Answer to Question #313956 in Physics for abby

Question #313956

The angular position of a point on a rotating wheel is given by θ = 3.04 + 4.83t2 + 2.99t3, where θ is in radians and t is in seconds. At t = 0, what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at t = 7.43 s? (d) Calculate its angular acceleration at t = 1.12 s. (e) Is its angular acceleration constant?


1
Expert's answer
2022-03-20T18:50:55-0400

Given:

θ=3.04+4.83t2+2.99t3\theta = 3.04 + 4.83t^2 + 2.99t^3


(a)

θ(0)=3.04rad\theta(0) = 3.04 \:\rm rad

(b)

ω=dθdt=9.66t+8.97t2\omega=\frac{d\theta}{dt} = 9.66t+ 8.97t^2

ω(0)=0\omega(0) =0

(c)

ω(7.43)=9.66×7.43+8.97×7.432=567rad/s\omega(7.43)= 9.66\times 7.43+ 8.97\times 7.43^2\\ =567\:\rm rad/s

(d)

α=dωdt=9.66+17.9t\alpha=\frac{d\omega}{dt} = 9.66+ 17.9t

α(1.12)=9.66+17.9×1.12=29.8rad/s2\alpha(1.12)= 9.66+ 17.9\times 1.12=29.8\:\rm rad/s^2

(e) no


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