Answer to Question #313956 in Physics for abby

Question

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?

Answer 1

Given:

$\theta = 3.04 + 4.83t^2 + 2.99t^3$

(a)

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

(b)

$$\omega=\frac{d\theta}{dt} = 9.66t+ 8.97t^2$$

$$\omega(0) =0$$

(c)

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

(d)

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

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

(e) no