Question 263: Work, Energy and Power & Circular Motion

Question

A particle of mass '$$\mathrm{m}$$' moves along a circle of radius '$$r$$' with constant tangential acceleration. If K.E. of the particle is '$$E$$' by the end of third revolution after beginning of the motion, then magnitude of tangential acceleration is

Accepted Answer

$$\begin{aligned}
& 	ext { Using } 3^{	ext {rd }} 	ext { equation of motion, } \
& v^2=u^2+2 a s \
	herefore \quad v^2 & =2 a_t s \quad 	ext{.... (For u = 0)}\
	herefore \quad & a_t=\frac{v^2}{2 s}
\end{aligned}$$
By the end of $$3^{	ext {rd }}$$ revolution, distance covered, $$\mathrm{s}=3(2 \pi \mathrm{r})=6 \pi \mathrm{r}$$
$$	herefore \quad a_{\mathrm{t}}=\frac{\mathrm{v}^2}{2 	imes 6 \pi \mathrm{r}}$$ .... (i)
Also, $$\frac{1}{2} \mathrm{mv}^2=\mathrm{E}$$
$$	herefore \quad v^2=\frac{2 \mathrm{E}}{\mathrm{m}}$$ ..... (ii)
Substituting equation (ii) in equation (i),
$$a_{\mathrm{t}}=\frac{2 \mathrm{E}}{2 	imes 6 \pi \mathrm{r} 	imes \mathrm{m}}=\frac{\mathrm{E}}{6 \pi \mathrm{rm}}$$

Answer

$$\frac{\mathrm{E}}{2 \pi \mathrm{rm}}$$

Answer

$$\frac{\mathrm{E}}{6 \pi \mathrm{rm}}$$

Answer

$$\frac{\mathrm{E}}{8 \pi \mathrm{rm}}$$

Answer

$$\frac{\mathrm{E}}{4 \pi \mathrm{rm}}$$

Work, Energy and Power & Circular Motion

A particle of mass '$\mathrm{m}' moves along a circle of radius 'r' with constant tangential acceleration. If K.E. of the particle is 'E$' by the end of third revolution after beginning of the motion, then magnitude of tangential acceleration is

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