91Ó°ÊÓ

$$\begin{gathered} \mathrm{F}=12\mathrm{N} \\ \mathrm{m}=10\mathrm{kg} \\ \mathrm{r}=0.1\mathrm{m} \end{gathered}$$

Use formula for torque to find angular acceleration. Use angular acceleration to find linear acceleration, and finally use Newton’s second law to find friction force.

Formula are as follow:

$$\begin{gathered} \mathbf{Ï„}\mathbf{=}\mathbf{\pm õÎ\pm }\mathbf{=}\mathbf{F}\mathbf{×}\mathbf{r} \\ \mathbf{a}\mathbf{=}\mathbf{Î\pm °ù} \end{gathered}$$

Where, is torque, is force,is moment of inertia, is radius, is angular acceleration and is acceleration.

(a)

To calculate acceleration of center of mass, first calculate angular acceleration as follows:

$\mathrm{Ï„}=\mathrm{\pm õÎ\pm }$

Moment of inertia about point of contact is given by parallel axis theorem as follows:

$$\begin{gathered} \mathrm{I}=0.5{\mathrm{mr}}^2+{\mathrm{mr}}^2 \\ \mathrm{I}=1.5{\mathrm{mr}}^2 \end{gathered}$$

Force is applied at the top of the cylinder, considering the bottom of the cylinder as pivot point and r as the radius of the cylinder,

$$\begin{gathered} 1.5{\mathrm{mr}}^2×\mathrm{Î\pm }=\mathrm{F}×2\mathrm{r} \\ 1.5×10×0.1^2×\mathrm{Î\pm }=12×2×0.1 \end{gathered}$$

So,

$\mathrm{Î\pm }=16\mathrm{rad}/{\mathrm{s}}^2$

Now, acceleration of center of mass is as follows:

$$\begin{gathered} \mathrm{a}=\mathrm{Î\pm °ù} \\ \mathrm{a}=16×0.1 \\ \mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, acceleration of center of mass is $1.6\mathrm{m}/{\mathrm{s}}^2$.

(b)

From above calculations,

$\mathrm{Î\pm }=16\mathrm{rad}/{\mathrm{s}}^2$

Hence, angular acceleration of center of mass is $16\mathrm{rad}/{\mathrm{s}}^2$.

(c)

Frictional force (F) in unit vector form is as follows:

Now, to find friction force, use Newton’s second law of motion,

So, friction force is as follows:

$$\begin{gathered} 12-\mathrm{F}=\mathrm{m}×{\mathrm{a}}_{\mathrm{cm}} \\ 12-\mathrm{F}=10×1.6 \end{gathered}$$

So,

$\mathrm{F}=4.0\mathrm{N}$

In unit vector notation,

$\stackrel{→}{\mathrm{F}}=\left(4.0\mathrm{N}\right)\hat{\mathrm{i}}$

Since, value F is positive, it is along the positive x axis.

Hence, frictional force in terms of unit vector notation is $\mathrm{F}=4\hat{\mathrm{i}}$.

Therefore, the formula for torque to find the angular acceleration can be used. Using angular acceleration, linear acceleration can be found. Newton’s second law can be used to find friction force.