Question

36) A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

Short answer

The moment of inertia of the wheel is $0.983kg.m^2$.

Step-by-step solution

Step:-1 explanation

In the given question, Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center.

Step:-2 Concept

We know that the final angular speed formula is

${\theta }_f-{\theta }_i={\omega }_{i}t+\frac{1}{2}\alpha t\_\_\_\_\left(i\right)$

We know that${\omega }_f=\frac{2{\theta }_f}{t}\_\_\_\_\_\_\_\_\_\_\_\left(ii\right)$

Moment of the inertia

$KE=\frac{1}{2}l{\omega }_f^2\_\_\_\left(iii\right)$

Step3:- here we find angular displacement.

${\theta }_f-{\theta }_i={\omega }_{i}t+\frac{1}{2}\alpha t$

Here we need to change the final angular displacement = 8.20revs

Here change revs to rad

=$8.20\times 2\pi$

We know that the value of the $\pi =3.14$.

We get,

$$\begin{gathered} 8.20\times 2\times 3.14 \\ =51.496rad \end{gathered}$$

Now,$$\begin{gathered} {\theta }_f-{\theta }_i={\omega }_{i}t+\frac{1}{2}\left({\omega }_f-{\omega }_i\right)t \\ {\theta }_f-0=0t+\frac{1}{2}\left({\omega }_f-0\right)t \\ {\theta }_f=\frac{1}{2}{\omega }_{f}t \end{gathered}$$

Step:-4 here we find a moment of the inertia

Put the value in the equation (ii),

$$\begin{gathered} {\omega }_f=\frac{2\times 51.496}{12.0} \\ =\frac{102.992}{2} \\ =8.582 \end{gathered}$$

Step:-5 calculate inertia

Put the value in equation (iii)

$$\begin{gathered} l=\frac{2K.E}{{\omega }_f^2} \\ l=\frac{2\times 36}{{\left(8.562\right)}^2} \\ =\frac{72}{73.30} \\ =0.983kg.m^2 \end{gathered}$$

The moment of inertia of the wheel is$0.983kg.m^2$.