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ParthKohli
 2 years ago
Classical Mechanics: circular motion
ParthKohli
 2 years ago
Classical Mechanics: circular motion

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0What is the thing written on the left side of the diagram?

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

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

anonymous
 2 years 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)

anonymous
 2 years 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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