Question 211: Work, Energy and Power & Circular Motion

Question

Two objects of masses ' $m_1$ ' and ' $m_2$ ' are moving in the circles of radii ' $r_1$ ' and ' $r_2$ ' respectively. Their respective angular speeds ' $\omega_1$ ' and ' $\omega_2$ ' are such that they both complete one revolution in the same time ' $t$ '. The ratio of linear speed of ' $m_2$ ' to that of ' $m_1$ ' is

Accepted Answer

We know that the linear speed $ v $ for an object in circular motion is given by:
$$ v = r \omega, $$
where $ r $ is the radius of the circle and $ \omega $ is the angular speed.
Given that both objects complete one revolution in the same time $ t $, their angular speeds are equal. In fact, the angular speed for one complete revolution is:
$$ \omega = \frac{2\pi}{t}. $$
For the two objects:
For mass $ m_1 $ with radius $ r_1 $:
$$   v_1 = r_1 \omega_1 = r_1 \left(\frac{2\pi}{t}\right)   $$
For mass $ m_2 $ with radius $ r_2 $:
$$   v_2 = r_2 \omega_2 = r_2 \left(\frac{2\pi}{t}\right)   $$
Taking the ratio of the linear speeds $ \frac{v_2}{v_1} $:
$$ \frac{v_2}{v_1} = \frac{r_2 \left(\frac{2\pi}{t}\right)}{r_1 \left(\frac{2\pi}{t}\right)} = \frac{r_2}{r_1}. $$
Thus, the ratio of the linear speed of $ m_2 $ to that of $ m_1 $ is:
$$ \frac{v_2}{v_1} = \frac{r_2}{r_1}. $$
The correct answer is therefore Option D.

Answer

$\omega_1: \omega_2$

Answer

$\mathrm{T}_2: \mathrm{T}_1$

Answer

$\mathrm{m}_1: \mathrm{m}_2$

Answer

$\mathrm{r_2: r_1}$

Work, Energy and Power & Circular Motion

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