(10%) Problem 9: Three children are riding on the edge of a merry-go-round that
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(10%) Problem 9: Three children are riding on the edge of a merry-go-round that is a disk of mass 106 kg, radius 1.8 m, and is spinning at 25 rpm. The children have masses of 21.2 kg,26.4 kg, and 34.2 kg Randomized Variables M=106 kg m1=212kg m12 = 264 kg m3 = 342 kg r=1.8m f 25 rpm If the child who has a mass of 264 kg moves to the center of the merry-go-round, what is the new magnitude of angular velocity in rpm? Grade Summary Deductions Potential 0% 100% sin0 tan() | | ( Submissions Attempts remaining: 7 cosO cotanO asin) acosO atanO acotan)sinh(0 cosh0 tanh) cotanh0 Degrees Radians % per attempt) detailed view 123 END CLEAR Submit Hint I give Hints: 1 % deduction per hint. Hints remaining: 2 Feedback: deduction per feedback.Explanation / Answer
Given,
M = 106 kg ; R = 1.8 m ; w1 = 18 rpm = 1.89 rad/s
m1 = 21.2 kg ; m2 = 26.4 kg and m3 = 34.2 kg
We need to find the new mangnitude of angular velocity 2.
From the law of conservation of angular momentum we know that
L(f) = L(i)
I1 w1 = I2 w2
w2 = I1 w1 / I2
The intial moment of inertia of the merry go round and children system will be, I1 will be:
I1 = 1/2 M R^2 + m1 R^2 + m2 R^2 + m3 R^2 = R^2 ( 0.5 M + m1 + m2 + m3 )
I1 = 1.8^2 ( 0.5 x 106 + 21.2 + 26.4 + 34.2 ) = 436.75 kg-m^2
After the m2 = 26.4 goes into the center the moment of inertia I2 becomes:
I2 = 1/2 M R^2 + m1 R^2 + m2 x 0 + m3 R^2 = R^2 ( 0.5 M + m1 + m3 )
I2 = 1.8^2 ( 0.5 x 106 + 21.2 + 34.2) = 351.22 kg-m2
w2 = I1 w1/I2
w2 = 436.75 x 25 rpm/351.22 = 31.09 rpm
Hence, w2 = 31.01 rpm
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