Question

A heavy mass is attached to a thin wire and is whirled in a vertical circle with constant speed. The wire is most likely to break.

• A. When the mass is at height B.
• B. When the mass is at the lowest point.
• C. When the wire is horizontal.
• D. At an angle of ${30}^{\circ }$ from the upward vertical.

Answer 1

The correct option is B When the mass is at the lowest point.

Let us consider a vertical circular motion of speed $v$.


Let $A,B,C\text{and}D$ be the points on a vertical circle as shown in the figure.
By applying $\sum F=ma$ along radial direction at $A$, we get
$T-mg=\frac{m{v}^2}{r}\Rightarrow T=\frac{m{v}^2}{r}+mg............\left(1\right)$
at $B$, we get
$T=\frac{m{v}^2}{r}...............\left(2\right)$
at $C$, we get
$T+mg\cos {30}^{\circ }=\frac{m{v}^2}{r}\Rightarrow T=\frac{m{v}^2}{r}-\text{mg cos}{30}^{\circ }----\left(3\right)$
at $D$, we get
$T+mg=\frac{m{v}^2}{r}\Rightarrow T=\frac{m{v}^2}{r}-mg----\left(4\right)$

From $\left(1\right),\left(2\right),\left(3\right)\text{and}\left(4\right)$, we can see that the value of tension in the string is maximum at point A i.e at lowest point.
$\therefore$ wire has more chance to break at this point.
Hence, option (b) is the correct answer.