determine the postimpact velocities of the two cars if the impact’s coefficient of restitution

is eD0:22. In addition, determine the percentage of kinetic energy loss.

Solution

whose solution is

vC

Ax DmBŒe.v

Bx v

Ax/Cv

BxçCmAv

Ax

mACmBD13:64 m=s;(4)

1128 Solutions Manual

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Dynamics 2e 1129

Problem 5.184

The two spheres,

A

and

B

, with masses

mAD1:35 kg

and

mBD

2:72 kg

, respectively, collide with

v

AD26:2 m=s

and

v

BD

22:5 m=s

. Let

˛D45ı

. Compute the value of

ˇ

if the component

of the postimpact velocity of

B

along the LOI is equal to zero and

if the COR is eD0:63.

AD26:2 m=s

BD22:5 m=s

postimpact component of velocity along the LOI for particle Bto be equal to zero, i.e.,

vC

Bx D0: (7)

Computation.

Substituting the first of Eqs. (5), the first of Eqs. (6), and Eq. (7) into Eqs. (1) and (4) yields

two equations in the two unknowns vC

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