PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 9, Problem 3P
To determine
The additional revolutions required by the disk to reach an angular speed of
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Chapter 9 Solutions
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
Ch. 9 - Prob. 1PCh. 9 - Prob. 2PCh. 9 - Prob. 3PCh. 9 - Prob. 4PCh. 9 - Prob. 5PCh. 9 - Prob. 6PCh. 9 - Prob. 7PCh. 9 - Prob. 8PCh. 9 - Prob. 9PCh. 9 - Prob. 10P
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- A disk 8.00 cm in radius rotates at a constant rate of 1200 rev/min about its central axis. Determine (a) its angular speed in radians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s.arrow_forwardFigure P10.18 shows the drive train of a bicycle that has wheels 67.3 cm in diameter and pedal cranks 17.5 cm long. The cyclist pedals at a steady cadence of 76.0 rev/min. The chain engages with a from sprocket 15.2 cm in diameter and a rear sprocket 7.00 cm in diameter. Calculate (a) the speed of a link of the chain relative to the bicycle frame, (b) the angular speed of the bicycle wheels, and (c) the speed of the bicycle relative to the road, (d) What pieces of data, if any, are not necessary for the calculations?arrow_forwardA turntable (disk) of radius r = 26.0 cm and rotational inertia0.400 kg m2 rotates with an angular speed of 3.00 rad/s arounda frictionless, vertical axle. A wad of clay of mass m =0.250 kg drops onto and sticks to the edge of the turntable.What is the new angular speed of the turntable?arrow_forward
- A bicycle is turned upside down while its owner repairs a flat tire on the rear wheel. A friend spins the front wheel, of radius 0.381 m, and observes that drops of water fly off tangentially in an upward direction when the drops are at the same level as the center of the wheel. She measures the height reached by drops moving vertically (Fig. P10.74 on page 332). A drop that breaks loose from the tire on one turn rises h = 54.0 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel.arrow_forwardDuring a certain time interval, the angular position of a swinging door is described by = 5.00 + 10.0t + 2.00t2, where is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.arrow_forwardA record rotates through 5.6 radians as it slows down uniformly from 78.0rpm to 22.8 rpm. What is the magnitude of the angular acceleration of the record?arrow_forward
- A disk with a diameter of 30 cm speeds up from 20 rad/s to 40 rad/s in 20s. (a) How many revolutions will the disk go through during that time period? (b.) What is the average linear acceleration of the disk? A 96 rev; 0.30 m/s^2 B) 96 rev; 0.15 m/s^2 c) 85 rev; 0.30 m/s^2 85 rev; 0.15 m/s^2arrow_forwardThe crankshaft in a race car goes from rest to 3000 rpmrpm in 3.5 ss. What is the angular acceleration of the crankshaft? How many revolutions does it make while reaching 3000 rpmrpm?arrow_forwardPoint A of the circular disk is at the angular position 0 = 0 at time t = 0. The disk has angular velocity wo = 0.29 rad/s at t = 0 and subsequently experiences an angular acceleration a = 1.8t where t is in seconds, and a is in radians per second squared. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 2.7 s. Assumer = 145 mm. α Answers: VA = aд = 90 (i ( i 0 6.806 i + i + i i 1 0.705 j) m/s j) m/s²arrow_forward
- Hi, in physics I they have The earth's radius is 6.37×106m; it rotates once every 24 hours. I have already found the angular speed ω= 7.3×10−5 rad/s and the point on the equator v= 460 ms. What I'm trying to understand is what is the speed of a point on the earth's surface located at 2/5 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?)arrow_forwardA 76-cmcm-diameter wheel accelerates uniformly about its center from 150 rpmrpm to 390 rpmrpm in 3.5 ss . Determine its angular acceleration. Determine the radial component of the linear acceleration of a point on the edge of the wheel 1.2 ss after it has started accelerating. Determine the tangential component of the linear acceleration of a point on the edge of the wheel 1.2 ss after it has started accelerating.arrow_forwardSolution (a) Find the angular displacement after 2.00 s, in both radians and revolutions. 1 A0 = w;t + Use Equation 7.8, setting @; 1.60 rad/s, a = 3.50 rad/s2, and t = 2.00 s. 1 Að = (1.60 rad/s) (2.00 s) + (3.50 rad/s²) (2.00 s)² 2 A0 = 10.2 rad Convert radians to revolutions. A0 = (Am rad)(1.00 rev/2r rad) AO = 16.02 X rev Your response differs from the correct answer by more than 100%. (b) What is the angular speed of the wheel at t = 2.00 s? Substitute the same values into Equation 7.7. @ = @; + at = 1.60 rad/s + (3.50 rad/s²)(2.00 s) W = rad/s Remarks The result of part (b) could also be obtained from Equation 7.9 and the results from part (a). - INarrow_forward
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