MHT CET · Physics · Motion In Two Dimensions
A particle starting from rest moves along the circumference of a circle of radius ' \(\mathrm{r}^{\prime}\) with angular acceleration ' \(\alpha^{\prime}\) '. The magnitude of the average velocity, in the time it completes the small angular displacement ' \(\theta\) ' is
- A \(r\left(\frac{2}{\alpha \theta}\right)^{2}\)
- B \(r\left(\frac{\alpha \theta}{2}\right)\)
- C \(r\left(\frac{\alpha \theta}{2}\right)^{2}\)
- D \(r\left(\frac{\alpha \theta}{2}\right)^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(D) \(r\left(\frac{\alpha \theta}{2}\right)^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
(A)
\(S=r \theta\)
\(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}\)
\(\therefore t=\sqrt{\frac{2 \theta}{\alpha}}\)
Average velocity \(=\frac{\text { displacement }}{\text { time }}=\frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}}\)
\(=r \sqrt{\frac{\alpha}{2 \theta}} \cdot \theta=r \sqrt{\frac{\alpha \theta}{2}}\)
\(S=r \theta\)
\(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}\)
\(\therefore t=\sqrt{\frac{2 \theta}{\alpha}}\)
Average velocity \(=\frac{\text { displacement }}{\text { time }}=\frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}}\)
\(=r \sqrt{\frac{\alpha}{2 \theta}} \cdot \theta=r \sqrt{\frac{\alpha \theta}{2}}\)
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