Chapter 12, Problem 85A

To determine

The linear acceleration of the block, the angular acceleration of the drum, the tension in the rope, and the angular velocity of the drum.

Answer to Problem 85A

The linear acceleration of the block is $a=0.98\text{m}/{\text{s}}^{\text{2}}$ .

The angular acceleration of the drum is $\alpha =3.92\text{rad}/{\text{s}}^{\text{2}}$ .

The tension in the rope is $F_{\text{T}}=35.3\text{ N}$ .

The angular velocity of the drum after the block has fallen $5.00\text{ m}$ is $\omega =12.5\text{rad}/\text{s}$ .

Explanation of Solution

Given:

The radius of the drum is $r=0.250\text{ m}$ .

The moment of inertia is $I=2.25\text{ kg}\cdot {\text{m}}^2$

The mass of the block is $m=4.00\text{ kg}$ .

The fallen height is $x=5.00\text{ m}$ .

Formula used:

The expression for Newton’s second law for the block is,

$mg-F_{\text{T}}=ma$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $F_{\text{T}}$ is the tension in the rope, and $a$ is the acceleration.

The expression for Newton’s second law for the drum is,

$F_{\text{T}}r=\frac{Ia}{r}$

Here, $I$ is the moment of inertia and $r$ is the radius.

The expression for the angular acceleration is,

$\alpha =\frac{a}{r}$

The expression for the angular velocity is,

  $\omega =\alpha t$

Here, $\alpha$ is the angular acceleration and $t$ is the time.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The linear acceleration of the block is,

$\begin{array}{l}mg-F_{\text{T}}=ma\\ mg-\frac{Ia}{r^2}=ma\\ mg=ma+\frac{Ia}{r^2}\\ a=\frac{mg}{ma+\frac{Ia}{r^2}}\end{array}$

$\begin{array}{l}a=\frac{\left(4.00\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(4.00\text{ kg}\right)+\frac{\left(2.25\text{ kg}\cdot {\text{m}}^2\right)}{{\left(0.250\text{ m}\right)}^2}}\\ a=0.98\text{m}/{\text{s}}^{\text{2}}\end{array}$

The angular acceleration of the drum is

$\begin{array}{l}\alpha =\frac{a}{r}\\ \alpha =\frac{0.98\text{m}/{\text{s}}^{\text{2}}}{0.25\text{ m}}\\ \alpha =3.92\text{rad}/{\text{s}}^{\text{2}}\end{array}$

The tension in the rope is,

$\begin{array}{l}{F}_{\text{T}}=\frac{I\alpha }{r}\\ F_{\text{T}}=\frac{\left(2.25\text{ kg}\cdot {\text{m}}^2\right)\left(3.92\text{rad}/{\text{s}}^{\text{2}}\right)}{\left(0.250\text{ m}\right)}\\ F_{\text{T}}=35.3\text{ N}\end{array}$

The time taken for the block has fallen $5.00\text{ m}$ is,

$\begin{array}{l}x=\frac{1}{2}a{t}^2\\ t=\sqrt{\frac{2\times \left(5.00\text{ m}\right)}{\left(0.98\text{m}/{\text{s}}^{\text{2}}\right)}}\\ t=3.19\text{ s}\end{array}$

The angular velocity of the drum after the block has fallen $5.00\text{ m}$ is,

$\begin{array}{l}\omega =\alpha t\\ \omega =\left(3.92\text{rad}/{\text{s}}^{\text{2}}\right)\left(3.19\text{ s}\right)\\ \omega =12.5\text{rad}/\text{s}\end{array}$

Conclusion:

Thus, the linear acceleration of the block is $a=0.98\text{m}/{\text{s}}^{\text{2}}$ , the angular acceleration of the drum is $\alpha =3.92\text{rad}/{\text{s}}^{\text{2}}$ , the tension in the rope is $F_{\text{T}}=35.3\text{ N}$ and the angular velocity of the drum after the block has fallen $5.00\text{ m}$ is $\omega =12.5\text{rad}/\text{s}$ .