A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t
ime if the angular acceleration is constant?
1 answer:
Answer:
Explanation:
Given that the grand stone has initial angular velocity of
w(ini)= 6rad/
And it has a final angular velocity of
w(fin)=12.20rad/sec
Time taken is t=16s
Using equation of angular motion
To get angular acceleration (α)
w(fin)=w(ini)+αt
12.20=6+16α
16α=12.20-6
16α=6.2
α=6.2/16
α=0.3875rad/sec²
The angular acceleration is 0.39rad/s²
Angle that he turn using
w(fin)²=w(ini)²+2αθ
12.2²=6²+2×0.3875θ
12.2²-6²=0.775θ
0.775θ=112.84
Then, θ=112.84/0.775
θ=145.6radian
The angular displacement is 145.6rad
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