Esession. masteringphysics.com C HW5.1 (Chapter 9, 10) Problem 9.77 K)5 of 35 Co

Question

Esession. masteringphysics.com C HW5.1 (Chapter 9, 10) Problem 9.77 K)5 of 35 Constants Part A A 2.6-m-diameter merry-go-round with a mass of 240 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 34 kg What is the merry-go-round's angular speed, in rpm, after John jumps on? Express your answer using two significant figures rpm Submit Previous Answers Request Answer Incorrect: Try Again: 5 attempts remaining Next > Provide Feedback MacBook Air

Explanation / Answer

Here, Initial angular momentum = final angular momentum

Initially, angular speed w1 = 2*pi*20/60 = 2.094 /s ; moment of inertia of merry go round = 0.5*240*1.3^2 = 202.8kgm^2 ; mass of john = Mj = 34kg ; speed of john = 5m/s

Final angular speed = w2 ; momennt of inertia of john about center of merrgy go round = Mj*r^2 = 57.46

So. 202.8*2.094 + 34*5*1.3 = (202.8+57.46)*w2

So, w2 = 2.48 /s

So,merry go round's angular speed after john jumps in is 60*w2 / 2*pi = 23.69 rpm

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