Question

a body of mass m starts to move on a circular path of radius r. at any instant it's speed is u and tangential acceleration cx where c is a constant and x is distance travelled by the particle . the gain in k.e in one revolution is?​

Answer 1

Answer:

I think you mean initial speed is u instead of instantaneous , because since tangential acceleration is not zero speed won't be uniform .

Now

at x = 0 :- v = u

at x = 2πr :- v = v ( let after one revolution )

Since

$a = v \frac{dv}{dx} \\ \\ adx = vdv \\ \\ \int _{0}^{2\pi r}cx.dx = \int_{u}^{v} vdv \\ \\ c {(2\pi r)}^{2} = {v}^{2} - {u}^{2} \\ \\ {v}^{2} - {u}^{2} = 4\pi {}^{2} {r}^{2} c \\ \\ gained \: kinetic \: energy \: \\ = change \: in \: kinetic \: energy \\ \\ so \\ \\ \red{k = \frac{1}{2}m {v}^{2} - \frac{1}{2}m {u}^{2} } \\ \\ k = \frac{1}{2} m( {v}^{2} - {u}^{2} ) \\ \\ k = \frac{m}{2} \times 4 {\pi}^{2} {r}^{2} c \\ \\ \huge\purple{k = 2 {\pi}^{2} c \: m{r}^{2} }$