Question

A mass of 0.28 kg is fiixed to the end of a 0.93 m long string that is fixed at the other end. Iniitially at rest, the mass is made to rotate around the fixed end with an angular acceleration of 4.43 rad/s. After how many revolutions is the cetripetal force acting on the mass 502 N ?

Answer 1

Let the revolutions required is n to reach centripetal force acting on the mass 502 N.

Let the final velocity of the mass is v.

Let the final angular velocity is $\omega$ .

Given that the mass m=0.28 kg.

The length of the string l=0.93 m

The angular acceleration is $\alpha$ =4.43 rad/s².

Now we from centripetal force formula

  

mu F=

  

Fl от ,2 U* m

     

F1 or U= 1 m

$or\: \: \: \: \: \: \: v=\sqrt{\frac{502\times 0.93}{0.28}}$

$\therefore \: \: \: \: \: \: \: v=40.83328474\: \: \: \: \: m/s$

Now the final angular speed is

$\omega =\frac{v}{l}$

$or\: \: \: \: \: \: \omega =\frac{40.83328474}{0.93}$

  $\therefore \: \: \: \: \: \: \omega =43.90675779\: \: \: \: rad/s$

Now let the total angular displacement is $\Theta$ .

We know

$\omega ^{2}=2\alpha \Theta$

$or\: \: \: \: \: \: \: \Theta =\frac{\omega ^{2}}{2\alpha }$

   $or\: \: \: \: \: \: \: \Theta =\frac{\left ( 43.90675779 \right ) ^{2}}{2\times 4.43 }$

$\therefore \: \: \: \: \: \: \: \Theta =217.5850315\: \: \: \: rad$

Now we know

$\Theta =2\pi n$

$or\: \: \: \: \: \: \: n=\frac{\Theta }{2\pi }$

$or\: \: \: \: \: \: \: n=\frac{217.5850315 }{2\pi }$

$\therefore \: \: \: \: \: \: \: n=34.62973331\approx 34.63$

$\therefore$ After 34.63 revolutions the centripetal force acting on the mass is 502 N.

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