Problem P8: The disk in a computer hard drive spins at 7200 rpm At the radius of 0 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 Apply Equation (81) 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 2, 1 μm = 10 6 m Angular velocity in consistent units (Table 81): ω = ( 7200 rpm) 01047 = 758 Velocity of a point on the disk: rad m v = ( 00 m) 758 = 22 62 s s Rate at which bits pass by the read/write head: 22 62 m s 5 bits Data bit rate = 9 05 10-6 25 10 m s 905 10 5 bits each second 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 20
Problem P86: 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 Apply Equation (81), 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 Convert angular velocity to consistent units (Table 81): ω = ( 280 rpm) 01047 = 29 Velocity of point A: rad m v = ( 0 mm) 29 = 880 s s mm m m v = 880 = 0 88 s 1000 mm s m 0 88 s 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 While the magnitude of the velocity of point A is constant, the magnitude of the velocity of point B changes This mechanism translates rotational motion to translation motion and is known as a Crank-Rocker mechanism It is common in many mechanical systems 2
Problem P88: A small automobile engine produces 260 m of torque at 2100 rpm Determine the engine's power output in the units of kw and hp Apply Equation (85), P = Tω, for instantaneous power, where ω has units rad/s Use Table 81 to convert speed into dimensionally consistent units, and Table 72 to convert power units Convert angular velocity to consistent units: ω = ( 2100 rpm) 01047 = 219 9 Calculate power in SI units (Equation (85): rad 4 m P = ( 260 m) 2199 = 5 717 10 s s = 5717 10 4 W = 5717 kw Convert to USCS units: 4 - hp P = ( 5717 10 W) 141 10 = 76 6 hp W 572 kw or 767 hp 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 25
Problem P811: 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 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 Apply Equation (48) to find the torque applied and then apply Equation (85), P = Tω, to find the instantaneous power, where ω has units rad/s Use Tables 6 and 81 to convert quantities into dimensionally consistent units Assume that the child is exerting the force exactly tangential to the merry-go-round platform Calculate torque using Equation (48) and convert the distance using Table 6: m T = ( 90 )( 4 ft) = 109 8 m 28 ft Convert speed to rad/s using Table 81: ω = 40 rpm 0 1047 = 4 2 Calculate power using Equation (85): rad P = ( 109 8 m) 4 2 = 461 2 W s 461 2 W This is equivalent to 062 hp and is a significant amount of power for a child to generate and would only be feasible for short instances of time This is possible because the child would not be constantly pushing the merry-go-round but instead would be pushing once every few seconds using one of the handles on the platform 28
Problem P817: 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 1 2 4 5 Apply Equation (88) ω g = ( p / g )ω p at each mesh point 60 = 2 70 ( 125 rpm) = 107 1rpm 2 2 70 2 = 1 45 ( 1071 rpm) = 166 7 rpm 1 60 = 4 5 4 5 4 = 214 rpm 5 85 ( 125 rpm) = 214 rpm 4 ( ) = 88 2 rpm 5 Gear 1 2 4 5 Speed, rpm Direction 167 CW 107 CCW 125 CW 214 CCW 88 CW 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 246
Problem P820: For the compound geartrain, obtain an equation for the velocity and torque ratios in terms of the numbers of teeth labeled The overall velocity ratio is the product of velocity ratios at each mesh point (Equation (822)), since there are two gears per shaft Since the input and output power levels must be the same (Equation (812)), the torque ratio is the inverse of the velocity ratio 1 5 7 ωoutput = 2 4 6 8 ωoutput 1 5 7 VR = = ωinput 2 4 6 8 1 2 4 6 8 TR = = VR 1 5 7 input The expression for the velocity ratio illustrates that if gears 1,, 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 251