Physics Help Forum Difficult Merry-Go-Round Question?

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Feb 23rd 2011, 01:11 AM #1 Junior Member   Join Date: Aug 2009 Posts: 18 Difficult Merry-Go-Round Question? A merry-go-round of radius 9.7m is rotating at 3.2 rpm. It slows uniformly to a stop in 17s. At the time it begins to slow down, Harry is sitting on a horse at the edge of the merry-go-round situated on the positive x axis. a Write expressions for Harry's position, velocity and acceleration vectors (x and y components) as a function of time. b Find Harry's displacement vector and his angular displacement for the time that the merry-go-round is slowing down. I have no clue how to start this problem. Please help.
 Feb 23rd 2011, 07:42 AM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 You know that this will involve the trigonometric functions cosine and sine? Hint: x is first maximum, then goes to zero, then to a minimum, then back to zero and the maximum. y is first at zero, to a maximum, to zero again, then a minimum and back to zero. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Feb 23rd 2011, 09:53 AM #3 Junior Member   Join Date: Aug 2009 Posts: 18 here is what I got so far. The way this has to be done is to find the angle as a function of time: In particular, θ'' = (d/dt)^2 (θ) = - α = - ω0/17 θ' = dθ/dt = ω0 - αt θ = 0 + ω0*t - αt^2/2 Therefore: θ(t) = ω0*t - αt^2/2 where ω0 = (2π*3.2/60) = (π*32/300) = (π*8/75) α = ω0/17 = 8π/(17*75) Harry's displacement is: (x,y) = 9.7*(cosθ, sinθ) = 9.7*(cos(ω0*t - αt^2/2), sin(ω0*t - αt^2/2)) Harry's velocity is: (v_x, v_y ) = (d/dt)(x,y) = 9.7*(ω0 - αt)(-sin(ω0*t - αt^2/2), cos(ω0*t - αt^2/2)) Harry's acceleration is: (a_x, a_y) = (d/dt)(v_x, v_y)
 Feb 23rd 2011, 11:50 AM #4 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 It seems right to me. Now the displacement and angular displacement should be easier. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Feb 23rd 2011, 12:12 PM #5 Junior Member   Join Date: Aug 2009 Posts: 18 Where do you start to find the displacement vector and angular displacement? I have no clue.
 Feb 23rd 2011, 12:20 PM #6 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 On your sketch, the starting point is at 9.7 m on the x-axis. The last point is at the position (x, y) at t = 12s. Then, the distance between the starting point and the last point gives the magnitude of the displacement, and then, you have to find the angle (say between a vertical line to the last point). You will need to know some circular measure for that. The angular displacement is simply given by: θ(t) = ω0*t - αt^2/2 when t = 12 s __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?