homework and exercises - Particle sliding on a sphere

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?

Answer

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.