Question

Two cars of masses \( m_1 \) and \( m_2 \) are moving in the circles of radii \( r_1 \) and \( r_2 \) respectively. Their 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 \( m_1 \) to the linear speed of \( m_2 \) is

• A. \( r_1 : r_2 \)
• B. \( T_1^2 : T_2^2 \)
• C. \( \omega_1^2 : \omega_2^2 \)
• D. \( m_1 : m_2 \)

Hint:

The linear velocity in circular motion is given by \( v = r \omega \), where \( r \) is the radius and \( \omega \) is the angular velocity. Use this relationship to find the ratio of linear velocities.

Answer 1

The Correct Option is A

Step 1: Relationship between linear and angular velocity.
The linear velocity \( v \) is related to the angular velocity \( \omega \) by the formula: \[ v = r \omega \] where \( r \) is the radius of the circular path and \( \omega \) is the angular velocity.
Step 2: Applying the given conditions.
Since both cars complete one revolution in the same time \( t \), their angular velocities are related by: \[ \omega_1 = \frac{2\pi}{T_1} \quad \text{and} \quad \omega_2 = \frac{2\pi}{T_2} \] Since the time \( t \) is the same for both, the linear velocities of the cars are: \[ v_1 = r_1 \omega_1 \quad \text{and} \quad v_2 = r_2 \omega_2 \]
Step 3: Ratio of linear speeds.
The ratio of linear speeds is: \[ \frac{v_1}{v_2} = \frac{r_1 \omega_1}{r_2 \omega_2} = \frac{r_1}{r_2} \]
Step 4: Conclusion.
Thus, the ratio of the linear speeds is \( r_1 : r_2 \). Therefore, the correct answer is (A).