This exercise was intended to demonstrate many aspects of the concepts and constructions contained within a system with conservative forces. In particular the rotating table top was intended reintroduce rotational motion with the idea of working from the inertial laboratory frame and the non-inertial rotational frame. The ``potential'' for interchangeability between kinetic energy, potential energy and work are all embodied in this exercise. Hence it can serve as conceptual framework for the more subtle aspects of the discussion in Chapter 8.
 
 

QUESTION: An object of mass m is mounted in on a horizontal, massive rotating table within a tight-fitting (but frictionless) recessed groove as shown in the figure below. [NOTE: The necessity of the groove is to limit all acceleration to the radial direction. As the mass changes position in the track there will be a tangential forces acting against the mass but since it is constrained and the table is massive the rotation rate will remain unchanged (hence NO work will be done with respect to this action).]

To the mass a massless and inflexible string is tied and then passed through a frictionless hole in the center of the table. The other end of the string is tied to a second mass, M, which is hanging freely. The position the first mass is initially at a point, R, which produces an equilibrium configuration so that the second mass remains stationary. An infinitesimal radial displacement of the first mass (m) towards the center causes the second mass to descend.

What is the velocity of the first mass when it just reaches the hole in rotating table?

1.

To do this problem with forces is rather difficult since the radial acceleration is function of radial position. This is something that isn't overly stressed in the class and will not be a major testing point.

2.

Since there are no frictional forces or any other non-conservative attributes the sum of the kinetic energy and the potential energies are a constant of the motion (KE+PE = constant). With respect to the laboratory frame there is only kinetic energy of 1/2 mv² of the first mass and, by construction, Mgh of potential energy in the second mass at the starting point. Equilibrium demands that F_c = mv²/R = F_g = Mg so that v²=MgR/m (specifying the kinetic energy.)

At the end point the velocity of the first mass must go to zero since the the tangential velocity in the track is proportion to the radius. However since the two masses are tied together their final velocities must be the same. The second mass (M) will have fallen to a height of h-R.

The final relationship is therefore:
 

$$KE+ PE = Mgh + \frac{1}{2} m MgR/m = Mg(h-R) + \frac{1}{2} (M+m)v_f^2$$

or

$$v_f = \sqrt{3MgR/(M+m)} ~~.$$

 

3.

Now that you have the answer you should delve a bit more and see where this takes us. With respect to an observer in the rotating frame of the table there is no apparent velocity of the first mass so that it appears to be initially at rest. If you reproduce the calculation of last step you incorrectly solve the problem to give the claimed result

$$KE + PE = Mgh = \frac{1}{2} (M+m)v_f^2$$

which is clearly wrong. The upshot of this is that non-inertial reference frames do not universally allow you to use the fact that a system experiencing conservative forces has the sum of the kinetic and the potential energies as a constant of the motion. (This is perhaps obvious if you cut the massless string and watch the first mass (m) fly off the rotating table gaining kinetic energy the entire time it is in the track.)

4.

You can actually solve this problem in the rotating non-inertial frame if you invent a potential energy expression which represents the action of the centripetal acceleration. In this case we will have constructed a potential energy representation which, is in some sense, a function of the kinetic energy (which is a function of position as well). For this we return to the net work integral:

$$W_{net}= \int_{r_1}^{r_2}F_c~dr$$

with F_c = mv²/r. Notice that, for the track, v (tangential) is a function of radial position. This is not good. As a pre-step we need to relate v to the rotation motion of the table. To do we need to relate the angular displacement of the table, $\theta$,to the velocity v. This is a ``two'' step argument:

(a)

The relationship between a circular arc s and an angle $\theta$ (in radians) is defined by a proportionality constant, the radius r, so that $s = r \theta$.

(b)

Differentiating with respect to time give $ds/dt =r d\theta/dt$.

(c)

We recognize ds/dt as an instantaneous tangential velocity v and we define $d\theta/dt \equiv \omega$ the angular velocity (in radians per seconds with $2\pi$ radians equaling one revolution.

(d)

Finally we have $v =\omega r$ and $\omega$ is independent of r.

With the above expression $F_c = mv^2/r = m\omega^2r$ and this can be substituted into the above integral giving
 

$$W_{net}= \int_{r_1}^{r_2} (-m \omega^2r)dr$$

It is convenient to reference this integral to the center of the circle (were F_c is zero)

$$W_{net}= \int_{0}^{r} (-m \omega^2r)dr = -m\omega^2 r^2/2$$

and this sign convention signifies that centripetal ``potential'' energy is lost as an object moves from the center outward (the opposite of a gravitational potential).

For the two tethered masses we can, finally, write down a potential energy expression with two terms $U(r) = U_c + U_g = -m\omega^2 r^2/2 + M g r$.

5.

Although this last step took some time this form helps show the utility of the relationship between potential energy and the forces acting on an object with, in this case, F(r) = - dU(r)/dr. The final expression of this exercise becomes:

$$F(r) = -[-m\omega^2 r +Mg]= m\omega^2 r -Mg$$

At the initial equilibrium position (at R) F(r)=-dU(r)/dr=0 but the second derivative $d^2 U(r)/dr^2=-m\omega^2$ is negative implying an unstable equilibrium. With this U(r) above we can obtain the velocity of the two masses (radial component in the case of mass m) at any radius along the track correctly in the non-inertial reference frame.  This expression can also be used in the inertial frame as long as we are careful to not include tangential componets of the velocity in the kinetic energy
term.