Question 181: Work, Energy and Power & Circular Motion

Question

The figure shows two masses ' $m$ ' and ' $M$ ' connected by a light string that passes through ${ }_a$ small hole ' $O$ ' at the centre of the table. Mass ' $m$ ' is moved round in a horizontal circle with ' $O$ ' as the centre. The frequency with which ' $m$ ' should be revolved so that ' $M$ ' remains stationary is
( $\mathrm{g}=$ gravitational acceleration)

Accepted Answer

Step 1: Forces Acting on the Masses
When $m$ moves in a circle, it pulls on the string with a force. That same force pulls up on mass $M$, balancing its weight ($Mg$).
Step 2: Centripetal Force for Mass $m$
To keep moving in a circle of radius $L$, mass $m$ needs a force equal to $mr\omega^2$, where $r=L$ and $\omega$ is the angular speed.
Step 3: Set Forces Equal
The upward force on $M$ from the string is the same as the force needed by $m$ to move in a circle.

$$ mr\omega^2 = Mg $$
Step 4: Substitute $r = L$
$$ mL\omega^2 = Mg $$

So,

$$ \omega^2 = \frac{Mg}{mL} $$
Step 5: Relate $\omega$ (angular speed) to $f$ (frequency)
We use $\omega = 2\pi f$.

$$ \omega^2 = (2\pi f)^2 = 4\pi^2 f^2 $$
Step 6: Solve for $f$
Set the expressions for $\omega^2$ equal:

$$ 4\pi^2 f^2 = \frac{Mg}{mL} $$

So,

$$ f = \frac{1}{2\pi}\sqrt{\frac{Mg}{mL}} $$

Answer

$\frac{1}{\pi} \sqrt{\frac{\mathrm{ML}}{\mathrm{mg}}}$

Answer

$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Mg}}{\mathrm{mL}}}$

Answer

$\frac{1}{\pi} \sqrt{\frac{\mathrm{Mg}}{\mathrm{mL}}}$

Answer

$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{ML}}{\mathrm{mg}}}$

Work, Energy and Power & Circular Motion

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