Question

a wheel at rest is rotated with a uniform angular acceleration such that the particle on the rim off the wheel experiences a tangential acceleration of 20π ms-² the radius of the well is 2 metre the instantaneous period of rotation at the end of 4 second is
A- 0.08 second
B- 0.01 second
C- 0.02 second
D- 0.05 second​

Answer 1

Explanation:

Explanation:

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Answer 2

Answer:

given = angular accelaration = 2₀πms⁻²

            radius = 2m

to find =  period of rotation i.e. t=?

solution =

angular accelration(∝) =  a a1 / k = 20/ 100 = 1/5rad s²

∝=  dw /dt = ωdω/dt = ∫dθ = 2₀∫ωdω....[put the limit 0 to 2π and 0to ω]

[θ] = 20 [ ω²/2] = ω=√π/5

ω= 2π/T

T = 2π/ω

T= 2πX √5/ √π X5

T = 0.08

so the option a is correct answer.