Sliding Along a Rotating Rod

A 1 m \SI{1}{\meter} rod of uniform composition and total mass 1 kg \SI{1}{\kilo\gram} has one end hinged at the origin. A 1 kg \SI{1}{\kilo\gram} bead is placed at the other end and is free to slide along the rod (if it reaches the open end, it will separate from the rod).

The bead and rod are prepared so that the bead sits near the open end of the rod which makes an angle of θ = 6 0 \theta=60^\circ with the horizontal.

If this system is released from rest at time zero, at what time (in seconds) does the bead slide off of the rod?

Assumptions:

  • g = 10 m / s 2 . g = \SI[per-mode=symbol]{10}{\meter\per\second\squared}.


The answer is 0.650.

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2 solutions

Steven Chase
Jul 30, 2017

Here is the energy / Lagrange way. We will operate in polar coordinates ( r , θ ) (r,\theta)

Bead kinetic energy:

E b e a d = 1 2 m ( r 2 θ ˙ 2 + r ˙ 2 ) E_{bead} = \frac{1}{2} m (r^2 \dot{\theta}^2 + \dot{r}^2)

Rod kinetic energy: E r o d = 1 2 m L 2 3 θ ˙ 2 E_{rod} = \frac{1}{2} \frac{m L^2}{3} \dot{\theta}^2

Bead potential energy:

U b e a d = m g r s i n θ U_{bead} = mgr \, sin \theta

Rod potential energy:

U r o d = 1 2 m g L s i n θ U_{rod} = \frac{1}{2} mgL \, sin \theta

Lagrangian:

L = E t o t a l U t o t a l L = 1 2 m ( r 2 θ ˙ 2 + r ˙ 2 ) + 1 2 m L 2 3 θ ˙ 2 m g r s i n θ 1 2 m g L s i n θ L = E_{total} - U_{total} \\ L = \frac{1}{2} m (r^2 \dot{\theta}^2 + \dot{r}^2) + \frac{1}{2} \frac{m L^2}{3} \dot{\theta}^2 - mgr \, sin \theta - \frac{1}{2} mgL \, sin \theta

Equations of motion:

d d t L r ˙ = L r d d t L θ ˙ = L θ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{r}}} = \frac{\partial{L}}{\partial{r}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}

There is nothing special about the math from here on, so I'll just post the results. They match what @Mark Hennings derived via Newtonian mechanics. Having expressions for the "double-dot" terms for the two variables, we can readily solve numerically.

r ¨ = r θ ˙ 2 g s i n θ θ ¨ = 2 r r ˙ θ ˙ g r c o s θ 1 2 g L c o s θ r 2 + L 2 3 \ddot{r} = r \dot{\theta}^2 - g \, sin{\theta} \\ \ddot{\theta} = \frac{-2 r \dot{r} \dot{\theta} - gr \, cos{\theta} - \frac{1}{2} g L \, cos{\theta}}{r^2 + \frac{L^2}{3}}

Here are some interesting plots of the behavior (time is on the horizontal axis). We see that when θ \theta is positive, the bead slides toward the origin as expected. At this time, the angle is decreasing, also as expected. When the angle turns negative, gravity begins to pull the bead toward the end of the rod, causing the radius to increase once more. It eventually slides off at around t = 0.650 t = 0.650 .

So, as the problem is stated, the bead won't fall off at all. The rod and the bead will hit the ground first, which was my intuition about the thing.

This is a physics problem, not fundamentally a math problem, so the answer has to comport with the physicality of the given situation. If the hinge isn't on the ground, you need to say so, and I'd avoid using the term "ground" and substitute "horizontal" to be clear.

Ban Yan - 3 years, 9 months ago

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Yes, I know. When I posted this problem, there was no mention of "ground". That was added later. I will report it.

Steven Chase - 3 years, 9 months ago

Want an understanble explanation for a class 10 student. All these are beyond my brain's capacity. I am bad at playing bomcer balls

Swapan Das - 3 years, 9 months ago

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I'm afraid there's no way to tackle this one without some heavy math.

Steven Chase - 3 years, 9 months ago

There should be a hint that the solution is numerical, so that it is clear that the task is to find the correct differential equations, not to attempt to solve them.

Darko Simonovic - 3 years, 9 months ago

Don't you have to account for the two components of the kinetic energy of the bead, i.e. the translational and the rotational? You didn't consider the latter. (But maybe you are assuming point mass.)

Eric Zeng - 3 years, 9 months ago

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Indeed, the bead is a point-mass. This can be inferred from the fact that no size was given for the bead.

