Question

In Fig.$\mathbf{10}\mathbf{-}\mathbf{42}$ , a cylinder having a mass of$\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$ can rotate about its central axis through point$\mathbf{\text{O}}$ . Forces are applied as shown: ${\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N}}\mathbf{}$,${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N}}$ ,${\mathbf{F}}_{\mathbf{3}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N}}$ , and${\mathbf{F}}_{\mathbf{4}}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}N}}$ . Also,$\mathbf{r}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$ and$\mathbf{R}\mathbf{=}\mathbf{}\mathbf{12}\mathbf{\text{\hspace{0.17em}cm}}$ . Find the (a) magnitude and (b) direction of the angular acceleration of the cylinder. (During the rotation, the forces maintain their same angles relative to the cylinder.)

Short answer

1. Magnitude of the angular acceleration of the cylinder is$9.7\text{\hspace{0.17em}rad}/{\text{s}}^2$.
2. Direction of the angular acceleration of the cylinder is counterclockwise.

Step-by-step solution

Understanding the given information

1. The mass of cylinder is,$\mathrm{m}=2.0\text{\hspace{0.17em}kg}$
2. ${\mathrm{F}}_1=6.0\text{\hspace{0.17em}N}$
3. ${\mathrm{F}}_2=4.0\text{\hspace{0.17em}N}$
4. ${\mathrm{F}}_3=2.0\text{\hspace{0.17em}N}$
5. $\mathrm{R}=0.12\text{\hspace{0.17em}m}$
6. $\mathrm{r}=0.050\text{\hspace{0.17em}m}$

Concept and formula used in the given question 

From the given figure and information, it is possible to find the net torque acting on the cylinder. Using this net torque and value of moment of inertia of a cylinder, you can find the magnitude of the angular acceleration of the cylinder. The formulas used for above are given below.

1. Net Torque, ${\mathbf{\tau }}_{\mathbf{net}}\mathbf{=}{\mathbf{F}}_{\mathbf{1}}\mathbf{R}\mathbf{-}{\mathbf{F}}_{\mathbf{2}}\mathbf{R}\mathbf{-}{\mathbf{F}}_{\mathbf{3}}\mathbf{r}$
2. Angular acceleration,$\mathbf{\alpha }\mathbf{=}{\mathbf{\tau }}_{\mathbf{net}}\mathbf{/}\mathbf{I}$
3. Moment of inertia,$\mathbf{I}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}{\mathbf{MR}}^{\mathbf{2}}$

(a) Calculation for the magnitude of the angular acceleration of the cylinder

Calculating net torque by the above formula,

${\mathrm{\tau }}_{\mathrm{net}}={\mathrm{F}}_1\mathrm{R}-{\mathrm{F}}_2\mathrm{R}-{\mathrm{F}}_3\mathrm{r}$

Substitute all the value in the above equation

$\begin{array}{c}{\mathrm{\tau }}_{\mathrm{net}}=\left(6.0\text{\hspace{0.17em}N}\right)\left(0.12\text{\hspace{0.17em}m}\right)-\left(4.0\text{\hspace{0.17em}N}\right)\left(0.12\text{\hspace{0.17em}m}\right)-\left(2.0\text{\hspace{0.17em}N}\right)\left(0.050\text{\hspace{0.17em}m}\right)\\ =71\text{\hspace{0.17em}Nm}\end{array}$

Also, Moment of inertia will be

$\begin{array}{c}\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2\\ =\frac{1}{2}\left(2.0\text{\hspace{0.17em}kg}\right)\times {\left(0.12\text{\hspace{0.17em}m}\right)}^2\\ =0.0144{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Now, substituting these values to find the angular acceleration of the cylinder

We have,

$\mathrm{\alpha }=\frac{{\mathrm{\tau }}_{\mathrm{net}}}{\mathrm{I}}$

Substitute all the value in the above equation

$\begin{array}{c}\mathrm{\alpha }=\frac{71\text{\hspace{0.17em}Nm}}{0.014{\text{\hspace{0.17em}kg.m}}^{\text{2}}}\\ =9.7\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Hence the angular acceleration is, $9.7\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.

(b) Calculation for the direction of the angular acceleration of the cylinder 

Since the value of acceleration is position, the direction of the angular acceleration of the cylinder is, counter clockwise.