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ParthKohli
 one year ago
Classical Mechanics: circular motion
ParthKohli
 one year ago
Classical Mechanics: circular motion

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370882795176:dwFind the time when the particles collide.

kutabs
 one year ago
Best ResponseYou've already chosen the best response.0What is the thing written on the left side of the diagram?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0@kutabs 2 rad/s So this time I tried to do with the fact that \(\theta_1 + \theta_2 = \dfrac{3\pi}{2}\) and using the equations of motion.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That isn't a problem at all. The problem is that there are more questions. I am not able to figure out \(a_R\) of each particle at the time of collision.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1is teh acceleration given for the 2rad/sec speed?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Ah, no worries, I get it. It's easy to find \(v\) for each particle after I get the \(t\). Thank you for the time guys.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1this is then the same as the last question really\[\frac12a_1t^2v_1t+d_1=\frac12a_2t^2+v_2t+d_2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\theta_1 = 2t \]\[\theta_2 = t + t^2\]Thank you, I can manage after that.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I was just having this little confusion that I cleared myself. :

kutabs
 one year ago
Best ResponseYou've already chosen the best response.0Let them meet at the angle theta as shown dw:1370883098101:dw For 1: Covering pi/2+(pi/2theta)= omega*t For 2: Covering pi/2+theta= omega*t+1/2*alpha*(t^2)

kutabs
 one year ago
Best ResponseYou've already chosen the best response.0Now from these two equations 1&2there are 2 variables theta and time (t). Solve them, and you'll get the result.
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