🚀 go-pugleaf

Gagan_Bhullar@dwc.com (Gagan Bhullar) wrote:

>Could someone help me with some physics problems please i would really
>appreciate it.
I will give you my solutions to these problems.  Please assume that I
am just dabbling and check other solutions.
Note that a letter or number in brackets after a quantity would
normally be written as a subscript.

>1)  a child hunting for his favorite wooden horse, is running on the ground
>around trhe edge of a stationary merry-go-round.  the angular speed of the
>child has a constant value of 0.250 rad/s.  at the instant the child spots
>the horse, one-quarter of a turn away, the merry-go-round began to move (in
>the direction of the child running) with a constant angular acceleration of
>0.0100 rad/s^2.  what is the shortest time it takes for the child to catch up
>to the horse?
1.  w(c) = 0.25 rad/s  
     an(o) = 1/4 revolution = 1/4(2 pi) = pi/2 radians
     a = 0.01 radians/s^2
    for the child                   an(c) = w(c) t = 0.25  t
    for the merry-go-round   an(m) = an(o) + 1/2 a t^2
    thus (to catch up)          0.25 t = an(o) + 1/2 a t^2
    and so                           0.25 t = pi/2 + 1/2  .01  t^2
                                          0.01 t^2  -  0.5 t  + pi = 0
    using quadratic:       t = 42.6306   or   t =  7.36932
    thus shortest time is 7.36932  or about  7.40 seconds



>2)a solid ball of mass M and radius R starts at rest at height h above the
>bottom of a path with a positive parabolic shape.  it rolls without slipping
>down the left side of the path.  the right side of the path, starting at the
>bottom,  is frictionless.  to what height does the ball rise on the right?
2.  Since the ball has no friction on the way up  it cannot convert   
     the spin energy it received on the way down to height on the way 
     up.        
     Let h = initial height and let H = height to which it will rise.
     For a sphere, rotational inertia = 2/5 m r^2
     Rolling down:  mgh = 1/2 m v^2 + 1/2 (2/5 m r^2) w^2
     Thus               mgh  = 1/2 m v^2 + 1/2(2/5 m r^2) (v^2 / r^2)
                            mgh  = 1/2 m v^2 + 1/5 m v^2
                            mgh  = 7/10 m v^2  = 0.7 m v^2
     and so            v^2  =  gh / 0.7

     Rolling up       mgH = 1/2 m v^2  
     thus                mgH = 1/2 m (gh / 0.7)
                              gH = gh / 1.4
                                H = h/1.4 = 5/7 h
     and so  the ball will slip up to a position 5/7 of its original 
     height.

>3)  a skater's body has rotational inertia 4.2 (kg)m^2  with his fists held
>to his chest and 5.7 kg(m^2) with arms outstreched.  the skater is twirling
>at 3.0 rev/s while holding 2.5 kg weights in each outstretched hand; the
>weights are 76 cm from his rotation axis.  if he pulls his hands in to his
>chest how fast wil he be twirling?
3.   I(1)  = 5.7    (skater's initial rotational inertia)
      I(2)  = 4.2   (skater's final rotational inertia)  
      w(1) = 3 rev/s = 18.84876 radians/s
      R = 76 cm = 0.76 m
      M = 5 kg   (Since there are two 2.5 kg  weights.
     
      I(3) = M R^2  = 2.888  (rotational inertia of weights)
      P(1) = I(1) w(1) = 107.4379  (skater's initial momentum)
      P(3) = I(3) w(1) =  54.43521 (initial momentum of weights)
      P(T) = P(1) + P(3) = 161.8731 (total angular momentum)

     The problem is solved by conservation of angular momentum:

      P(T) = I(2) w(2)
     Thus  w(2) = P(T)/ I(2) = 161.8731/4.2 = 38.54122  radians/second
     or  w(2) = 6.134286 rev/s = approx.  6.1 rev/s  
 
>4)  a physics student is standing on an initially motionless, frictionless
>turntable with rotational inertia 0.31 kgm^2.  he's holding a wheel of
>rotational inertia 0.22 kgm^2 spinning at 130 rpm about a vertical axis. 
>when he turns the wheel upside down, student and turntable begin rotating at
>70 rpm.  a) what is the students' mass, considering him to be a cylinder 30
>cm in diameter?  b)  how much work did he do in turning the wheel upside
>down?  neglect the distance between the axes of the turntable and wheel.
4.  I(t) = .31 kg m^2
     I(w) = .22 kg m^2
     w(w1) = 130 rpm = 13.61357 radians/s  (angular speed of wheel)
     w(st) = 70 rpm = 7.330383 radians/s (student & turntable)
     R(s) = 1/2 (30 cm) = 15 cm = 0.15 m  (student radius)

     P(w1) = I(w) w(w1) = 2.994985  (initial angular momentum of wh.)
     P(t) = I(t) w(st) = 2.272419 (final angular momentum of t-table) 
     
     final angular speed of wheel is: 
     w(w2) = -[w(w1) - w(st)] = -6.283187 rad/s

     P(w2) = I(w) w(w2) = -1.382301 (final angular momentum of wh.)

     final angular momentum of student of mass M(s) is:
     P(s) = I(s) w(st) = 1/2 M(s) R(s)^2 w(st) = 0.082466 M(s)

     from conservation of angular momentum:

     P(w1) = P(s) + P(t) + P(w2)
     Thus   P(s) = P(w1) - P(t) - P(w2) = 2.104868
     and M(s) = P(s)/0.082466 = 25.52381 kg

     and so the mass of the student is about 26 kg

     for part (b)   Energy before = 20.38622 joules
                        Energy after = 20.38622 joules
     and so the student did no work.
>thanks

I hope these are at least partly right
Dave Stock