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

`6.5\; {\rm N}`.

Step-by-step 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 &= (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} * 5\; {\rm m\cdot s^(-2)} \\ &= 6.5\; {\rm N} \end{aligned}`.