Question

The shaft of a motor has an angular displacement θ that is a function of time given by the equation: θ(t) = 4.40 t 3 rad/s3 + 1.40 t2 rad/s2 . At time t = 0.00 s the wheel is at rest and is oriented at θ = 0.00 rad. a) Derive the equation that specifies the angular velocity of the shaft as a function of time. b) Derive the equation that specifies the angular acceleration as a function of time.

Answer 1

a)
`\omega = 13.2t^2(rad)/(s^3)+2.80t(rad)/(s^2)`

b)
`\alpha=26.4t(rad)/(s^3)+2.80(rad)/(s^2)`

Step-by-step explanation:

You have that the angular displacement is given by:

`\theta=4.40t^3(rad)/(s^3)+1.40t^2(rad)/(s^2)`

a) the angular velocity is given by the derivative in time, of the angular displacement, that is:

`\omega=(d\theta)/(dt)=(d)/(dt)[4.40 t^3 rad/s^3 + 1.40 t^2 rad/s^2]\\\\\omega=(d\theta)/(dt)=13.2t^2(rad)/(s^3)+2.80t(rad)/(s^2)`

b) the angular acceleration is the derivative, in time, of the angular velocity:

`\alpha=(d\omega)/(dt)=(d)/(dt)[13.2t^2(rad)/(s^3)+2.80t(rad)/(s^2)]\\\\\alpha=26.4t(rad)/(s^3)+2.80(rad)/(s^2)`