Question

A particle slides down from the top outside smooth surface of a fixed sphere of radius $a=10m$. The initial horizontal velocity to be imparted to the particle 'at the top' is $5Km/s$ if it leaves the surface at a point whose vertical height above the center of sphere is $3a/4$ Find the value of $K$.

Answer 1

Given: A particle slides down from the top outside smooth surface of a fixed sphere of radius a=10m. The initial horizontal velocity to be imparted to the particle 'at the top' is 5K m/s if it leaves the surface at a point whose vertical height above the center of sphere is $\frac{3a}{4}$

To find the value of K.

Solution:

Let "O" be the sphere's center point, and say the particle is at some point P on the sphere, so that the radius OP forms an angle "$\theta$" with the vertical. Let "N" be the normal force acting on the particle at that point. Then the centripetal components of the forces acting on the particle are:

1. Centripetal component of gravity force: $+\left(mg\right)\cos \theta$

2. Centripetal component of normal force: -N (negative because the force acts radially outward)

The net centripetal force on the particle at P is therefore:

$F_{centrip}=\left(mg\right)\cos \theta -N......\left(i\right)$

We can also calculate the particle's speed at P, by the work-energy theorem:

Change in KE = Total work done by all forces:

$\frac{1}{2}m(v_p{)}^2-\frac{1}{2}m(v_{initial}{)}^2=W$

The normal force "N" does no work, because it always points perpendicular to the direction of motion. Therefore, gravity is the only force that does any work, and the work it does is (force $\times$ distance)=(Weight $\times$ drop in height) = $mga(1-\cos \theta )$

Therefore:

$\frac{1}{2}m(v_p{)}^2-\frac{1}{2}m(v_{i}nitial{)}^2=mga(1-\cos \theta )$

From which:

$(v_p{)}^2=(v_{initial}{)}^2+2ga(1-\cos \theta ).......\left(ii\right)$

By Newton's 2nd Law, we know that

$F_{centripetal}=\left(m\right)a_{centripetal}.........\left(iii\right)$

And we know from circular kinematics that:

$a_{centripetal}=\frac{v^2}{R}=\frac{(v_p{)}^2}{a}......\left(iv\right)$

Combine equations (i) through (iv), we get:

$\left(mg\right)cos\theta -N=\frac{m\left[\right(v_{initial}{)}^2+2ga(1-cos\theta )]}{a}$

When the particle leaves the surface:

N = 0 (because no contact), and:

height = $\frac{3a}{4}$ (given in problem)

Since height also clearly equals $\left(a\right)\cos \theta$, this means:

$\frac{3a}{4}=\left(a\right)\cos \theta ⟹\cos \theta =\frac{3}{4}$

(at point of leaving)

Substituting in the above, we get

$\left(mg\right)\times \frac{3}{4}-0=\frac{m\left[\right(v_{initial}{)}^2+2ga(1-3/4)]}{a}$

Solving for v, we get

$v=\frac{1}{2}\sqrt{ag}$

Since this equals 5Km/s (given), we have:

$5K=\frac{1}{2}\sqrt{ag}⟹K=\frac{1}{2}\times \frac{1}{5}\times \sqrt{10\times 10}⟹K=\frac{1}{10}\times 10⟹K=1$