A stone with mass $0.8 \; kg$ is attached to one end of a string $0.9 \; m$ long. The string will break if its tension exceeds $600 \; N$. The stone is whirled in a horizontal circle, the other end of the string remains fixed. Find the maximum speed, the stone can attain without breaking the string.

Given,

Mass of stone $(m) = 0.8\;kg$

Length of string $(r) = 0.9\;m$

Maximum tension $(T_{max}) = 600\;N$

Maximum speed $(v_{max}) = \;?$

For maximum speed at which the string will not break,

$T_{max} = \frac{m\;v^2_{max}}{r}$

$v^2_{max} = \frac{T_{max}\;r}{m} = \frac{600\;*\;0.9}{0.8} = 675$

⇒ $v = 25.98\;m/s$

∴ The maximum speed $(v_{max}) = \;25.98\;/m/s$

Return to  Main menu

An object of mass $0.5 \; kg$ is rotated in a horizontal circle by a string $1 \; m$ long. The maximum tension in the string before it breaks is $50 \; N$. What is the greatest number of revolutions per second of the object?

Given,

Mass of an object $(m) = 0.5\;kg$

Radius of horizontal circle $(r) = 1\;m$

Tension in the string $(T) = 50\;N$

Greatest number of revolution $(f) = \;?$

In case of horizontal circle, there is no maximum and minium tension. So we can write,

$T = m\; \omega^2\;r = m\;4\pi^2\;f^2\;r$

$f^2 = \frac{T}{4\;\pi^2\;m\;r}$

$f = \frac{1}{2\;\pi}\;\sqrt{\frac{T}{m\;r}} = \frac{1}{2\pi}\;\sqrt{\frac{50}{0.5\;*\;1}} = 1.6\;rev/sec$

⇒ Greatest number of revolution $(f) = \;1.6\;rev/sec$

Return to Main Menu

A bob of mass $200 \; gm$ is whirled in a horizontal circle of radius $50 \; cm$ by a string inclined at $30^{\circ}$ to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle.

Given, (need figure)

Mass of the bob $(m) = 200 \; gm = 0.2\;kg$

Radius of horizontal circle $(r) = 50 \; cm = 0.5\;m$

Angle of inclination with vertical $(\theta) = 30^{\circ}$

Tension in the string $(T) =\;?$

Speed of the mass $(v) = \;?$

In case of the horizontal circle, we have

(i) To balance the weight of the body,           $T\;Cos\;\theta = mg$

$T = $ $\frac{m\;g}{Cos\; \theta} = \frac{0.2 \;*\;10}{Cos\;30}$ $= 2.3\;N$

Tension in the string $T = 2.3\;N$

(ii) To provide the necessary centripetal force,        $T\;Sin\;\theta = $ $\frac{M\;v^2}{r}$

$v^2 = $ $ \frac{T\;Sin\;\theta \;*\;r}{m}$

$v = $ $ \sqrt{\frac{2.3 \; * \; Sin\;30\;*\;0.5}{0.2}}$ $ = 1.69\;m/s$

Speed of mass $(v) = 1.69\;m/s$

Return to Main Menu

A certain string breaks when a weight of $25\;N$ acts on it. A mass of $500\;gm$ is attached to one end of the string of $1\;m$ long and is rotated in a horizontal in a horizontal circle. Find the greatest number of revolutions per minute which can be made without breaking the string?

Given,

 

Maximum string $(T_{max}) = 25 \; N$

Mass of the object $(m) = 500 \; gm = 0.5 \; kg$

Radius of circle $(r) = 1\;m$

Frequency $(f) = \; ?$

We have,

$T_{max} = $$\frac{mv^2}{r}$$ = m\;\omega^2\;r$ = Centripetal Force

or, $T_{max} = m\;(2\;\pi\;f)^2\;r$

$25 = 0.5 * 4\pi^2 * f^2 * 1$

$f^2 = $$\frac{25}{0.5\; * \;4\; *\; \pi^2 \;*\; 1}$$ = 1.266$ rev/sec

$f = 1.125\;$ rev/sec

⇒ $f = 1.125 * 60\;rpm = 67.5\;rpm$

Return to Main Menu