Q.A stone of mass $0.2\ \text{kg}$, tied to a string of length $1\ \text{m}$, is whirled in a horizontal circle at a constant speed of $4\ \text{m/s}$. Assuming the string stays horizontal, find the tension in it.

Given: mass $m = 0.2\ \text{kg}$, string length (radius) $r = 1\ \text{m}$, speed $v = 4\ \text{m/s}$, string assumed to stay exactly horizontal.\n\nFor the stone to move in a horizontal circle, a net centre-directed (centripetal) force of magnitude $mv^2/r$ is required, and here the only horizontal force available to supply it is the string's tension $T$ (the idealisation that the string stays perfectly horizontal means gravity's effect on the string's own alignment is …

Set the string's tension equal to the required centripetal force, $T = mv^2/r$, since tension is the only force available t …

• Adding the stone's weight $mg$ into the tension calculation, which would be needed only if the string's small dip below horizontal (a conical pendulum) were being modelled, not this idealised purely-horizontal ca …