Chapter 8.2, Problem 29PP

To determine

The moment of inertia of the pulley.

Answer to Problem 29PP

  $8.98\text{kg}\cdot {\text{m}}^{\text{2}}$

Explanation of Solution

Given:

The initial rotational speed of the pulley is ${\omega }_i=0\text{}\text{rad/s}$ .

The final rotational speed of the pulley is ${\omega }_f=17\text{}\text{rev/min}$ .

The time taken by pulley to increase the speed is, $\Delta t=5.0\text{}\text{s}$

The force with which the chain is pulled from the pulley, $F=16.0\text{N}$

The radius of the pulley, $r=0.20\text{m}$

Formula used:

The angular acceleration can be obtained by:

  $\alpha =\frac{\Delta \omega }{\Delta t}=\frac{{\omega }_f-{\omega }_i}{t_f-t_i}$

Where, ${\omega }_f$ is the final angular velocity, ${\omega }_i$ is the initial angular velocity, and the $\Delta t$ is the time taken.

The torque can be obtained by:

  $\tau =Fr\sin \theta$

Where, F is the applied force, r is the distance of the point of application of force from the axis of rotation, and $\theta$ is the angle between F and r .

And, angular acceleration can be written as,

  $\alpha =\frac{{\tau }_{net}}{I}$

Where, ${\tau }_{net}$ is the net torque on a rotating object, and $I$ is the moment of Inertia of the object.

Calculation:

Initial angular velocity of the rum is, ${\omega }_i=0\text{}\text{rad/s}$ .

And, the final angular velocity is,

  $\begin{array}{c}{\omega }_f=17\text{rev/min}\\ \text{= 17}\text{rev/min}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\text{min}}{60\text{s}}\\ =1.78\text{rad/s}\end{array}$

Time taken, $\Delta t=5.0\text{s}$

Then, the angular acceleration will be,

  $\begin{array}{c}\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}\\ =\frac{1.78-0}{5}\\ =0.356{\text{rad/s}}^{\text{2}}\end{array}$

Now, the torque generated on the pulley will be,

  $\begin{array}{c}{\tau }_{net}=Fr\sin 90°\\ =\left(16\right)\left(0.20\right)\\ =3.2\text{}\text{Nm}\end{array}$

Then, it is known that the angular acceleration can be written as,

  $\alpha =\frac{{\tau }_{net}}{I}$

Using calculated values in above, the moment of inertia of the pulley will be,

  $\begin{array}{c}0.356=\frac{3.2}{I}\\ I=8.98\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the moment of inertia of the pulley is $8.98\text{kg}\cdot {\text{m}}^{\text{2}}$ .