Steven Chase - 3 years, 9 months ago
Mark Hennings
Jul 30, 2017

If the rod makes an angle θ \theta above the horizontal, and the bead is a distance x x from the hinge, then the position vector of the bead is r = x ( cos θ sin θ ) \mathbf{r} \; = \; x\binom{\cos\theta}{\sin\theta} and so the equation of motion of the bead is ( x ¨ x θ ˙ 2 ) ( cos θ sin θ ) + ( x θ ¨ + 2 x ˙ θ ˙ ) ( sin θ cos θ ) = ( 0 10 ) + N ( sin θ cos θ ) (\ddot{x} - x\dot{\theta}^2)\binom{\cos\theta}{\sin\theta} + (x\ddot{\theta} + 2\dot{x}\dot{\theta})\binom{-\sin\theta}{\cos\theta} \; = \; \binom{0}{-10} + N\binom{-\sin\theta}{\cos\theta} where N N is the magnitude of the normal reaction between the bead and the rod. Taking radial and tangential components, we obtain the equations x ¨ x θ ˙ 2 = 10 sin θ x θ ¨ + 2 x ˙ θ ˙ = N 10 cos θ \begin{aligned} \ddot{x} - x\dot{\theta}^2 & = -10\sin\theta \\ x\ddot{\theta} + 2\dot{x}\dot{\theta} & = N - 10\cos\theta \end{aligned} On the other hand, the equation of motion of the rod is 1 3 θ ¨ = ( N x + 5 cos θ ) \tfrac13\ddot{\theta} \; = \; -(Nx + 5\cos\theta) Eliminating N N from these equations, we obtain the simultaneous differential equations x ¨ x θ ˙ 2 = 10 sin θ x 2 θ ¨ + 2 x x ˙ θ ˙ + 10 x cos θ = 1 3 θ ¨ 5 cos θ \begin{aligned} \ddot{x} - x\dot{\theta}^2 & = -10\sin\theta \\ x^2\ddot{\theta} + 2x\dot{x}\dot{\theta} + 10x\cos\theta & = -\tfrac13\ddot{\theta} - 5\cos\theta \end{aligned} together with the initial conditions x ( 0 ) = 1 x(0) = 1 , x ˙ ( 0 ) = 0 \dot{x}(0) = 0 , θ ( 0 ) = 1 3 π \theta(0) = \tfrac13\pi , θ ˙ ( 0 ) = 0 \dot\theta(0) = 0 .

Solving these equations numerically, I obtain that x = 1 x=1 for the first time after t = 0 t=0 at time t = 0.650123 t = \boxed{0.650123} .

Thanks for a challenging and illuminating problem. I was looking through the data and came to this description: The transition between the "phases" and "events" are of course completely smooth.

Phase 1: For the first 300ms the bead drops and the rod begins rotating clockwise when the rod gets to about 45 deg above level the centrifugal reaction to the bead's circular path and reduced centripetal component of gravity begin decelerating the bead's inward motion. The bead is about a third the way in from the end.

Event 2: at about 464ms the bead stops all radial motion, the rod is about 10 degrees below level and the bead has traveled just over half the length of the stick.

Event 3: At about 504ms despite gravity acting on the bead and rod, to accelerate the rod clockwise the Coriolis forces are starting to dominate, and these forces want to slow the rod, and they are taking over. The rod is about 30degrees below horizontal.

Phase 4: The effect of the Coriolis force is immense, the bead is accelerating outwards and peaks at 3.5G, the rod is decelerating rapidly and loses half its speed before the bead leaves the rod. That's a massive energy transfer to the bead far larger than the gravity effect during this phase.

Event 5: At about 650ms the bead leaves the rod.

In summary: The beads drops down the rod about half way, the bead and rod swing to about 30degress below level and then the rod flings the bead off (at over 3G) the bead leaves nearly vertically downwards.

Ed Sirett - 3 years, 9 months ago

Thanks. I got a notification that you filed a report, but I can't see anything there. Was there a problem?

Steven Chase - 3 years, 10 months ago

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I made a small mistake (incorrect minus sign), which meant I thought your answer was wrong. Having found my error, I deleted the report...

Mark Hennings - 3 years, 10 months ago

Can you elaborate your second step?

Rohit Gupta - 3 years, 10 months ago

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All I am doing is differentiating r \mathbf{r} twice with respect to time, using the chain and product rules as appropriate, then applying Newton's Second Law to the bead, then taking components. That gets me down to "On the other hand..."

Mark Hennings - 3 years, 10 months ago

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Ok, thanks.

Rohit Gupta - 3 years, 10 months ago

I was assuming that the X-axis is some sort of hard surface (like a table) in which case the bead would not slide off. The question (to me) was somewhat misleading.

Paul Mak - 3 years, 9 months ago

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Yeah, I notice that. The wording about "ground" wasn't there when I originally posted it. Please report. Thanks

Steven Chase - 3 years, 9 months ago

0 pending reports

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