# A rotating dumbbell

On a horizontal table there is a dumbbell placed vertically. The ends of the dumbbell are two small spheres linked by a massless thin rod of length $l$. The upper ball is struck, receiving a horizontal velocity $v$. Find the magnitude of $v$ such that the bottom ball immediately detaches from the table.

In the picture I've drawn the set up and equations at the equilibrium state, right before $v$ takes precedence and violates equilibrium. The above system writes like this:

\begin{cases} N = mg \\ N_0 = 2mg \end{cases}

Now, for the lower ball A to detach from the table, the direction of N acting on A should become opposite and be at least equal to $mg$. That is $$N\ge mg$$ But is this possible physically? I mean, can the change in linear momentum cause N to change direction? If so explain please rigorously step by step

More than that, I am not sure how the system behaves in this circumstances. My intuition says that we should have a rotation of B around the instantaneous point A. Another stackexchange post says that the system actually has a rotation around the CM and a horizontal translation of it. Are those two equivalent?

we can write (using Chasles' formula):

$$\vec{v_B} = \vec{v_A}+\vec{\omega}\times\vec{r_{AB}}=\omega\vec{k}\times l\vec{i}=\omega l\vec{j}$$

this quite clearly indicates to me that actually the velocity of B is horizontal. Probably the above statement is only for one instant after the system actually starts to move.

I am a bit confused and I don't know exactly what questions to ask and consequently what responses to search for. I am trying to learn physics on my own and is difficult to construct your own model. A rigorous mathematical treat of this problem should be much appreciated.

• See this post for relevant math on offset impulses. – John Alexiou Jan 1 '18 at 19:24
• The issue here is that there is a frictional impulse at the contact acting horizontally which changes the rate of rotation. Since this impulse is purely a function of sliding friction, one must know the duration of sliding to estimate the total magnitude. I am not convinced enough information is given to solve this problem. – John Alexiou Jan 2 '18 at 5:51

When there is no fixed axis-of-rotation, your best choice is to separately analyze motion of center-of-mass and rotation about center-of-mass. Your link, and in particular the answer by P.S. emphasizes this point.

Following is not a rigorous mathematical treat, but an effort to built a physics model.

That link handles a dumbbell flat on a horizontal table:

The lower ball accelerates upwards and shifts from its initial position. Upward acceleration at t=0 is radial hence: $a_u=\omega^2r=\frac{v_{rot}^2}{r}=\frac{v_{rot}^2*2}{l}$.

In the current problem, the system is vertical and we need to add gravitation which accelerates the whole system downwards with $a_d=g$. For ball B to detach, We need to make sure that $a_u>a_d$ or $v_{rot}^2>\frac{lg}{2}$.

Going back to P.S. answer in the link, $v_{rot}=\frac{V}{2}$ where V is the velocity of ball A (the other half rises from $v_{cm}$). We conclude that:

$$V>\sqrt{2lg}$$

There are some problems in the setting of the question!

If we assume the upper sphere has any velocity, V, it means already some time has passed from the equilibrium because V= A.t. A is acceleration and t time. Therefor asking about what V is wrong, for t to be zero( immediately per question), the immediate V must be zero!

Also if we change the question into an impulse or impact to upper sphere case, the other problem is the lower sphere is not a zero radius mass and before it can start to move vertically it has to rotate enough under combination of torque and tension. This is assuming that the impact is elastic and goes through the CG of upper sphere otherwise the system becomes a bit complicated!