Question

You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed of 3 m/s, what is the centripetal force action on the ball?

Answer 1

Answer:

$6.5\; {\rm N}$.

Explanation:

When an object travel at a speed of $v$ in a circle of radius $r$, the (centripetal) acceleration of that object would be $a = (v^{2} / r)$.

In this question, the ball is travelling at $v = 3\; {\rm m\cdot s^{-1}}$ in a circle of radius $r = 1.8\; {\rm m}$. The (centripetal) acceleration of this ball would be:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}$.

By Newton's Laws of Motion, for an object of mass $m$, if the acceleration of that object is $a$, the net force on that object would be $m\, a$. Since the acceleration of this ball is $a = 5\; {\rm m\cdot s^{-2}}$, the net force on this ball would be:

$\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}$.