# Two masses connected by vertical rod is displaced

The following question is from 2016 $F=ma$ exam. My question is with regards to one step in the reasoning of the solution.

Original question: A small ball of mass $3m$ is at rest on the ground. A second small ball of mass $m$ is positioned above the ground by a vertical massless rod of length $L$ that is also attached to the ball on the ground. The original orientation of the rod is directly vertical, and the top ball is given a small horizontal nudge. There is no friction; assume that everything happens in a single plane. Determine the horizontal displacement $x$ of the second (originally top) ball just before it hits the ground.

The given answer is $3L/4$.

I understand that the C.M. does not move horizontally, so the horizontal displacements of the balls are always in $3:1$ ratio. However, how do we prove that the heavier ball never lifts off the ground? To me it seems plausible that while the C.M. is moving down and the rod is pivoting around the C.M., the heavier ball could lose contact with the ground. Hence, it is totally plausible that when the lighter ball hits the ground, the rod is not in a horizontal position. How do I eliminate this possibility?

• @JohnForkosh The normal force is $4mg$ at the start while weight is only $3mg$... – suncup224 Sep 16 '17 at 10:54
• @suncup224 Shouldn't you also consider the additional downwards force provided by the tension in the rod acting on the lower ball? – EigenFunction Sep 16 '17 at 10:58
• @EigenFunction You do NOT as this those internal forces do not contribute any force outside as it is a rigid body. – Tausif Hossain Sep 16 '17 at 11:01
• @TausifHossain Wouldn't that only be true for the net force acting on the entire rigid body? It is true that, if we consider the body as the rod and both balls, then the forces acting on the body is only the weight of the body and the reaction force. However, WITHIN the body, the force acting on each ball is clearly not 4mg for each, therefore shouldn't the tension contribute to the force acting on each ball? – EigenFunction Sep 16 '17 at 11:07
• Oh yes then you are correct and if you read my answer you'll find that I have actually(unknowingly)used it as I calculated the acceleration of the bottom end of the rod, although I have not explicitly stated the forces that cause the acceleration. – Tausif Hossain Sep 16 '17 at 11:15

I've deleted my previous answer as it has some flaws in logic. I will attempt to show you how this situation works below. The center of mass of the object is $$L/4$$ from the bottom. As the only forces acting on the object is vertical( gravity and normal reaction force at bottom) so it's center of mass will travel only vertically.

Now if we can show that there can exists a plausible normal reaction force which may vary( and that is not acting downwards, that is, it is positive) so that the end of the rod does not move vertically (after plugging in all the other accelerations of motion) then we can prove that the rod's end does not lift up.

First we find the moment of intertia about the center of mass(COM) of the rod. You'll find it is $$\frac{3}{4}mL^2$$. Then we take moments about COM. We will arrive at the following equation: $$\alpha=\frac{Nsin(\theta)}{3mL}$$ Where $$N$$ is the normal reaction force and $$\theta$$ is the angle rotated. Now the linear acceleration at bottom is given by $$a=\frac{L}{4}\alpha$$ so thus the vertical component of the acceleration on the bottom is given by : $$a_v=asin(\theta)$$( refer to diagram below). Thus finally: $$a_v=(N/12m)sin^2(\theta)$$ so the vertical force on the end of the rod is given by $$3ma_v$$.

Now we can write an equation so that the vertical acceleration at bottom is zero and find the expression for $$N$$: $$N+3ma_v-3mg=0$$ So finally: $$N=\frac{12mg}{3+sin^2(\theta)}$$ Here you can see that for $$\theta=0-to-90$$, $$N$$ remains positive( situation is hence plausible as there is no problem with the varying positive magnitudes of $$N$$). Thus the bottom of the rod does not lift up. Note: The horizontal component of the linear acceleration at the bottom of the rod still exists thus the bottom of the rod moves to the left from the initial position. • Thanks for the detailed analysis. I understand your approach so I did the calculations separately after that and got different results (but same conclusion). Not sure if I missed something. For example, for me the moments about CM is $\sin\theta(\frac{3}{4}Lmg-\frac{1}{4}L3mg + \frac{1}{4}LN) = N\frac{1}{4}L\sin\theta$, so dividing by $I$ gives $\alpha = \frac{N\sin\theta}{3mL}$. Therefore $a_v$ in the frame of CM is $(\frac{1}{4}L\alpha)\sin\theta = \frac{N\sin ^2\theta}{12m}$. We want this balanced by $a_v$ of CM which is $g-\frac{N}{4m}$. Solving, $N = \frac{12mg}{\sin ^2\theta +3}$ – suncup224 Sep 16 '17 at 11:15
• To check, I substituted in the special values of $\theta=0$ and $\pi/2$, and they yield $N=4mg, 3mg$ respectively, which seems to be correct, while your suggested solution yields $N=3mg, \frac{11}{5}mg$ – suncup224 Sep 16 '17 at 11:17
• I seem to have treated the problem just for the bottom mass separately but for the whole body your answer seems absolutely correct. – Tausif Hossain Sep 16 '17 at 11:21
• I missed out the tension in the rod due to the mass at the bottom, I'll correct it wait. – Tausif Hossain Sep 16 '17 at 11:22
• Thanks so much for having this discussion with me. It enlightened me alot! – suncup224 Sep 16 '17 at 12:01

The ground doesn't push you back with a greater force

The Normal force and the force with which $3m$ mass is pushing the ground form action-reaction pairs. The Normal force has only the job to prevent the end $3m$ of rod from sinking into the ground.

Also, the rod is rotating about $CM$ and there is no other force other than Normal force that provides torque. If there would have been an other force which provided torque, then there would have been a possibility of end $3m$ lifting up.