I believe most of you probably solved the following problem using energy conservation as shown here. It states

A particle starts from rest at the top of a frictionless sphere of radius R and slides on the sphere under the force of gravity. How far below its starting point does it get before flying off the sphere?

I've be trying to solve this problem using only Newton's laws without energy conservation. I would like to know if it is possible and, if it is, if you could give me some ideas of how to solve it. The problem I am currently having is that I believe that the Normal force in this problem is not a constant, but a function of the angle.

I believe it is clear that the block's trajectory is a curve before it falls from the sphere. If it is a curve, we have a centripetal force given by

$$ m\frac{v^2}{R} = mg\cos\theta - N(\theta) $$

Where I believe $N$ is a function of $\theta$.

When the blocks gets off the sphere, there is no normal force anymore, so at this instant the centripetal resultant is just

$$ m\frac{v^2}{R} = mg\cos\theta $$

One can also see that in the $y$ axis, the resultant force is given by

$$ ma_{y} = P - N(\theta)\cos\theta $$

And the acceleration is

$$ a_{y} = g - \frac{N(\theta)}{m}\cos\theta $$

Now I could try to solve

$$ \frac{dv_y}{dt} = g - \frac{N(\theta)}{m}\cos\theta $$

to get the velocity in the $y$ axis and somehow figure it the height where the normal force is zero... Anyway, this is what I know from the problem and I am lost. Any tips on how to solve it?

  • $\begingroup$ Comparing the last 2 equations, $a_y=dv/dt$ but this is not correct. $\endgroup$ – sammy gerbil Jun 13 '16 at 0:52
  • $\begingroup$ @sammy it was typo. I wrote dv instead of dv_y $\endgroup$ – hdhzero Jun 13 '16 at 1:27
  • $\begingroup$ Too many variables : $v$, $v_y$, $t$, $\theta$, $N$. Can you eliminate some of them? $\endgroup$ – sammy gerbil Jun 13 '16 at 1:31
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    $\begingroup$ Your first mistake is trying to do it in rectangular coordinates. Work it in rotational terms (torque, moment of inertia, and angular coordinates) because that makes the differential equation part of the problem one dimensional. Next, integrating the differential equation requires a trick to eliminate the nasty and unnecessary presence of time. After that it's plug-n-chug. $\endgroup$ – dmckee --- ex-moderator kitten Jun 13 '16 at 2:36
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    $\begingroup$ No. I found the equations of motion on the surface by force considerations and then solved them to get the same answer without ever computing an energy. It just takes deeper mathematics. And it helps to recognize that as long as the particle remains on the surface treating it as a rotation is easier than working it in rectangular coordinates. $\endgroup$ – dmckee --- ex-moderator kitten Jun 13 '16 at 3:09

We put the circular orbit of the particle on a straight line and convert the motion to a 1-dimensional rectilinear motion as follows : The arc length, the natural parameter $\:s(t)\:$ is the distance travelled on the straight line till time $\:t\:$. The speed $\:v(t)\:$ on the straight line is the magnitude of the tangent to the circle velocity. Now, on the straight line the particle is moving like under the influence of the tangent force which is $\:f_{t}=mg\sin(\theta)\:$ so under a variable acceleration $\:a_{t}=g\sin(\theta)\:$. But $\:\theta=s/R\:$ so the differential equation of motion is

\begin{equation} \dfrac{\mathrm{d}^{2} s}{ \mathrm{d}t^{2} }-g\sin\left(\dfrac{s}{R}\right)=0, \qquad \left[\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right]_{t=0}=0, \qquad s(0)=0 \tag{01} \end{equation}

since the particle starts at rest on the origin.

