Question 239: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

At end points of the half revolution magnitude of the velocity is same but it directs in opposite direction.
$$\begin{array}{ll}
	herefore & \Delta \mathrm{V}=\mathrm{V}-(-\mathrm{V}) \
	herefore & \Delta \mathrm{V}=2 \mathrm{~V}
\end{array}$$
Time taken to complete the half revolution is
$$\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}$$
Average acceleration is, $$a=\frac{\Delta V}{t}=\frac{2 V}{\frac{\pi r}{V}}$$
$$	herefore \quad a=\frac{2 V^2}{\pi r}$$

Answer

$$\frac{\mathrm{V}^2}{\mathrm{r}}$$

Answer

$$\frac{2 V^2}{r}$$

Answer

$$\frac{2 V^2}{\pi r}$$

Answer

$$\frac{\mathrm{V}^2}{\pi \mathrm{r}}$$

Work, Energy and Power & Circular Motion

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

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