Results 1 to 4 of 4

Thread: [SOLVED] Rotation?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    [SOLVED] Rotation?

    A man of mass 80 kg mass is standing on the rim of a circular platform rotating about its axis. The platform with the man on it rotates at 12 r.p.m. How will the system rotate, if the man moves to the platform's centre? What work will the man perform in changing his position? The mass of the platform is 200 kg and the radius is 1.2 m.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA


    If I recall correctly, moment of inertia follows the rule of superposition. So, the inertia contained in the spinning platform is I_p=\frac12MR^2, where M is the platform's mass and R its radius. The inertia of the man on the edge is I_m=mR^2, where m is the man's mass. Summing, the initial kinetic energy of the system is then K_i=\frac12I_i\omega_i^2 and the final kinetic energy, when the man has walked to the center, is K_f=\frac12I_f\omega_f^2, where I_i=\frac12MR^2+mR^2 and I_f=\frac12(M+m)R^2. Because of conservation of angular momentum, no energy is lost, so K_f=K_i, and solving for \omega_f, we arrive at 16.1 rpm, a 34% increase in angular speed. This at least answers the first part of the question. The question about the man's work is more complicated, as he must work against a centripetal force vector of m\frac{v^2}r, where the platform's tangential velocity, v, and the man's distance from its center, r, is constantly changing.

    This is the exact same phenomenon we learn in grade school about why an ice skater's spin speeds up as they pull their arms in.
    Last edited by Media_Man; Jun 2nd 2009 at 07:16 AM. Reason: right math, wrong answer
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Apr 2009
    Atlanta, GA

    Running in circles

    As per my last post, the platform's angular velocity increases as a function of the man's distance from its center. It can be shown, by making the appropriate substitutions of my last post, that that function is this: \omega=\omega_0\sqrt{\frac{(\frac12M+m)R^2}{\frac1  2MR^2+mr^2}} , where again, \omega_0 is the initial angular velocity, M is the mass of the platform, m is the mass of the man, R the radius of the platform, and r the man's distance from the center.

    The man's work is the integral of the force he must work against over the distance he travels, so W=\int_R^0 F(r)dr, the bounds of integration of course being the edge of the platform to the platform's center. The force in question here is centripetal force, F=m\frac{v^2}r=mr\omega^2. Remember, the platform is speeding up but the man is getting closer to the center, so F will actually be a decreasing function. Substituting, F(r)=\omega_0^2(\frac12M+m)R^2\frac r{\frac{M}{2m}R^2+r^2} . Remember, everything in this function is a constant except r, making this a relatively simple integral. In fact, we can substitute some constants for simplicity's sake, W=\int_R^0 \frac{kr}{a+r^2}dr .

    Making a simple substitution ( u=a+r^2\rightarrow du=2rdr) this is a pretty easy integral yielding W=\omega_0^2(\frac12M+m)R^2\frac12\ln(\frac{M}{2m}  R^2+r^2)|_R^0 . Now just convert everything to standard units and evaluate numerically.
    Last edited by Media_Man; Jun 2nd 2009 at 07:23 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    May 2009

    The correct method.

    Here's how to solve this correctly:


    The mechanics of this problem are that the chap's actions cannot result in any net external torque on the system and so angular momentum must be conserved. Since his movements must alter the moment of inertia the angular velocity will change. In fact it increases and so the kinetic energy of the system is increased and it is this KE GAIN that is provided for by the work he does! Hence note that KE is NOT conserved!

    Overall energy is conserved because the extra KE is provided by the work he does which is provided by the chemical energy stored in his body.

    So let I and \omega be the moment of inertia and angular speed respectively when he is at radius r on the disc. Furthermore let I_0 and \omega_0 correspond to r=R i.e. when he's at the rim.

    Then by conservation of angular momentum we have

    I \omega = I_0 \omega_0

    and I(r) = mr^2 + \frac{1}{2}M R^2

    where m is the mass of the man and M is the mass of the disc.

    so we have

    \omega (r) = \frac{I_0}{I(r)} \omega_0


    \color[rgb]{0,0,1} \boxed{\omega (r) = \frac{(M +2 m) R^2}{MR^2 +2 m r^2} \omega_0}

    Now to compute the work done there are two ways.

    One way is to consider the system being equivalent to a non-rotating system with gravitational force outwards (centrifugal force) and so the work done is to overcome this force and is hence converted to potential energy (in this frame of reference).

    The second way, which is what I'll do, is just work out the KE gained in the system as it doesn't involve any integration.


    W = \frac{1}{2}(I_f \omega_f^2 - I_0 \omega_0^2 )

    where the f subscripts denote the final value of the variable when r=R.

    So using the fact that I_f \omega_f = I_0 \omega_0 from conservation of angular momentum we have that the work, W, is given by

    W = \frac{1}{2} I_0 \omega_0 \left( \omega_f - \omega_0 \right)

    = \frac{1}{2} I_0 \omega_0^2 \left(   \frac{MR^2 +2 mR^2}{MR^2} -1       \right)

    = \frac{m}{M} I_0 \omega_0^2.

    So in explicit terms we have

    \color[rgb]{0,0,1} \boxed{W =  \frac{m(2m+M)}{2M} R^2  \omega_0^2}.

    I'll leave it to you to put in the numerical values!

    Last edited by the_doc; Jun 3rd 2009 at 04:09 PM. Reason: Made an equation look neater!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Jan 14th 2012, 08:02 PM
  2. Rotation in 3d
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 30th 2011, 03:36 PM
  3. [SOLVED] Rotation and Magnetic field (Physics)
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: Mar 5th 2009, 03:41 AM
  4. [SOLVED] Trig integral w/ volume of rotation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Feb 9th 2009, 01:58 AM
  5. [SOLVED] Trig integral in volume of rotation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 8th 2009, 08:23 PM

Search Tags

/mathhelpforum @mathhelpforum