Consider a particle moving in a vertical circle of radius $r$. The particle is acted by gravity and it is connected to the center of the circle by a massless rod.

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If I decompose the gravity force into its components along the radial and tangential directions to the circle I get

$$-mg \hat{j}=-mg(\cos\theta \hat{\theta}+\sin\theta \hat{r}) $$

where $\hat{j},\hat{\theta},\hat{r}$ are unit vectors along the $y,\theta,r$ directions (the positive $\theta$ direction is counterclockwise, and the positive $r$ direction is pointing out of the circle)

If I then apply Newton's second law I get the 2 equations (the first being the tangential component of the force, the second the radial component)

$$-mg\cos\theta =mr\frac{d^{2}\theta}{dt^{2}}\tag{1}$$

$$N-mg\sin\theta = -mr\left(\frac{d\theta}{dt}\right)^{2}\tag{2}$$

where $N$ is just the reaction of the rod on the particle.

Integrating the first of these equations I get $$-mg\sin\theta=\frac{1}{2}mr\left(\frac{d\theta}{dt}\right)^{2}\tag{3}$$ where I assume that $d\theta/dt=0$ when $\theta=0$ (so the integration constant is zero).

Replacing in the second equation I get



which seems wrong since it seems to imply the reaction of the rod to the particle also points to the center of the circle like the radial gravity component.

I must have done a mistake in a sign somewhere but don't see where.

  • $\begingroup$ that sounds right, the rod is radial, and constraints the mass radially, therefore the force it exerts are in the radial direction. Why do you think it is wrong? (btw I haven't checked you maths) $\endgroup$
    – JMLCarter
    Dec 24, 2016 at 20:10
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    $\begingroup$ @JMLCarter I think it is wrong because N points inward toward the center of the rod just like the radial component of the gravity force does. If both radial forces point towards the center, I think the particle could not stay on the circle but would move towards the center. $\endgroup$ Dec 24, 2016 at 20:18
  • $\begingroup$ Your first integration is wrong because you integrated the left side with respect to $\theta$ and the right side with respect to time, $t$, but you didn't coordinate the two variables. There isn't any differential relationship specified that justifies that integral. $\endgroup$
    – Bill N
    Dec 25, 2016 at 2:16
  • $\begingroup$ @BillN If you differentiate equation (3) with respect to time you get $-mg\cos\theta d\theta/dt=mr d\theta/dt d^{2}\theta/dt^{2}$ which is the same as equation (1). This shows that (3) is an integral of (1). $\endgroup$ Dec 25, 2016 at 2:34
  • $\begingroup$ Yep, that derivative works. OTOH when you have an indefinite integral you need to include a constant of integration. Haven't checked all the possibilities, but don't assume it's zero without justification. $\endgroup$
    – Bill N
    Dec 25, 2016 at 2:43

2 Answers 2


It is sometimes better to find an answer by a different route and then consider what went wrong or right using the initial problem solving equations.

Let $N$ be the force exerted by the rod, length $l$, radially inwards and the speed of the mass $m$ be $v$ when the angle between the rod and the horizontal is $\theta$.

N2L gives $mg \sin \theta+N = \frac {mv^2}{l}$.

Now consider energy conservation with the speed of the mass equal to $v_o$ when $\theta =90^\circ$.

$\frac 12 m v^2_o + mgl(1-\sin \theta) =\frac 12 m v^2$

Combining the two equations and eliminating $v$ gives

$N= \frac {mv^2_o}{l}+mg(2-3\sin \theta)$.

This equation shows that the force exerted on the mass by the rod $N$ depends on the angle $\theta$ and the speed at the top $v_o$ and can be positive or negative.

You might have done a similar solution with a string which can only exert an inward radial force and if the mass is not moving fast enough it fails to execute a circular.
The rod can exert an outward radial force to ensure the mass executes a circular path.

My concern is with your integration of the first equation where you must have included an unspecified initial condition otherwise you would have finished up with a constant of integration. What was your initial condition?

  • $\begingroup$ My initial condition was that for $\theta=0$, $d\theta/dt=0$. Is this inconsistent? $\endgroup$ Dec 24, 2016 at 23:55
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    $\begingroup$ @user2175783 I think that with your initial condition the mass can only be oscillating when $\theta$ is negative i.e. Below the horizontal line that you have drawn in your diagram. So it cannot move in a complete vertical circle. When the mass is at $\theta = 0$ it is not moving, it has no kinetic energy so cannot rise any higher. $\endgroup$
    – Farcher
    Dec 25, 2016 at 20:20
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    $\begingroup$ @user2175783 You would then have to adapt my conservation of energy equation and have that $mgl\sin \theta = \frac 12mv^2$ with $\theta$ being measured from the horizontal line and being positive below it. $\endgroup$
    – Farcher
    Dec 25, 2016 at 20:30
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    $\begingroup$ The rod is massless all it does is keep the mass at a fixed distance from the pivot point. The only two forces acting on the mass are gravitational attraction and the radial force exerted by the bar. My aim was to point you to a solution to help you find a solution using your method. See what you get by putting one of the limits of your integration as ${d\theta}{dt}=\omega_o$ when $\theta=90^\circ$. $\endgroup$
    – Farcher
    Dec 25, 2016 at 21:23
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    $\begingroup$ Then compare the result of your integration with the energy conservation method remembering $l\frac{d\theta}{dt}=v$ and $l\omega_o=v_o$ $\endgroup$
    – Farcher
    Dec 25, 2016 at 21:54

OK it looks like you have
1) integrated a tangential expression for acceleration due to gravity to get
2) an expression for tangential velocity due to gravity.
The change from cos to sin is due to the integration, not a change in direction. Which you have
3) then substituted into a radial expression for forces, replacing the term for radial gravitational acceleration with tangential velocity due to gravity, which I think is your mistake.

Just because two terms have the same form doesn't mean they represent the same thing. maybe clearer to say that radial and tangential should be resolved distinctly they are orthogonol.


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