Circular Motion probem 2

Instead of using a=dv/dt use a=vdv/ds.

The given condition is |a_T|=|a_N|

i.e -vdv/ds=v²/R .

Integrate with proper limits and you will get the answer.

 

You got the answer with your approach or the one i asked you to do ?

 

The particle was decelerating, so your first equation should have been a=-v²/R. The other error was, that when you integrated v with respect time, you forgot the lower limit of integration. So your result is dimensionally incorrect. Tanya's solution is very elegant and simple, but yours is also all right if you do it properly.

ehild

 

Yes, but you have ln(R-vot), it is not zero at t=0.

 

After integration when you put t=0,the term doesn't vanish.You have erroneously assumed it to be 0.