Question

Review. An object with a mass of $\mathbf{m}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{10}\mathbf{kg}$ is attached to the free end of a light string wrapped around a reel of radius role="math" localid="1663743188635" $\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{m}$ and mass $\mathbf{M}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}$ kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in Figure $\mathbf{10}\mathbf{.}\mathbf{55}$ . The suspended object is released from rest $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ above the floor. Determine

(a) the tension in the string,

(b) the acceleration of the object, and

(c) the speed with which the object hits the floor.

(d) Verify your answer to part (c) by using the isolated system (energy) model.

Short answer

a). The tension in the string is, $11.4\mathrm{N}$.

b). The acceleration of the object is $7.572\frac{\mathrm{m}}{{\mathrm{s}}^2}$

c). The speed with which the object hits the floor is $9.53\frac{\mathrm{m}}{\mathrm{s}}$.

d). Using the isolated system (energy) model is $9.53\frac{\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Tension in the string.

a).

Given us,

$\begin{array}{l}\mathrm{m}=5.10\mathrm{kg}\\ \mathrm{M}=3\mathrm{kg}\\ \mathrm{R}=0.25\mathrm{kg}\\ {\mathrm{y}}_{\mathrm{i}}=6\mathrm{m}\\ {\mathrm{y}}_{\mathrm{f}}=0\end{array}$

Using Newton’s second law of motion,

$\begin{array}{c}\sum {\mathrm{F}}_{\mathrm{y}}=\mathrm{mg}-\mathrm{T}={\mathrm{ma}}_{\mathrm{y}}\\ 5.10(9.8)-\mathrm{T}=5.10\left(\mathrm{a}\right)\\ 49.98-\mathrm{T}=5.1\mathrm{a}\mathrm{...........}\left(1\right)\end{array}$

Now, MI of pulley of the solid disk is given as,

$\begin{array}{l}\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2\\ \mathrm{I}=\frac{1}{2}\left(3\right){(0.25)}^2\\ \mathrm{I}=0.0938{\mathrm{kgm}}^2\mathrm{.........}\left(2\right)\end{array}$

Model pulley as a rigid object under net torque,

$\begin{array}{l}\mathrm{\tau }={\mathrm{I}}_{\mathrm{\alpha }}\\ \mathrm{\alpha }=\frac{{\mathrm{TR}}^2}{\mathrm{I}}=\frac{\mathrm{a}}{\mathrm{r}}\\ \mathrm{a}=\frac{{\mathrm{TR}}^2}{0.0988},\mathrm{........}\mathrm{by}\left(2\right)\end{array}$

Now, by putting these values in equation one,

$\begin{array}{c}49.98-\mathrm{T}=5.10\left(\frac{0.{25}^2}{0.0938}\right)\mathrm{T}\\ 49.98-\mathrm{T}=3.4\mathrm{T}\\ \\ \mathrm{T}=11.4\mathrm{N}\end{array}$

Hence, the tension in the string is,$11.4\mathrm{N}$ .

Acceleration of the object.

b).

$\begin{array}{c}\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{ma}}_{\mathrm{y}}\\ \\ \mathrm{mg}-\mathrm{T}=\mathrm{ma}\\ \\ \mathrm{T}=\mathrm{mg}-\mathrm{ma}\\ \\ \mathrm{a}=\frac{2\mathrm{T}}{\mathrm{M}}\end{array}$

$\begin{array}{c}\frac{1}{2}\mathrm{Ma}=\mathrm{mg}-\mathrm{ma}\\ \\ \mathrm{a}\left(\mathrm{m}+\frac{\mathrm{M}}{2}\right)=\mathrm{mg}\\ \\ \mathrm{a}=\frac{\mathrm{m}}{\mathrm{m}+\frac{\mathrm{M}}{2}}\mathrm{g}\\ \\ \mathrm{a}=\frac{5.10\mathrm{kg}}{5.10\mathrm{kg}+\frac{3}{2}\mathrm{kg}}\end{array}$

$\mathrm{a}=7.572\frac{\mathrm{m}}{{\mathrm{s}}^2}$

Hence, the acceleration of the object is $\mathrm{a}=7.572\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Speed with which the object hits the floor.

c).

$\begin{array}{c}{\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{aH}\\ \\ {\mathrm{v}}^2-{(0.0)}^2=2\left(7.572\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(6\mathrm{m}\right)\\ \\ \mathrm{v}=9.53\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the speed with which the object hits the floor is $9.53\frac{m}{s}$.

Isolated system (energy) model.

d).

$\begin{array}{c}{\mathrm{v}}_{\mathrm{f}}=\sqrt{\frac{\mathrm{mgh}}{\frac{\mathrm{m}}{2}+\frac{1}{2}.\frac{\mathrm{x}}{{\mathrm{R}}^2}}}\\ \\ =\sqrt{\frac{(5.10\mathrm{kg})(9.8)(6.00\mathrm{m})}{\frac{1}{2}(5.10)+\frac{1}{2}\left(\frac{0.0009375}{0.{025}^2}\right)}}\\ \\ =9.53\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, Using the isolated system (energy) model is $9.53\frac{\mathrm{m}}{\mathrm{s}}$.