WileyPLUS -> C D edugen.wileyplus.com/edugen/student/mainfr.un. :: Apps HP Conne

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WileyPLUS -> C D edugen.wileyplus.com/edugen/student/mainfr.un. :: Apps HP Connected Watch NHL Live Stre- Watch NBA Live Strea × D Solution Manual Wiley.pc × Chegg Study l Guided So. ABP Practice Assignment GradebookORION ssignment FULL SCREEN PRINTER VERSION BACK ES Chapter 11, Problem 063 016 In the figure here, a 25 kg child stands on the edge of a stationary merry-go-round of radius 2.0 m. The rotational inertia of the merry-go-round about its rotation axis is 160 kg m2. The child catches a ball of mass 0.6 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of magnitude 11 m/s, at angle = 50With a line tangent to the outer edge of the merry-go-round as shown. What is the angular speed of the merry-go-round just after the ball is caught? 41 28 067 46 13 41 063 71 Ball Child dy I'm Cortana. Ask me anything 11:52 PM 4/27/2016

Explanation / Answer

Fairly basic problem on conservation of angular momentum
Lf = L0.
Lf = If*f , with If = IMGR + (25.6 kg)(2 m)^2 = 102.4+160 = 262.4 kg m2
The 25.6 kg term represents the additional rotational inertia due to the child-plus-ball at the edge of the MGR (Merry-Go-Round).

Lf = L0 = |~r ×~p|

= (2 m)(11 kg m/s)(sin50)

= 16.853 kg m^2/s.

Note that the angle between ~r and ~p is 90 = 40

f = L0/If = (16.853 kg m2/s)/( 262.4 kg m2) = 0.0642 rad/s

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