Question

The small ball of mass m = 0.5 kg is attached to point A via string and is moving at constant speed in a horizontal circle of radius R = 0.16 m. The rate of rotation of the ball around the circle is ω = 2 rad/s. If the gravitational acceleration constant g is 10 m/s2 , find the value of height d.

Answer 1

Answer:

d = 2.45 meters

Explanation:

Mass of the ball, m = 0.5 kg

Radius of the circle, r = 0.16 m

The angular speed of the ball around the circle is, $\omega=2\ rad/s$

The attached figure shows the whole scenario. Let $F_t$ is the force acting on the ball in tangential direction. The forces will balanced each other at equilibrium.

In horizontal direction,

$T\ sin\theta=F_t=mr\omega^2$................(1)

In vertical direction,

$T\ cos\theta=mg$...............(2)

From equation (1) and (2) :

$tan\theta=\dfrac{r\omega^2}{g}$

Also,

$tan\theta=\dfrac{r}{d}$

$d=\dfrac{g}{\omega^2}$$d=\dfrac{9.8}{2^2}$

d = 2.45 meters

So, the value of d is 2.45 meters. Hence, this is the required solution.