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A particle starts from rest and moves on a curve with constant angular acceleration of 3.0 rad/s square units .An observer starts his stopwatch at a certain instant and record that the particle covers an angular span of 120 rad at the end of 4th second.How long had the particle moved when the observer started his stopwatch?

A particle starts from rest and moves on a curve with constant angular acceleration of 3.0 rad/s square units .An observer starts his stopwatch at a certain instant and record that the particle covers an angular span of 120 rad at the end of 4th second.How long had the particle moved when the observer started his stopwatch?

Grade:11

3 Answers

Kuldeep Pal
53 Points
4 years ago
Let The total displacement be when stpwatch starts be “X”
so, the total displacent =
   X=0.t + (1/2).3.4.4
   X=12rad.
 
So. The Angular displacement when the stopwatch starts=
   X= (1/2).3.T2 _______(1)
And Also it is given that total time is 4s and Total Angular Displacent is 12rad. 
So,    12-X=(1/2).3(4-T)2 ______(2)
Hence Two equations And two unknown , We Can Find ‘T’ and ‘X’.
Leaf
10 Points
4 years ago
I don’t think that’?s right.I’l?l try making the question more clear.The 4th    ?second is 4 seconds after the stopwatch was started and the total displacement was 120 rad throughout its entire motion.We’ re supposed to find the amount of time it travelled until the observer started his stopwatch.
Kuldeep Pal
53 Points
4 years ago
Whats problem in this ..This is more simpler than previous one.
The Angular displacent in ‘n’th second is given by:
     (X)n = u + (a/2)(2n-1)
     Where, u = the angular velocity when the stopwatch start.
                 a = The Angular Acceleration of The Motion.
                 n = ‘n’th second.
  So,    120 = u + (3/2)(7)
            u = 219/2.
        Hence , We have The Angular velocity when Stopwatch Starts.
   And By,   v2 = u2 + 2aX. we Can find the angular Displacement.
                 {(219/2)2 -(0)2}/2.3 = X
  The Method Is right.

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