Firefox Mastering Physics: HW 20 x C A Merry-go-round ls A Co x G sin 0 Google S

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Firefox Mastering Physics: HW 20 x C A Merry-go-round ls A Co x G sin 0 Google Search x https://session mastering physics com offset next&assignmentProblemID-77674850; PHY 191 Spring 2017 HW 20 Problem 12.81 Problem 12.81 Part A A merry-go-round is a common piece of playground equipment. A What is the merry-go-round's angular velocity, in rpm, after John jumps on? 3.0-m-diameter merry-go-round with a mass of 260 kg is Express your answer to two significant figures and include the appropriate units. spinning at 23 rpm John runs tangent to the merry-go-round at 4.0 m/s, n the same direction that it is tuming, and jumps onto the outer edge. John's mass is 30 kg. Value Units Submit My Answers Give Up ABC Fri 1:09 PM previous l 3 of 3 l return to assignment Continue Provide Feedback

Explanation / Answer

radius of merry go round r = 3/2 = 1.5m mass M= 260kg

moment of inertia I = 0.5Mr2 = 0.5 x 260 x 1.5 x 1.5 = 292.5 Kgm2

Initial angular velocity of merry go round w1 = 23rpm = 23 x 2x 3.14 rad/60s = 2.41rad/s

Initial angular momentum of the merry go round Lm = I w1 = 292.5 x 2.41 =704.14  Kgm2/s

initia angular momentum of the john Lj = mvr = 30 x 4 x1.5 = 180 Kgm2/s

Initial angular momentum of the system Li =Lm + Lj = 704.14 + 180 =884.14 Kgm2/s

Moment of inertia of john Ij = mr2 = 30 x 1.5x 1.5 = 67.5  Kgm2

Let the final angular momentum of the system be w2

from conservation of angular momentum Li = Lf

884.14 = ( I + Ij ) w2

=(292.5 + 67.5) w2

= 360 w2

w2 = 884.14/360 = 2.46 rad/s

= (2.46 x 60) 2Pi rpm

= 2.46 x 60/6.28

= 23.46rpm

Hence angular velocity of merry go round after john jums = 23.46 rpm

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