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Q5.22:

If, A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :

Solution :

Part 1: Calculating Tension in the String

Given Variables:

  • Mass of the stone, $m = 0.25\text{ kg}$
  • Radius of the circle, $r = 1.5\text{ m}$
  • Angular frequency, $\nu = 40\text{ rev/min} = \frac{40}{60}\text{ rev/s} = \frac{2}{3}\text{ rev/s}$

Step 1: Determine Angular Velocity ($\omega$)

The relation between angular frequency and angular velocity is given by $\omega = 2\pi\nu$ [Definition of Angular Velocity].

$\omega = 2 \times \pi \times \frac{2}{3} = \frac{4\pi}{3}\text{ rad/s} \approx 4.189\text{ rad/s}$

Step 2: Calculate Centripetal Force ($T$)

The tension in the string provides the necessary centripetal force for horizontal circular motion [Per Newton's Second Law].

$T = F_c = m\omega^2r$

$T = 0.25 \times \left(\frac{4\pi}{3}\right)^2 \times 1.5$

$T = 0.25 \times \frac{16\pi^2}{9} \times 1.5$

$T = 0.25 \times 1.777 \times 9.87 \times 1.5 \approx 6.58\text{ N}$

Part 2: Calculating Maximum Speed ($v_{max}$)

Given: $T_{max} = 200\text{ N}$

Step 3: Relate Tension to Linear Speed

Using the centripetal force formula $T = \frac{mv^2}{r}$, we solve for $v$ [Centripetal Force Axiom]:

$v_{max} = \sqrt{\frac{T_{max} \times r}{m}}$

$v_{max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{\frac{300}{0.25}} = \sqrt{1200}$

$v_{max} \approx 34.64\text{ m/s}$

Part 3: Trajectory Analysis

When the string breaks, the centripetal force vanishes [Per Newton's First Law of Motion]. According to the law of inertia, the stone will continue to move in the direction of its instantaneous velocity at the point of release.

Conclusion: The stone will fly off along the tangent to the circular path at the point where the string breaks.

O Stone Path of release
Parameter Calculated Value
Tension ($T$) $6.58\text{ N}$
Max Speed ($v_{max}$) $34.64\text{ m/s}$

Final Solution: The tension in the string is 6.58 N, the maximum speed is 34.64 m/s, and the trajectory of the stone is along the tangent to the circular path at the point of breakage.


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