Question

a particle of mass m is initially situated st the point p inside a hemispherical surface of radius r . A horizontal acceleration of magnitude a0 is suddenly produced on the particle in the horizontal direction. if gravitational acceleration is neglected, the time taken by particle to touch the sphere again is:

Answer 1

Answer on Question #52583, Physics, Mechanics | Kinematics | Dynamics

A particle of mass $m$ is initially situated at the point $p$ inside a hemispherical surface of radius $r$ . A horizontal acceleration of magnitude $a_0$ is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is:

Solution:

From figure $\mathrm{OP} = \mathrm{r}$

The kinematics equation is

$$x = x _ {0} + v _ {0} t + \frac {1}{2} a _ {0} t ^ {2}$$

where

$x_0 = P$ is initial position

$v_{0} = 0m / s$ is initial speed

$a_0$ is acceleration

$x = B$ is final position.

Thus,

$$t = \sqrt {\frac {2 (x - x _ {0})}{a _ {0}}}$$

From figure

$$x - x _ {0} = P B = 2 P A$$ $$P A = O P \cos \alpha = r \cos \alpha$$

Hence,

$$t = \sqrt {\frac {2 * 2 * r \cos \alpha}{a _ {0}}} = \sqrt {\frac {4 r \cos \alpha}{a _ {0}}}$$

Answer: $\sqrt{\frac{4r\cos\alpha}{a_0}}$

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