Circular Motion Problem 2: Help with Solving | Physics Forum

[thunderhadron]

Hi friend the problem is as follows:



Attempt:





Please friends help me in this.
Thank you all in advance

 

[Tanya Sharma]

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.

 

[thunderhadron]

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.

Thank you very much tanya. I got the answer.

But Was I doing it in wrong manner?

 

[Tanya Sharma]

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

 

[thunderhadron]

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

By your approach.

 

[thunderhadron]

By your approach.

Tanya please see this issue also,

https://www.physicsforums.com/showthread.php?p=4491282#post4491282

 

[ehild]

Thank you very much tanya. I got the answer.

But Was I doing it in wrong manner?

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

 

[thunderhadron]

The other error was, that when you integrated v with respect time, you forgot the lower limit of integration.

ehild

The lower limit of time should be zero. The question states that.

 

[ehild]

The lower limit of time should be zero. The question states that.

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

 

[Tanya Sharma]

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

 

[thunderhadron]

Thank you very much friends. I got the answer. Problem has been cleared.