On the other hand the condition for the particle to leave the sphere is the normal force to be zero \begin{equation} N=mg\cos(\theta)-ma_{c}=mg\cos(\theta)- \dfrac{mv^{2}}{R}=0 \tag{02} \end{equation} that is \begin{equation} \boxed { \bbox[#FFFF88,8px]{\:\:\left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}-gR\cos\left(\dfrac{s}{R}\right)=0 \:\:}} \tag{03} \end{equation}

Now, we must solve (01) to find at which point the condition (03) is satisfied. But it'll proved to be not necessary. So, multiplying (01) by $ \dfrac{ \mathrm{d} s }{\mathrm{d} t} $ we have \begin{equation} \dfrac{ \mathrm{d} s }{\mathrm{d} t}\dfrac{\mathrm{d}^{2} s}{ \mathrm{d}t^{2} }-g\dfrac{ \mathrm{d} s }{\mathrm{d} t}\sin\left(\dfrac{s}{R}\right)=0 \tag{04} \end{equation} or \begin{equation} \dfrac{ \mathrm{d} s }{\mathrm{d} t} \dfrac{ \mathrm{d} }{\mathrm{d} t} \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right) +\dfrac{ \mathrm{d} }{\mathrm{d} t}\left[gR\cos\left(\dfrac{s}{R}\right)\right]=0 \tag{05} \end{equation} that is \begin{equation} \dfrac{ \mathrm{d} }{\mathrm{d} t} \Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) \Biggr]=0 \tag{06} \end{equation}

This means that we have found a constant of integration of (01) and more explicitly using the initial conditions

\begin{equation} \Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2g\cos\left(\dfrac{s}{R}\right) \Biggr]=\text{constant}=\Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) \Biggr]_{t=0}=2gR \tag{07} \end{equation} or \begin{equation} \boxed { \bbox[#FFFF88,8px]{\:\: \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) =2gR \:\:}} \tag{08} \end{equation}

Substructing equations (08) and (03) side by side we have finally

\begin{equation} \cos\left(\theta\right)=\cos\left(\dfrac{s}{R}\right) =\dfrac{2}{3} \tag{09} \end{equation}

Notes :

  1. The differential equation of motion (01) is identical to that in the Dvij's answer but with respect to $\:s(t)=\theta(t)R\:$ instead of $\:\theta(t)\:$.

  2. I find the constant of integration (07) of equation (01) motivated by the fact that there exists a constant : the energy. I inserted the energy conservation through the back door.

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If a 'law' of Physics can be really uncared for and still you can predict the outcome of an experiment completely accurately then it is not a law of Physics. So if energy conservation is a physical fact here then either implicitly or explicitly, we are going to use that fact - otherwise, we must not be able to predict the complete outcome. So I assume your question is to calculate the trajectory of the ball without any explicit use of energy conservation but via (as you have mentioned) Newton's equations cold.

Since the radius of the sphere is constant it is easy to use the angular motion equations rather than using rectangular equations with two components. I am measuring $\theta$ from the vertical.

$Rmg\sin\theta$ $=$ $mR^2\dfrac{d^2\theta}{dt^2}$

Or, $g\sin\theta=R\dfrac{d^2\theta}{dt^2}$

This is the equation of motion. We will put the initial conditions $\theta=0$ and $\dfrac{d\theta}{dt}=0$. And we will get more than one solutions to this differential equations! ( It is strange in a way and why that happens is a long discussion. But it does not suggest that Newtonian Mechanics is probabilistic or only partially deterministic. It only suggests that the initial state, in some cases, is not completely described via the derivatives up to the first order in time - we need to specify something more.) Out of those solutions, we will pick the solution in which $\theta$ increases with time. So essentially we now have a known function of time, $f(t)$, so that $\theta=f(t)$.

Having known this much, we can simply write an equation for the Normal Reaction force as follows:



To find out the $\theta$, at which the ball leaves the surface, we will write $N=0$. And that yields

$\theta=\cos^{-1} \bigg(\dfrac{R(f'(t))^2}{mg}\bigg)$

Or, $f(t)=\cos^{-1} \bigg(\dfrac{R(f'(t))^2}{mg}\bigg)$

This is again an equation in $t$ and it can be solved. Note that it is not a differential equation. Becuase the function $f$ is known in the explicit terms of $t$ and thus the equation is just an equation in $t$. Solving it will give the value of time at which the ball leaves the sphere. Call that time $t=T_k$.

Thus, the angle at the time of leaving $\theta_k$$=$$f(T_k)$.

