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71P

Expert-verifiedFound in: Page 326

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 11-60, a constant horizontal force of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10kg, its radius is 0.10 m****, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?**

- Acceleration of center of mass is $1.6\mathrm{m}/{\mathrm{s}}^{2}$.
- Angular acceleration of center of mass is$16\mathrm{rad}/{\mathrm{s}}^{2}$.
- Frictional force in terms of unit vector notation is $\mathrm{F}=4\hat{\mathrm{i}}$.

$\mathrm{F}=12\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{m}=10\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{r}=0.1\mathrm{m}$

**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: **

${\mathbf{\tau}}{\mathbf{=}}{\mathbf{I\alpha}}{\mathbf{=}}{\mathbf{F}}{\mathbf{\times}}{\mathbf{r}}\phantom{\rule{0ex}{0ex}}{\mathbf{a}}{\mathbf{=}}{\mathbf{\alpha r}}$

**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:

$\mathrm{\tau}=\mathrm{I\alpha}$

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

$\mathrm{I}=0.5{\mathrm{mr}}^{2}+{\mathrm{mr}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{I}=1.5{\mathrm{mr}}^{2}$

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,

$1.5{\mathrm{mr}}^{2}\times \mathrm{\alpha}=\mathrm{F}\times 2\mathrm{r}\phantom{\rule{0ex}{0ex}}1.5\times 10\times 0.{1}^{2}\times \mathrm{\alpha}=12\times 2\times 0.1$

So,

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

Now, acceleration of center of mass is as follows:

$\mathrm{a}=\mathrm{\alpha r}\phantom{\rule{0ex}{0ex}}\mathrm{a}=16\times 0.1\phantom{\rule{0ex}{0ex}}\mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^{2}$

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

(b)

From above calculations,

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

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:

$12-\mathrm{F}=\mathrm{m}\times {\mathrm{a}}_{\mathrm{cm}}\phantom{\rule{0ex}{0ex}}12-\mathrm{F}=10\times 1.6$

So,

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

In unit vector notation,

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

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.

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