Chapter 9, Problem 82P

(a)

To determine

To Calculate: The moment of inertia of combination of platform-drum.

Answer to Problem 82P

The moment of inertia of combination of platform-drum is $1.2\text{ kg}\cdot {\text{m}}^2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10\text{ cm}$

  $\begin{array}{l}=\left(10\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\\ =0.100\text{ m}\end{array}$

Mass of the hanging object, $M=2.50\text{ kg}$

Distance through which the object falls, $D=1.80\text{ m}$

Time, $t_1=4.2\text{s}$

Formula used:

Torque, $\tau =I\alpha$

Where, I is the moment of inertia and $\alpha$ is the angular acceleration.

From Newton’s second law of motion:

  ${\displaystyle \sum F_i=m_i{a}_i}$

Where, F represents force, m represents mass and a represents acceleration.

Second equation of motion is:

  $s=v_{0}t+\frac{1}{2}a{t}^2$

Where, s is the displacement, t is the time, a is the acceleration and $v_0$ is the initial velocity.

Calculation:

Apply the Newton’s second law of motion to the platform

  $\begin{array}{l}{\displaystyle \sum \tau =TR}\\ =I_0\alpha \mathrm{.......}\left(1\right)\end{array}$

Apply the Newton’s second law of motion to the weight:

  $\begin{array}{l}{\displaystyle \sum F_y=Mg-T}\\ =Ma\mathrm{.......}\left(2\right)\end{array}$

Relation between angular acceleration and acceleration is $\alpha =\frac{a}{R}$

Substituting for $\alpha$ in equation $\left(1\right)$ then the tension in the string is

  $T=I_0\frac{a}{R^2}$

Now, substituting for the tension in equation $\left(2\right)$ , and solve for the moment of inertia,

  $\begin{array}{l}Mg-I_0\frac{a}{R^2}=Ma\\ I_0=\frac{M{R}^2\left(g-a\right)}{a}\\ =M{R}^2\left(\frac{g}{a}-1\right)\mathrm{..........}\left(3\right)\end{array}$

The relation among the distance, acceleration and time is given by

  $s=v_{0}t+\frac{1}{2}a{t}^2$

The intial velocity $\left(v_0\right)$ of the object is zero.

Substitute for the initial velocity $\left(v_0\right)$ of the object and distance in the above equation:

  $\begin{array}{l}D=\frac{1}{2}a{t}^2\\ a=\frac{2D}{t^2}\end{array}$

The moment of inertia of combinaiton of platform-drum is: $\begin{array}{l}{I}_0=M{R}^2\left(\frac{g{t}^2}{2D}-1\right)\\ =\left(2.50\text{ kg}\right){\left(0.100\text{ m}\right)}^2\left(\frac{\left(9.80{\text{m/s}}^2\right){\left(4.20\text{ s}\right)}^2}{2\left(1.80\text{ m}\right)}-1\right)\\ =1.18\text{ kg}\cdot {\text{m}}^2\\ \approx \boxed{1.2\text{ kg}\cdot {\text{m}}^2}\end{array}$

Conclusion:

The moment of inertia of combination of platform-drum is $1.2{\text{ kg/m}}^2$ .

(b)

To determine

To Calculate: The total moment of inertia.

Answer to Problem 82P

The total moment of inertia is $3.1\text{ kg}\cdot {\text{m}}^2$ .

Explanation of Solution

Given data:

Radius of the concentric drum, $R=10\text{ cm}$

  $\begin{array}{l}=\left(10\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\\ =0.100\text{ m}\end{array}$

Mass of the hanging object, $M=2.50\text{ kg}$

Distance fall by the object, $D=1.80\text{ m}$

Time, $t_2=6.8\text{s}$

Formula used:

From the previous part:

  $I_{total}=M{R}^2\left(\frac{g{t}^2}{2D}-1\right)$

Calculation:

Substitute the values and solve for total moment of inertia:

  $\begin{array}{l}{I}_{total}=M{R}^2\left(\frac{g{t}^2}{2D}-1\right)\\ =\left(2.50\text{ kg}\right){\left(0.100\text{ m}\right)}^2\left(\frac{\left(9.80{\text{m/s}}^2\right){\left(6.80\text{ s}\right)}^2}{2\left(1.80\text{ m}\right)}-1\right)\\ =\boxed{3.1\text{ kg}\cdot {\text{m}}^2}\end{array}$

Conclusion:

Total moment of inertia is $3.1\text{ kg}\cdot {\text{m}}^2$ .

(c)

To determine

To Calculate: The moment of inertia of the object.

Answer to Problem 82P

The moment of inertia of the object is $1.9\text{ kg}\cdot {\text{m}}^2$ .

Explanation of Solution

Given data:

The total moment of inertia is $3.1\text{ kg}\cdot {\text{m}}^2$ .

The moment of inertia of combination of platform-drum is $1.2\text{ kg}\cdot {\text{m}}^2$ .

Formula used:

The moment of inertia of the object can be calculated by using the formula:

  $I=I_{total}-I_0$

Calculation:

Moment of inertia of the object, $I=I_{total}-I_0$

  $\begin{array}{l}=3.12\text{kg}\cdot {\text{m}}^2-1.18\text{ kg}\cdot {\text{m}}^2\\ =\boxed{1.9\text{kg}\cdot {\text{m}}^2}\end{array}$

Conclusion:

Moment of inertia of the object is $1.9\text{ kg}\cdot {\text{m}}^2$ .