008 part 1 of 2 100 points A merry go round rotates at the rate of 47 rev s

# 008 part 1 of 2 100 points a merry go round rotates

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008 (part 1 of 2) 10.0 points A merry-go-round rotates at the rate of 0 . 47 rev / s with an 98 kg man standing at a point 2 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center? Consider the merry-go-round is a solid 57 kg cylinder of radius of 2 m. Correct answer: 13 . 1076 rad / s. Explanation: Given : ω i = 0 . 47 rev / s , m = 98 kg , r i = 2 m , r f = 0 m , M = 57 kg , and R = 2 m . The merry-go-round can be modeled as a solid disk with angular momentum L d = I d ω = parenleftbigg 1 2 M R 2 parenrightbigg ω and the man as a point mass with angular momentum L m = I m ω = ( m r 2 ) ω . Angular momentum is conserved, so L f = L i L m,f + L d,f = L m,i + L d,i I m,f ω f + I d ω f = I m,i ω i + I d ω i parenleftbigg m r 2 f + 1 2 M R 2 parenrightbigg ω f = parenleftbigg m r 2 i + 1 2 M R 2 parenrightbigg ω i ω f = 1 2 M R 2 + m r 2 i 1 2 M R 2 + m r 2 f ω i = 1 2 (57 kg) (2 m) 2 + (98 kg) (2 m) 2 1 2 (57 kg) (2 m) 2 + (98 kg) (0 m) 2 × (0 . 47 rev / s) · 2 π rev = 13 . 1076 rad / s . 009 (part 2 of 2) 10.0 points What is the change in kinetic energy due to this movement? Correct answer: 7586 . 77 J. Explanation: The moment of inertia of the system about the axis of rotation at any moment is I = I d + I m = 1 2 M R 2 + m r 2 , kuruvila (lk5992) – HW 11 – opyrchal – (11113) 5 and the kinetic energy at any time is K rot + K trans = 1 2 I ω 2 = 1 2 parenleftbigg 1 2 M R 2 + m r 2 parenrightbigg ω 2 = 1 4 M R 2 ω 2 + 1 2 m r 2 ω 2 Thus the change in kinetic energy of the sys- tem is K = K f K i = parenleftbigg 1 4 M R 2 + 1 2 m r 2 f parenrightbigg ω 2 f parenleftbigg 1 4 M R 2 + 1 2 m r 2 i parenrightbigg ω 2 i = bracketleftbigg (57 kg)(2 m) 2 4 + (98 kg)(0 m) 2 2 bracketrightbigg × (13 . 1076 rad / s) 2 bracketleftbigg (57 kg)(2 m) 2 4 + (98 kg)(2 m) 2 2 bracketrightbigg × (0 . 47 rev / s) 2 (2 π rad / rev) 2 = 7586 . 77 J . 010 10.0 points A star of radius 7 . 8 × 10 5 km rotates about its axis with a period of 44 days. The star undergoes a supernova explosion, whereby its core collapses into a neutron star of radius 23 km. Estimate the period of the neutron star (assume the mass remains constant). #### You've reached the end of your free preview.

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