Question

A wheel whose moment of inertia is $12{\mathrm{kgm}}^2$ has an initial angular velocity of $40\mathrm{rad}/\mathrm{s}$. A constant torque of $20\mathrm{Nm}$ acts on the wheel. The time in which the wheel is accelerated to $100\mathrm{rad}/\mathrm{s}$ is

Answer 1

Step 1: Given information

Torque, $\mathrm{T}=20\mathrm{Nm}$

Moment of inertia, $\mathrm{I}=12{\mathrm{kgm}}^2$

Step 2: To find

We have to find the time to which the wheel is accelerated to $100\mathrm{rad}/\mathrm{s}$.

Step 3: Calculation

Calculate the acceleration using the formula,

$\mathrm{T}=\mathrm{I\alpha }...\left(1\right)$

Where, $T$ is the torque, $\alpha$ is the acceleration, and

Substitute the values in equation (1).

$$\begin{gathered} 20=12\times \mathrm{\alpha } \\ \mathrm{\alpha }=\frac{20}{12} \end{gathered}$$

Use the first equation of motion,

$\mathrm{\omega }={\mathrm{\omega }}_{\mathrm{o}}+\mathrm{\alpha t}$

Where, ${\mathrm{\omega }}_{\mathrm{o}}$ is the initial velocity, and $\mathrm{\omega }$ is the final velocity, and $t$ is the time.

$$\begin{gathered} {\mathrm{\omega }}_0=40 \\ \mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t} \\ 100=40+\frac{20}{12}\mathrm{t} \\ \mathrm{t}=\frac{60\times 12}{20} \\ \mathrm{t}=36\sec \end{gathered}$$

Therefore, the time in which the wheel is accelerated to $100\mathrm{rad}/\mathrm{s}$ is $36\mathrm{seconds}$.