Question 270: 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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