So, I'm studying for an entrance exam and came across this question. We hit the bottom block of mass $M=2.5 kg$ with a hammer. This block is attached to a massless rod of length $\ell=0.6m$, pivoted on its midpoint. On the other end of this rod sits a point mass of $m=0.5kg$, as shown in the figure: enter image description here

I need to figure out the minimum impulse due to striking the bottom block with the hammer if the bar gets to the horizontal position. Consider $g=10\:\: m/s^2$.

Here is my attempt: I used the conservation of energy theorem to figure out the initial speed of the blocks, that is: $$\frac{2.5\cdot v^{2}}{2}+\frac{0.5\cdot v^{2}}{2}+0.5\cdot g\cdot 0.6=\left ( 2.5+0.5 \right )\cdot g\cdot 0.3$$ With that we find $v=2\:\: m/s$, implying $v_{C.O.M.}=\frac{4}{3}\:\:m/s$ and $\omega=\frac{20}{3}\:\:rad/s$.

Now, let's say the impulse due to the hammer is $\overrightarrow{J}$. Using the fact that $$\int \overrightarrow{\tau}\times dt = \Delta\overrightarrow{L}$$ we get: $$J\cdot 0.3 = I_{cm}\cdot \omega \:\:+r_{cm}\cdot p_{cm}$$ replacing the values i get $J = 6\:\:N\cdot s$.

The book, however, states that $J$ should be $8\:\: N\cdot s$

I attempted to do the problem from the $C.O.M$'s perspective ass well (using it as the pole of rotation), so when we hit the block, there's also an impulse due to the pivot, let's call it $J'$. Then: $$J - J' = \Delta p_{cm} \rightarrow J - J' = 4\:\: N\cdot s \:\:\:\: (1).$$ Using again the fact that $$\int \overrightarrow{\tau}\times dt = \Delta\overrightarrow{L}$$ we now get: $$ J\cdot 0.1 + J'\cdot 0.2 = I_{cm}\cdot \omega \rightarrow J + 2\cdot J' = 10\:\:N\cdot s \:\:\:\:(2) .$$ Again, using $(1)$ and $(2)$, I get $J=6\:\:N\cdot s$.

What am I missing? I can't find my mistake and I would appreciate any kind of help.

  • $\begingroup$ How is your $I_{\rm cm}$ value calculated? $\endgroup$ – John Alexiou Oct 2 '19 at 23:32
  • $\begingroup$ $2.5\cdot (0.1)^2 + 0.5\cdot (0.5)^2$ @ja72 $\endgroup$ – Leonardo V. Sailer Oct 2 '19 at 23:36
  • $\begingroup$ It is hard to tell what is going on because you are working with the numbers. Try to solve this using the variables names and only plug in the numbers in end. This way you can tell if the equations are correct and the units work out. $\endgroup$ – John Alexiou Oct 2 '19 at 23:41
  • $\begingroup$ What is the value of $g$ here? $\endgroup$ – John Alexiou Oct 2 '19 at 23:56
  • 1
    $\begingroup$ Well, I actually did that on paper, but I'm awful with MathJax and it took me a long time to write those few equations. If I had more time available I would have written it step-by-step. Sorry if it's hard to understand. @ja72 $\endgroup$ – Leonardo V. Sailer Oct 2 '19 at 23:57

I'd like to consider the system of two masses as a single body with mass

$$ m_C = m + M = 3\,{\rm kg} \tag{1}$$

center of mass from the pivot

$$c = \frac{ \tfrac{\ell}{2} m - \tfrac{\ell}{2} M }{m+M} = -0.2\, {\rm m} \tag{2}$$

and combined mass moment of inertia

$$I_C = m \left( \tfrac{\ell}{2}-c \right)^2 + M \left( -\tfrac{\ell}{2}-c \right)^2 = 0.15\,{\rm kg\,m^2} \tag{3}$$

The energy balance between vertical (initial) and horizontal configuration is found from the velocity of the center of mass $v_C = (-c) \omega$ and the rotational velocity $\omega$

$$ m_C g (0-c) = \tfrac{1}{2} m_C ( -\omega c )^2 + \tfrac{1}{2} I_C \omega^2 \tag{4} $$

Which is solved for the final rotational velocity

$$ \omega = \sqrt{ \frac{ 2 (-c) m_C g}{I_C + m c^2} } = \tfrac{20}{3}\,{\rm rad/s} \tag{5} $$

So far so good, up to now.

Now we have the equation of motion, where applied momentum (impulse) causes motion. There is the linear momentum equation and the angular momentum equation. As mentioned, the hammer impulse is $J$ and the pivot reaction is $J'$

$$ \begin{aligned} J - J' & = m_C v_C = -m_C c \omega \\ \left(\tfrac{\ell}{2}+c \right) J - c J' &= I_C \omega \end{aligned} \tag{6}$$

The above is solved for the two impulses as

$$ \begin{aligned} J &= \frac{2 \omega \left( I_C + m_C c^2 \right)}{\ell} = 6 \,{\rm N s} \\ J' & = \frac{2 \omega \left( I_C + m_C c^2 + m_C c \tfrac{\ell}{2} \right)}{\ell} = 2 \,{\rm N s} \end{aligned} \tag{7} $$

The answer matches the posting, so go figure.

  • $\begingroup$ Thanks for your answer, but I found $J$ to be $6\:\: N\cdot s$, the book gives another value. I think you meant $N\cdot s$ at the last two lines instead of $J\cdot s$. Thank you, now I can be sure that my answer is correct. $\endgroup$ – Leonardo V. Sailer Oct 3 '19 at 0:40
  • $\begingroup$ I see that. I also corrected the units for $J$. my bad. Did you get the same value for $J'$ also? Then you could be for sure for sure correct. $\endgroup$ – John Alexiou Oct 3 '19 at 0:48
  • $\begingroup$ Yes, I just verified that. Thank you very much. $\endgroup$ – Leonardo V. Sailer Oct 3 '19 at 0:57

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