MHT CET · Physics · Rotational Motion
A disc at rest is subjected to a uniform angular acceleration about its axis. Let \(\theta\) and \(\theta^1\) be the angle made by the disc in \(2^{\text {nd }}\) and \(3^{\text {rd }}\) second of its motion. The ratio \(\frac{\theta}{\theta^1}\) is
- A \(2: 3\)
- B \(1: 2\)
- C \(2: 3\)
- D \(4: 5\)
Answer & Solution
Correct Answer
(D) \(4: 5\)
Step-by-step Solution
Detailed explanation
Kinematic equation for rotational motion,
\(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
Disk is initially at rest, \(\omega_0=0\)
\(\Rightarrow \theta=\frac{1}{2} \alpha t^2...(i)\)
Angle described in \(2^{\text {nd }}\) second is,
\(\theta_1=\frac{1}{2} \alpha(2)^2=2 \alpha\)
Angle described in first 3 seconds will be,
\(\theta_2=\frac{1}{2} \alpha(3)^2=4.5 \alpha\)
Angle described in \(3^{\text {rd }}\) second will be, \(\theta^{\prime}=\theta_2-\theta_1\)
\(=4.5 \alpha-2 \alpha=2.5 \alpha\)
\(\therefore \frac{\theta}{\theta^{\prime}}\)
\(=\frac{2}{2.5}=\frac{4}{5}\)
\(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
Disk is initially at rest, \(\omega_0=0\)
\(\Rightarrow \theta=\frac{1}{2} \alpha t^2...(i)\)
Angle described in \(2^{\text {nd }}\) second is,
\(\theta_1=\frac{1}{2} \alpha(2)^2=2 \alpha\)
Angle described in first 3 seconds will be,
\(\theta_2=\frac{1}{2} \alpha(3)^2=4.5 \alpha\)
Angle described in \(3^{\text {rd }}\) second will be, \(\theta^{\prime}=\theta_2-\theta_1\)
\(=4.5 \alpha-2 \alpha=2.5 \alpha\)
\(\therefore \frac{\theta}{\theta^{\prime}}\)
\(=\frac{2}{2.5}=\frac{4}{5}\)
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