Question

You have a ball with mass 2.5 kg tied to a rope, and you spin it in a circle of radius

1.2 m. You know that the rope can withstand a tension of 130 N before it breaks.
How fast can you safely spin the ball without the rope breaking?

Answer 1

Approximately
`7.2\; {\rm m\cdot s^(-1)}` (rounded up), assuming that this circle is vertical and
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Let
`v` denote the tangential speed of the ball, and let
`r` denote the radius of the circle. Since the ball is in a circular motion, the acceleration on this ball would be equal to the centripetal acceleration
`a = (v^(2) / r)`. The net force on this ball would be
`F_{\text{net}} = m\, a = (m\, v^(2) / r)`.

The net force on this ball is also the vector sum of the tension
`T` in the rope and the weight of the ball
`m\, g`:

`F_{\text{net}} = (\text{weight}) + T`.

`T = F_{\text{net}} - (\text{weight})`.

Note that:

`\| T \| = \|F_{\text{net}} - (\text{weight})\| \le \|F_{\text{net}} \| + \| (\text{weight})\|`.

In other words, the magnitude of tension
`T` is at most equal to
`\|F_{\text{net}} \| + \| (\text{weight})\| = (m\, v^(2) / r) + (m\, g)`, which happens when weight and net force are in opposite directions.

When the speed of the ball is maximized, the magnitude of tension
`T` would be at the largest possible value of
`130\; {\rm N}`. Rearrange the equation and solve for speed
`v`:

`\displaystyle (m\, v^(2))/(r) + m\, g = \|T\|`.

`\begin{aligned}v^(2) = (r)/(m)\, (\|T \| - m\, g) = (r\, \|T\|)/(m) - r\, g\end{aligned}`.

`\begin{aligned}v &= \sqrt(r\, \ \\ &= \sqrt{((1.2)\, (130))/(2.5) - (1.2)\, (9.81)}\; {\rm m\cdot s^(-1)} \\ &\approx 7.2\; {\rm m\cdot s^(-1)}\end{aligned}`.