Question

α=-10.0-5.00 tA shaft is turning at 65.0 rad/s at time $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}$Thereafter, its angular acceleration is given by$\mathit{\alpha }\mathbf{=}\mathbf{-}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{t}$where$\mathit{\alpha }$is in${\mathbf{\text{rad/s}}}^{\mathbf{2}}$and t is in seconds.

(a) Find the angular speed of the shaft at$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{s}$.

(b) Through what angle does it turn between$\mathit{t}\mathbf{=}\mathbf{0}$and$\mathit{t}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{s}$?

Short answer

(a). The angular speed of the shaft at $t=3.00 \mathrm{s}$is$\omega =12.5 \mathrm{rad}/\mathrm{s}$.

(b). The angle is$\theta =127.5 \mathrm{rad}$.

Step-by-step solution

Define 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.

Angular speed of the shaft at time t

(a)

Given,

At $t=0$angular speed is $\omega =65 \mathrm{rad}/\mathrm{s}$.

Angular acceleration is$\alpha =-10-5\mathrm{t}$.

Integrating angular acceleration with respect to time.

$\begin{array}{rcl}\omega & =& \int \alpha dt\\ & =& \int \left(-10-5t\right)dt\\ & =& -10t-5\frac{t^2}{2}+c\end{array}$

Substitute $t=0$,

role="math" localid="1663738964573" $\begin{array}{rcl}65& =& -10\times 0-5\frac{0^2}{2}+c\\ c& =& 65\end{array}$

Consider

$\omega =-10t-5\frac{t^2}{2}+65$ → (1)

Given$t=3 \mathrm{s}$,

$\begin{array}{c}\omega =-10\times 3-5\frac{3^2}{2}+65\\ \omega =-30-\frac{5\times 9}{2}+65\\ \omega =12.5 \mathrm{rad}/\mathrm{s}\end{array}$

The angular speed of the shaft at $t=3.00 \mathrm{s}$is$\omega =12.5 \mathrm{rad}/\mathrm{s}$.

Angle by which the shaft turns

Consider equation (1),

$\omega =-10t-5\frac{t^2}{2}+c$

For $t=0$,

$\begin{array}{rcl}65& =& -10\times 0-5\frac{0^2}{2}+c\\ c& =& 65\end{array}$

Substitute c in equation (1).

$\omega =-10t-5\frac{t^2}{2}+65$

Integrating angular velocity from zero to 3s.

$\begin{array}{c}{\int }_0^3\omega dt={\int }_0^3\left(-10t-\frac{5t^2}{2}+65\right)dt\\ \theta =-10{\int }_0^{3}tdt-\frac{5}{2}{\int }_0^3{t}^{2}dt+{\int }_0^{3}65dt\\ \theta =-10\left[\frac{3^2}{2}\right]-2.5\left[\frac{3^3}{3}\right]+65\times 3\\ \theta =-10\times \frac{9}{2}-2.5\times 9+195\\ \theta =127.5 \mathrm{rad}\end{array}$

The angle is$\theta =127.5\mathrm{rad}$.