Dumbbells: Newsletter Challenge (02/12/08)

The question as it appears in the 02/12 edition of Specs & Techs from GlobalSpec:

You have a dumbbell (similar to the ones used in weight training) consisting of a rod of a certain length (l) and two masses (m) at each end. The dumbbell is standing vertically in a corner formed by two frictionless planes (a wall and the floor). When you move the bottom end slightly away from the wall, the dumbbell begins to slide. Determine the speed of the bottom end at the exact moment when the top end loses contact with the vertical plane.

(Update: Feb 19, 8:31 AM EST) And the Answer is...

You have a dumbbell (similar to the ones used in weight training) that consists of a rod of length l and two masses (m) at each end. The dumbbell is standing vertically in a corner formed by to frictionless planes (a wall and the floor). You move the bottom end slightly away from the wall, the dumbbell begins to slide. Determine the speed of the bottom end at exactly the moment when the top end loses contact with the vertical plane.

(Feb 20: 11:05 AM) D'oh, sorry about that: Here's the Correct Answer:

At the moment when the top mass loses contact with the wall the normal force is zero (hence its horizontal acceleration is also zero). A free-body diagram of the system is shown below.

Let be the vertical velocity of the top mass, and let the horizontal velocity of the bottom mass. Knowing that we can write

(1)

We can express gravity as a function of the v_x , as follows

By substituting the value of v_y from Eq. (1), we have

(2)

From the above equation we can determine the horizontal speed, which is given by

(3)

But this equation is still a function of y, a variable. Let's see if we can find a function that relates y to known quantities.

Like all isolated systems this system has to conserve energy. Let's assume the frame of reference to be the horizontal axis (the floor). The total energy before the bottom mass starts to slide is simply given by the potential energy of the top mass, or

(4)

The total energy at the moment the top mass looses contact with the wall is the sum of the potential energy of the top mass and the kinetic energies of both masses. In equation form we have,

(5)

The law of conservation of mechanical energy states that , or in equation form

By substituting Eq. (1) into the above equation we get,

(6)

Solving this equation for y, we get

(7)

Now, substitute Eq. (3) into Eq.(7) and solve for y, or

and

Finally,

(8)

By substituting Eq.(8) into Eq.(3) we finally get the speed of the bottom mass at the moment the top mass looses contact with the vertical wall, or

And

(9)

Note 1: It is interesting to see that Eq.(8) gives the height at which the top mass of the dumbbell looses contact with the vertical wall. One might think that this will happen only when y=0. Remember that the rod is not a flexible device.

Note 2: According to Eq. (9) the speed of the bottom mass is only dependent on the length of the rod (r) that keeps the two masses at the same distance from each other. For a given dumbbell this is a constant, so the horizontal speed is a constant. This is expected because our two planes – the wall and the floor (as is clearly stated in the question) – are frictionless.