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  • $\begingroup$ Hi. May I ask: What would be the satisfied condition in Lagrangian formalism for getting the angle when the body detaches from the sphere? Thank you. $\endgroup$ – Constantine Black Jun 13 '16 at 7:02
  • $\begingroup$ Your answer is still a function of time. If you don't know $f(t)$, nor the time at which $N=0$, how can you find $\theta$? $\endgroup$ – sammy gerbil Jun 13 '16 at 10:17
  • $\begingroup$ Also, with the initial conditions $\theta=0$ and $\frac{d\theta}{dt}=0$ the block will not move at all. It is initially balanced unstably at the top of the hemisphere. $\endgroup$ – sammy gerbil Jun 13 '16 at 10:27
  • $\begingroup$ @sammygerbil $\theta$ is known as a function of time. And as I said $f(t)$ is also known. So the final equation gives you the time at which the contact is lost. Putting that time in $f(t)$ gives the ultimate answer. $\endgroup$ – Dvij D.C. Jun 13 '16 at 10:56
  • $\begingroup$ @sammygerbil With the initial conditions $\theta=0$ and $\dfrac{d\theta}{dt}=0$, the block does move. The reason is that the instability is owing to the non-vanishing higher derivatives of $\theta$ (wrt time). In a way, you can tell, the angular velocity is infinitely small initially, but in a strict mathematical sense, you ought to put $\dfrac{d\theta}{dt}=0$ since it is a limit which has to be non-infinitesimal by definition. $\endgroup$ – Dvij D.C. Jun 13 '16 at 11:01

@dvij gave the equation $$g\sin \theta =R\frac{d^2\theta}{dt^2}=R\frac{d\omega }{dt}$$ If we multiply this by omega, we obtain: $$g\sin \theta \frac{d\theta}{dt}=R\omega\frac{d\omega }{dt}$$ If we integrate this equation between 0 and t, we obtain: $$g(1-\cos \theta)=\frac{R}{2}\omega^2$$ So we have $$mg\cos\theta-2mg(1-\cos \theta)=N=mg(3cos\theta-2)$$ I don't know whether this counts as an energy method or not.

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  • $\begingroup$ Spurious 'm' in penultimate eqn. You could also point out where the last line ties in with Dvij's answer. $\endgroup$ – sammy gerbil Jun 13 '16 at 14:15
  • $\begingroup$ I don't think there is a spurious m in my last equation. @Dvij omitted an m in his last equation. My next-to-last equation had a spurious m, which I removed. Thanks. $\endgroup$ – Chet Miller Jun 13 '16 at 15:33
  • $\begingroup$ Penultimate means next-to-last. That was not Dvij's only mistake. $\endgroup$ – sammy gerbil Jun 13 '16 at 15:38
  • $\begingroup$ Forgive me, but don't you think that you have repeated (copied) my answer? $\endgroup$ – lucas Jun 13 '16 at 17:20
  • $\begingroup$ @lucas Actully, no. My final equation gives N as a function only of mg and $\theta$, and allows one to calculate $\theta$ directly. Also, you in no place multiplied the force balance in the tangential direction by $d\theta/dt$ and then integrated. This is key to getting an explicit equation for the value of $\theta$ at which separation takes place. $\endgroup$ – Chet Miller Jun 13 '16 at 18:48

As @dmckee said, you should use polar coordinates for this problem. enter image description here Equations of motion are below: ($\omega$ is angular velocity and $\alpha$ is angular acceleration) $$mg\cos \theta-N=mR\omega^2\;\tag 1$$ $$mg\sin \theta=mR\alpha\;\tag 2$$ From $(2)$, we have $\alpha=\large{\frac gR}\sin \theta$

On the other hand, we know: $$\alpha\mathrm d \theta=\omega\mathrm d\omega\;\tag3$$ So, you can find $\omega^2$ as a function of $\theta$.

Then, you can use equation $(1)$ for determining the angle that block loses its contact with the sphere. Not that, at that angle, we have $N=0$

Once, you find $\cos\theta_f$ ($\theta_f$ is the angle that block loses its contact with the sphere) you can find The $h$ by this formula: $\cos \theta_f= \large{\frac{R-h}R}$ enter image description here

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