Problem statment: A rod hinged at one end is released from the horizontal position. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then find the maximum angle $θ$ in degrees made by the hinged upper half with the vertical.


Let $w$ be angular velocity

Use energy conservation to get $w_1=(3g/l)^{1/2}$

Now, as there is no external torque acting when the rod is vertical, angular momentum must be conserved (according to me).

So $w_2 = 8*w_1$ ($Iw=const$)

Use energy conservation again.

We get $cos(\theta) = -3$ which is definitely not right

However, if we assume angular velocity to not change will land to the answer which is $60°$

I just need a reason for the angular velocity to not change, I know how to proceed.

PS: Sorry for bad formatting, I am new here.

  • $\begingroup$ "Now, as there is no external torque acting when the rod is vertical, angular momentum must be conserved (according to me)".Angular momentum of WHAT? The broken off piece doesn't stop dead; it flies off horizontally into a parabolic path... $\endgroup$
    – DJohnM
    Commented Jun 23, 2019 at 18:35

2 Answers 2


Let us consider an analogous situation for linear motion. Suppose two trains were coupled and were going with same velocity, with no net force on them. Suddenly they are uncoupled in such a way that no force is excerted between them. Then the velocities of the trains will not change just after the decoupling as the net force on them were zero.

Similarly, consider the upper half of the rod. At vertical position there is no torque due to gravity. Since the halves of the rod don't apply any force on each other during separation, the net torque on the upper half of the rod will be zero. Therefore its angular momentum (hence angular speed) will not change.

In your solution you missed the fact that the lower end of the rod also has some angular momentum and it will retain that after separation. The angular momentum of the upper end will remain constant before and after separation.

Similarly, the linear momentum of each train will be conserved after decoupling (not that the total momentum of the two trains will be transferred to a single train)

  • $\begingroup$ Consider a ballet dancer rotating with arms streched initially. Now, if the arms are brought closer, the angular velocity increases because of the decrease in moment of inertia although no net torque is acting.This is where the confusion began. $\endgroup$
    – Mathboi
    Commented Jun 23, 2019 at 17:38
  • $\begingroup$ That is because the moment of inertia decreases and the angular momentum remains constant, so the angular speed must increase. But if the arms simply got separated from the body (without any force between the body and the arm during separation) there would be no change in angular velocity. $\endgroup$ Commented Jun 23, 2019 at 17:39
  • $\begingroup$ @Mathboi The lower half of the rod will retain its angular momentum after separation (as there was no torque on it) and will not transfer it to the upper half. $\endgroup$ Commented Jun 23, 2019 at 17:42

L = original length of rod

Θ = angle above vertical the (L/2) rod travels

w = angular velocity at the vertical point

H = the amount the CM of the full rod drops to become vertical (= L/2)

h = height the short piece CM rises from vertical to attain angle Θ (h = (L/4)*(1 - cosΘ))

m = mass of full rod

I = mL²/3

KE = PE → ½Iw² = mgH → mL²w²/6 = mg(L/2) or w² = 3g/L

Using the same energy equation for the short piece, and noting that I = mL²/24,

½(mL²/24)(3g/L) = (m/2)gh = (m/2)g(L/4)(1 - cosΘ) which simplifies to

cosΘ = 0.5

Θ = 60°

  • $\begingroup$ This same solution is all over the internet. But why isn't the angular velocity changing? $\endgroup$
    – Mathboi
    Commented Jun 23, 2019 at 17:16
  • $\begingroup$ Because a chunk of the AM that you are trying to conserve is sailing off with the broken-off piece of rod in its parabolic path... $\endgroup$
    – DJohnM
    Commented Jun 23, 2019 at 18:32

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