Question

A mass M is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when?

• A. the mass is at the highest point
• B. the wire is horizontal
• C. the mass is at the lowest point
• D. inclined at an angle of 60° from vertical

Answer 1

The correct option is C

the mass is at the lowest point

Explanation for correct option

Option : C

1. We know that to move a mass in a circular route a centripetal force occurs
2. This centripetal force is caused by the mass's weight,
3. When the mass is at its lowest position, the tension in the string acts in the opposite direction as the mass's weight, allowing us to make the following statement.

The mass is at the lowest point

1. If M is the mass at the lowest point :
   $$\begin{gathered} T-Mg=\frac{M{v}^2}{l} \\ T=\frac{M{v}^2}{l}+Mg \end{gathered}$$ v is velocity , l is length of string, T is tension in string
2. As a result, the lowest point has the maximum tension, while the highest point has the lowest.
3. As a result, the wire is most likely to break when the mass is at its lowest.
   Hence, option C is the correct option.

The mass is at the highest point:

1. If M is the mass at the highest point
   :$$\begin{gathered} T+Mg=\frac{M{v}^2}{l} \\ T=\frac{M{v}^2}{l}-Mg \end{gathered}$$
2. The highest point has lower tension when compared to the mass at the lower point therefore the mass at the lower point tends to break

Hence, option C is correct. The wire is most likely to break when the mass is at the lowest point.