(20 points) Two children, 50.0 kg each, sit on opposite ends of a narrow board w
ID: 1964069 • Letter: #
Question
(20 points) Two children, 50.0 kg each, sit on opposite ends of a narrow board with length of 4.60 m and mass of 40.0 kg (which can be treated as a thin rod of rotational inertia of 1/12 mL2 where L is the length of the board). The board is supported by a pin in the center and the children and board spin in the horizontal plane.What is the rotational inertia of the board plus children about a vertical axis through the center of the board?
What is the angular momentum (magnitude and direction) of the system if it is rotating with an angular speed of 0.314 rad/s in a clockwise direction as seen from above?
While the system is rotating, the children pull themselves in toward the center of the board until they are 2/3 as far from the center as before. What is the resulting angular speed?
Explanation / Answer
rotational inertia= mL2/2 + 2 x 50 x (4.6/2)2
=40 x 4.62 / 2 +2 x 50 x (4.6/2)2
=952.2 kgm2
angular momentum = rotational inertia X angular speed
= 952.2 x 0.314 = 299 N m s ( direction : vertically downwards)
new rotational inertia = mL2/2 + 2 x 50 x (4.6/3)2
=658.31 kgm2
new angular speed = angular momentum/new rotational inertia = 299/658.31 = 0.454 rad/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.