Problem P8.3: The disk in a computer hard drive spins at 7200 rpm. At the radius of 30 mm, a stream of data is magnetically written on the disk, and the spacing between data bits is 25 μm. Determine the number of bits per second which pass by the read/write head.
Approach:
Apply Equation (8.1) v = rω, where ω has units rad/s, to find the speed of a point on the disk. Determine the number of 25 μm increments (the spacing between bits) that pass by each second. From Table 3.2, 1 μm = 10 –6 m.
Solution:
Angular velocity in consistent units (Table 8.1):
(
)
s rad 753.8 rpm
s rad 0.1047 rpm
7200
ω ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛ =
Velocity of a point on the disk:
(
)
s m 62 2 2 s rad 753.8 m
0.03 .
v ⎟=
⎠ ⎞ ⎜
⎝ ⎛ =
Rate at which bits pass by the read/write head: Data bit rate
s bits 10 05 9 m 10 25
s m 62
22 5
6
- = ×
× .
.
9.05 × 10 5 bits each second Discussion:
This rate is significantly lower than most data transfer rates over the internet or across a USB or FireWire interface. This is because of the mechanical hardware involved in spinning and controlling the hard drive disks and readers.
Problem P8.6: The driving disk spins at a constant 280 rpm clockwise. Determine the velocity of the connecting pin at A. Also, does the collar at B move with constant velocity? Explain your answer.
Approach:
Apply Equation (8.1), v = rω, where ω has units rad/s and r is the length from the center of the disk to point A. Then, visualize the movement of point B to determine its velocity behavior.
Solution:
Convert angular velocity to consistent units (Table 8.1):
(
)
s rad 3 9 2 rpm
s rad 0.1047 rpm
280 .
ω ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛ =
Velocity of point A:
(
)
s m 880 s
rad 29.3 mm
30 ⎟=
⎠ ⎞ ⎜
⎝ ⎛ =
v
s m 88 0 mm 000 1
m s
mm
880 .
v ⎟=
⎠ ⎞ ⎜
⎝ ⎛ ⎟ ⎠ ⎞ ⎜
⎝ ⎛ =
s m .88 0
Velocity of point B (collar):
The collar moves to the right when point A is moving down the right-hand side of the driving disk. Once point A moves through the bottom of the disk to the left-hand side, the collar then starts to move back to the left. To change directions, the collar must come to a stop. Once the collar changes direction, it accelerates in the opposite direction until it slows down to an instantaneous stop and then changes direction again. Therefore the collar does not move with constant velocity.
Problem P8.8: A small automobile engine produces 260 N ⋅ m of torque at 2100 rpm. Determine the engine's power output in the units of kW and hp.
Approach:
Apply Equation (8.5), P = Tω, for instantaneous power, where ω has units rad/s. Use Table 8.1 to convert speed into dimensionally consistent units, and Table 7.2 to convert power units.
Solution:
Convert angular velocity to consistent units:
(
)
s rad 9 19 2 rpm
s rad 0.1047 rpm
2100 .
ω ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛ =
Calculate power in SI units (Equation (8.5):
(
)
s m N 10 717 5 s rad 219.9 m
N
260 ⎟= × 4 ⋅
⎠ ⎞ ⎜
⎝ ⎛ ⋅
= .
P
= 5.717× 104 W = 5.717 kW Convert to USCS units:
(
)
766 hpW hp 10 1.341 W
10
5.717 4 -3 .
P ⎟=
⎠ ⎞ ⎜
⎝
⎛ × ×
=
57.2 kW or 76.7 hp Discussion:
Power is proportional to both torque and angular speed which means that if an engine transmits more torque or its rotational speed increases, it will produce a greater power output.
Problem P8.11: A child pushes a merry-go-round by applying a force tangential to the platform. To maintain a constant rotational speed of 40 rpm, the child must exert a constant force of 90 N to overcome the slowing effects of friction in the bearings and platform. Calculate the power exerted by the child in horsepower to operate the merry-go-round. The diameter of the platform is 8 ft.
Approach:
Apply Equation (4.8) to find the torque applied and then apply Equation (8.5), P = Tω, to find the instantaneous power, where ω has units rad/s. Use Tables 3.6 and 8.1 to convert quantities into dimensionally consistent units. Assume that the child is exerting the force exactly tangential to the merry-go-round platform.
Solution:
Calculate torque using Equation (4.8) and convert the distance using Table 3.6:
(
)( )
1098N mft 28 3
m ft 4 N
90 ⎟= ⋅
⎠ ⎞ ⎜
⎝ ⎛
= .
. T
Convert speed to rad/s using Table 8.1: s rad 2 4 rpm
s rad 1047 0 rpm
40 . .
ω ⎟⎟=
⎠ ⎞ ⎜⎜
⎝ ⎛ =
Calculate power using Equation (8.5):
(
)
4612Ws rad 2 4 m N 8
109. . .
P ⎟=
⎠ ⎞ ⎜
⎝ ⎛ ⋅ =
W 2 461.
Discussion:
Problem P8.17: The helical gears in the simple geartrain have teeth numbers as labeled. The central gear rotates at 125 rpm and drives the two output shafts. Determine the speeds and rotation directions of each shaft.
Approach:
Apply Equation (8.8) ωg = (Np/Ng)ωp at each mesh point. Solution:
(
125rpm)
1071rpm 7060 3 2 3
2 N .
N = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ω ω
(
107.1rpm)
1667rpm 45 70 2 1 2 1 . N N = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ω ω(
125rpm)
2143rpm 35 60 3 4 3 4 . N N = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ω ω(
214.3rpm)
882rpm 8535 4 5 4
5 N .
N = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ω ω
Gear 1 2 3 4 5
Speed, rpm Direction 167 CW 107 CCW 125 CW 214 CCW 88 CW Discussion:
The rotational speed of each gear is inversely proportional to its size. The smaller gears spin faster. Also, each pair of adjoining gears turns in opposite directions. If another gear was added to the right side of the geartrain, it would turn counterclockwise.
Problem P8.20: For the compound geartrain, obtain an equation for the velocity and torque ratios in terms of the numbers of teeth labeled.
Approach:
The overall velocity ratio is the product of velocity ratios at each mesh point (Equation (8.22)), since there are two gears per shaft. Since the input and output power levels must be the same (Equation (8.12)), the torque ratio is the inverse of the velocity ratio.
Solution:
3 5 7 1
2 4 6 8
output input
N N N
N
N N N N
ω =⎛⎜ ⎞⎛⎟⎜ ⎞⎟⎛⎜ ⎞⎛⎟⎜ ⎞⎟ω
⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
1 3 5 7 2 4 6 8
output input
N N N N VR
N N N N
ω ω
= =
2 4 6 8 1 3 5 7
1 N N N N
TR
VR N N N N
= = Discussion:
The expression for the velocity ratio illustrates that if gears 1, 3, 5, or 7 are increased in size, then the velocity ratio will increase. Conversely, if gears 2, 4, 6, or 8 are decreased in size, then the velocity ratio will decrease. The opposite relationships exist for the torque ratio.
