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A small particle of mass $m$ is atop of a semi-sphere as shown in the figure. A little push was given to the particle. Prove that the particle will leave the spherical surface at a height of (2/3)$R$.

Now, I've managed to get this result (i.e, (2/3)$R$) by equating the potential energy with the kinetic energy.But to be honest, it was a mere hunch. My question is why is that true?

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A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy.

The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal force, the mass will leave the surface.

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  • $\begingroup$ And the reaction force is equal to the centripetal force ? $\endgroup$ – Fadi Apr 6 '14 at 21:22
  • $\begingroup$ The sum of two forces cannot be an acceleration. $\endgroup$ – Tobias Apr 6 '14 at 21:24
  • $\begingroup$ @Fadi The reaction force is equal to the normal component of the gravitational force reduced by the centripetal force. If it becomes zero the mass leaves the surface of the sphere. (No reaction of the surface anymore.) In other words the normal comp. of grav. force has two "tasks" 1st it keeps the mass on the circular path 2nd it lets the mass exert pressure on the surface which the surface counters with the reaction force. $\endgroup$ – Tobias Apr 6 '14 at 22:11
  • $\begingroup$ @Tobias good !.Now,try this :equate the potential energy with the kinetic energy : $mgh$= ($m/2$) (4g($R$-$h$)).Solving will give : $h$=2/3$R$. My question,is there any significance for the point at which the potential energy is equal to the kinetic energy, because it it the same point at which the object detach !,or it's a mere chance ? $\endgroup$ – Fadi Apr 8 '14 at 20:01
  • $\begingroup$ @Fadi 1st: How do you get $E_{\rm kin}=\frac{m2}(4g(R-h))$? 2nd: The potential energy with $E_{\rm pot} = mgh$ is an arbitrary choice. The potential energy is only defined up to an arbitrary constant. $\endgroup$ – Tobias Apr 8 '14 at 20:24
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There follows my try to decompose the solution into a minimal amount of calculation and apart from that only geometrical considerations.

The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center. We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure).

Definition of z-coordinate

The conservation of energy says $\frac{v^2}2 = g z$. Therefore, we have $2g z= a_c R$. That means $a_c\sim z$. Furthermore, if the mass was constrained to the sphere then we would have at the equator $a_c(z=R)=2g$.

Now, you can draw a diagram that relates the centripetal acceleration to the angular position of the mass. (Do not worry. In our consideration we do not need any specific numeric value for the angle.)

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At the equator the mass would have normal acceleration of $a_c=2g$. this is expressed by the big circle. Depending on $\varphi$ the centripetal acceleration is shrinking proportional to the distance of the top line to the dashed one. To get the direction of the centripetal acceleration right we have drawn in the small circle.

The mass detaches from the surface when the normal component of the gravitational acceleration (marked with $g$. equals the centripetal acceleration (line from the center of the small circle to the inner intersection with the small circle). These lines form two edges of the triangle for the decomposition of the gravitational acceleration into the normal and the tangential component. The triangle composed of the blue and the red line and the blue arc has also one edge with length equal to the centripetal acceleration, the angles are also equal. Therefore they are congruent and the red line has length $g$.

Finally, we can just use the intercept theorem for the $2g$-diameter the $3g$ stretch composed of the skew $2g$ diameter and the red line and the top and the dashed line to obtain that the mass detaches at $\frac 23$ height of the sphere.

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  • $\begingroup$ $v$^2 /2 = g$z$ how come ? this is not true for every point .besides, where is the $z$ coordinate , at the big or small circle ? $\endgroup$ – Fadi Apr 8 '14 at 19:02
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    $\begingroup$ Thanks for considering my answer. I have added a figure illustrating the definition of the z-axis. The figure with the small and the big circle does not represent the (x,z)-location of the mass but the angle and the relation between the gravitational acceleration and the centripetal acceleration at the angular position where the mass detaches. $\endgroup$ – Tobias Apr 8 '14 at 19:36

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