Calculating Velocity of Falling Dumbbell: Two Masses Connected by Rod

[Square47]

Homework Statement

Two spheres of mass m are connected by a massless rod or length L, and set leaning against a frictionless wall so that the rod is perpendicular to the floor (also frictionless). The bottom mass begins to slide perpendicularly away from the wall, so that the top mass falls perpendicularly towards the floor. Find the velocity of the bottom mass when the velocity of both masses are equal.

Homework Equations

v^2=2gx
x=0.5gt^2

The Attempt at a Solution

Because sin and cos have equal derivatives at 45 degrees, the masses would have equal velocities at that point. This gives me a velocity of 2.4sqrt(L), twice the answer I should get--1.2sqrt(L). Some reverse engineering tells me that the angle would be 67 degrees for the answer in the book.

However, when I try a (much) more complicated set up, letting Lsin = L-y, (where y is the distance the top mass has fallen), using the identity sin^2+cos^2=1, I find that the distance the bottom mass has moved away from the wall Lcos=sqrt(2Ly-y^2). Since y=0.5gt^2, the distance from the wall should be sqrt(Lgt^2-0.25(gt)^2). Taking the derivative with respect to time gives me the velocity of the bottom mass, which I set equal to the velocity of the top mass, gt. Solving for t (by completing the square), I get two answers for t, correlating to two velocities, 2.4sqrt(L) and 5.8sqrt(L). Of course, the first answer is the same as above, which again confirms the 45 degree angle. So which is it, 45 degrees or 67?

 

[ehild]

You are right, the speeds are equal when the dumbbell makes 45° angle with the wall. The magnitude of the velocity you got is wrong. As it is not free fall, you can not use y=0.5gt^2 or v^2=2gx. The walls are frictionless, use conservation of energy.

ehild

 

[ideasrule]

Because sin and cos have equal derivatives at 45 degrees, the masses would have equal velocities at that point. This gives me a velocity of 2.4sqrt(L), twice the answer I should get--1.2sqrt(L). Some reverse engineering tells me that the angle would be 67 degrees for the answer in the book.

I don't see how either 2.4sqrt(L) or 1.2sqrt(L) are correct, especially because sqrt(L) does not have units of speed.

If the two masses have equal speed at 45 degrees, the potential energy lost by the system is mgL(1-1/sqrt(2)), since Lcos(45 deg) is the new height of the falling mass. This lost potential energy translates to kinetic energy, so 2*1/2mv^2=mgL(1-1/sqrt(2)). Solving for v doesn't give you a factor of either 2.4 or 1.2.

However, when I try a (much) more complicated set up, letting Lsin = L-y, (where y is the distance the top mass has fallen), using the identity sin^2+cos^2=1, I find that the distance the bottom mass has moved away from the wall Lcos=sqrt(2Ly-y^2). Since y=0.5gt^2

This assumption isn't valid. The upper mass doesn't just feel gravity; it also feels an upward force from the bar connecting the two masses. The force from this bar partially counteracts gravity, causing it to accelerate downwards at less than "g".

 

[Square47]

ahh, much better. I had a feeling that it wasn't free fall or trig that make this problem interesting. How would one investigate the forces exerted on the masses by the rod?

 

[ehild]

It is a bit complicated to take all forces into account. And the dumbbell is a "rigid body"( so the two balls move together) you do not need to consider the force between its parts, but there is a normal force from the wall and the other one from the ground in addition to the force of gravity.
Use conservation of energy to get the velocity, as Ideasrule suggested.

ehild