Question

A ball attached to a string starts at rest and undergoes a constant angular acceleration as it travels in a horizontal circle of radius 0.30 m. After 0.65 sec, the angular speed of the ball is 9.7 rad/s. What is the tangential acceleration of the ball

Answer 1

Answer:

Tangential acceleration, $a_t=4.47\ m/s^2$      

Explanation:

Given that,

Radius of the circle, r = 0.3 m

Initial speed of the ball, u = 0

After 0.65 sec, the angular speed of the ball is 9.7 rad/s, $\omega_f=9.7\ rad/s$

Let $\alpha$ is the angular acceleration of the ball. It is given by the change in velocity per unit time as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

Here, $\omega_i=0$ (at rest)

$\alpha =\dfrac{\omega_f}{t}$

$\alpha =\dfrac{9.7\ rad/s}{0.65\ s}$

$\alpha =14.92\ rad/s^2$

The tangential acceleration is given by in terms of angular acceleration as :

$a_t=r\alpha$

$a_t=0.3\times 14.92$

$a_t=4.47\ m/s^2$

So, the tangential acceleration of the ball is $a_t=4.47\ m/s^2$. Hence, this is the required solution.