A particle is moving on a circle of radius R such that at every instant the radial and tangential accelerations are equal in magnitude.If velocity of the particle be $v_0$ at t=0,then what should be time for completion of half of the first revolution ?
I took $dv/dt=v^2/R$ and tried to solve the differential equation.But I cannnot understand how to take the limits of integration.Please help!!
$$ \frac { dv }{ dt } =\frac { { v }^{ 2 } }{ r } \\ \frac { dv }{ { v }^{ 2 } } \quad =\quad \frac { 1 }{ r } \quad dt\\ \int { \frac { dv }{ { v }^{ 2 } } } \quad =\quad \int { \frac { 1 }{ r } \quad dt } \\ \frac { -1 }{ v } \quad =\quad \frac { t }{ r } +C $$
Now you solve for $C$ using the initial conditions:
$$ \\ \\ C\quad +\frac{1}{r}\times 0\quad =\quad \frac { -1 }{ { v }_{ 0 } } \\ C\quad =\quad \frac { -1 }{ { v }_{ 0 } } $$
Finally the answer is:$\\ \\ \frac { -1 }{ v } \quad =\quad \frac{1}{r}t-\frac { 1 }{ { v }_{ 0 } } $
Or by rearranging you get:
$$ v\quad =\quad \frac { r{ v }_{ 0 } }{ r-{ v }_{ 0 }t } $$