Question

A 940-g rock is whirled in a horizontal circle at the end of a 1.30-m-long string. (a) If the breaking strength of the string is 120 N, what’s the minimum angle the string can make with the horizontal? (b) At this minimum angle, what’s the rock’s speed?

Answer 1

(a) The minimum angle the string can make with the horizontal is 4.4 ⁰.

(b) The rocks speed at the minimum angle is 165.7 m/s.

The given parameters;

• mass of the rock, m = 940 g = 0.94 kg
• length of the string, L = 1.3 m
• Tension on the string, T = 120 N

The net force on the string is calculated as follows;

$Tsin(\theta) = mg\\\\Tcos(\theta) = \frac{mv^2}{r} \\\\\frac{Tsin(\theta)}{Tcos(\theta)} = \frac{mg r}{mv^2} \\\\tan(\theta) = \frac{rg}{v^2} \\\\v^2 = \frac{rg}{tan (\theta)} \\\\v = \sqrt{\frac{rg}{tan (\theta)}}$

The minimum angle the string can make with the horizontal is calculated as follows;

$Tsin(\theta) = mg\\\\sin(\theta) = \frac{mg}{T} \\\\sin(\theta) = \frac{0.94 \times 9.8}{120} \\\\sin(\theta) = 0.0767\\\\\theta = sin^{-1} (0.0767)\\\\\theta = 4.4 \ ^o$

The rocks speed at the minimum angle is calculated as follows;

$v = \sqrt{\frac{rg}{tan(\theta)} } \\\\v = \sqrt{\frac{1.3 \times 9.8}{tan(4.4)} } \\\\v = 165.7 \ m/s$

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