A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while the angular position of point Q is α  with respect to point P. Which of the following graphs represents the correct relation between A and α when the ball goes from Q to R?

CONCEPT:

• Centripetal acceleration is defined as the ratio of the square of the velocity and radius and it is written as,

           ac = \(\frac{v^2}{R}\) 

           here we have a_c as the centripetal acceleration, v is the velocity and R is the radius.

CALCULATION:

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure below,

Using the trigonometry identity to find h we have;

\(sin\alpha = \frac{h}{R}\)

⇒ h = R sin\(\alpha\)

and the first law of motion we have;

v² - u² = 2gh

Here is the initial velocity, u = 0 m/s we have;

v2 = 2gh

⇒ v = \(\sqrt{2gh}\) 

⇒ v = \(\sqrt{2gRsin\alpha}\) -----(1)

The centripetal acceleration is written as;

a_c = \(\frac{v^2}{R}\) 

Now, on putting the value of equation (1) above we have,

ac = \(\frac{2gRsin\alpha}{R}\) 

⇒ a_c = 2g sin\(\alpha\) -----(2)

At point Q we can see that normal N and force F which mg sin\(\alpha\) we have;

N -  mg sin α  = ma_c

⇒ N = ma_c + mg sinα ----(3)

Now, on putting the value of equation (2) in equation (3) we have;

N = m × 2 g sin α + mg sin α 

N = 3 × mg sin α 

The ratio of the centripetal force and normal reaction is written as;

\(\frac{ma_c}{N} = \frac{ 2mgsin\alpha}{3mg sin\alpha}\)

⇒ \(\frac{ma_c}{N} = \frac{2}{3}\) = constant

Therefore, the ratio of the centripetal force and normal reaction is constant.

Hence, option (3) is the correct answer.