MHT CET · Physics · Motion In Two Dimensions
A particle moves around a circular path of radius ' \(r\) ' with uniform speed ' \(V\) '. After moving half the circle, the average acceleration of the particle is
- A \(\frac{\mathrm{V}^2}{\mathrm{r}}\)
- B \(\frac{2 \mathrm{~V}^2}{\mathrm{r}}\)
- C \(\frac{2 \mathrm{~V}^2}{\pi \mathrm{r}}\)
- D \(\frac{\mathrm{V}^2}{\pi \mathrm{r}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{~V}^2}{\pi \mathrm{r}}\)
Step-by-step Solution
Detailed explanation
At end points of the half revolution magnitude of the velocity is same but it directs in opposite direction.
\(\begin{array}{ll}\therefore & \Delta V=V-(-V) \\\therefore & \Delta V=2 V\end{array}\)
Time taken to complete the half revolution is \(\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}\)
Average acceleration is, \(a=\frac{\Delta V}{t}=\frac{2 \mathrm{~V}}{\frac{\pi \mathrm{r}}{\mathrm{V}}}\)
\(\therefore a=\frac{2 V^2}{\pi r}\)
\(\begin{array}{ll}\therefore & \Delta V=V-(-V) \\\therefore & \Delta V=2 V\end{array}\)
Time taken to complete the half revolution is \(\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}\)
Average acceleration is, \(a=\frac{\Delta V}{t}=\frac{2 \mathrm{~V}}{\frac{\pi \mathrm{r}}{\mathrm{V}}}\)
\(\therefore a=\frac{2 V^2}{\pi r}\)
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