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Find the angular displacement of a particle in 6 sec on a circle with a angular velocity 2π/3

Shruti jumde , 8 Years ago
Grade 12
anser 2 Answers
Vikas TU

Last Activity: 8 Years ago

Angular displacement of a particle in 6 sec on a circle with a angular velocity 2π/3 is:
thetha = w*t = > 2pi*6/3 = > 4pi rad.
Ashmak Moon

Last Activity: 8 Years ago

There is a difference between angular distance and angular displacement, angular displacement is a vector quantity with magnitude equal to the shortest distance between initial and final positions of a particle and the direction is from initial to final position. in our case, the distance moved is complete two circomference '4pi’ and hence displacement is a null vector (a vector with zero magnitude and undetermined (not necessarily) direction. In our example, the direction is same as the direction of the angular velocity,i.e., clockwise or anticlockwise. 
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