Question

A shaft is turning at angular speed$\mathit{\omega }$at timerole="math" localid="1663739500576" $\mathit{t}\mathbf{=}\mathbf{0}$. Thereafter, its angular acceleration is given by $\mathit{\alpha }\mathbf{=}\mathit{A}\mathbf{+}\mathit{B}\mathit{t}$

(a) Find the angular speed of the shaft at time t.

(b) Through what angle does it turn between$\mathit{t}\mathbf{=}\mathbf{0}$and t?

Short answer

(a). The angular speed of the shaft at a time is$\omega =At+B\frac{t^2}{2}+\omega$.

(b). The angle is$\mathrm{\Delta }\theta =\omega t+\frac{1}{2}A{t}^2+\frac{1}{6}B{t}^3$.

Step-by-step solution

Angular speed

Angular speed or rotational speed, also known as angular frequency vector, is a vector measurement of the rate of rotation that refers to how fast an object rotates or rotates relative to another point, i.e. over time, how fast the angular position or orientation of an object changes.

Find the angular speed of the shaft at time t:

(a)

Given,

At $t=0$angular speed is $\omega$.

Angular acceleration is

$\alpha =\mathrm{A}+\mathrm{Bt}$

Do differentiation of angular speed with respect to time.

$\begin{array}{c}\frac{d\omega }{dt}=A+Bt\\ d\omega =(A+Bt)\mathrm{dt}\end{array}$

The integration we get,

$\omega =At+B\frac{t^2}{2}+c$

When$t=0$ angular speed is$\omega$.

$\begin{array}{rcl}\omega & =& A\times 0+B\frac{0^2}{2}+c\\ c& =& \omega \end{array}$

The angular speed of the shaft at a time is$\omega =At+B\frac{t^2}{2}+\omega$.

Find the angle

(b)

As we found a solution in step 2,

$\omega =At+B\left(\frac{t^2}{2}\right)+\omega$

Differentiate,

$\frac{d\theta }{dt}=\omega +At+\frac{1}{2}B{t}^2$

Do integration we get,

$\mathrm{\Delta }\theta =\omega t+\frac{1}{2}A{t}^2+\frac{1}{6}B{t}^3$

So the angle is$\mathrm{\Delta }\theta =\omega t+\frac{1}{2}A{t}^2+\frac{1}{6}B{t}^3$.