91影视

We wills start with writing equilibrium equations for X and Y axes.

For Y-axis: $mg\cos 胃=N鈥�鈥�鈥�鈰�鈰�\left(1\right)$

We can鈥檛 write equilibrium equations for X axis, because it is not static.

Therefore, the Newton鈥檚 Second Law is given by,

$ma=mg\sin 胃-F鈥�鈥�鈥�鈥�鈰�鈰�\left(2\right)$, where,

$F={渭}_{s}mg\cos 胃$, where ${渭}_s$is the coefficient of static friction.

The Newton鈥檚 Second Law for rotational motion is,

$FR=\frac{2}{3}m{R}^2伪鈥�鈥�鈥�鈰�鈰�\left(3\right)$,

where $伪=\frac{a}{R}$, is an angular acceleration and R is the radius.

Thus, equation $\left(3\right)$becomes,

$$\begin{gathered} FR=\frac{2}{3}m{R}^2\frac{a}{R}鈥�鈥�鈥� \\ 鈬�F=\frac{2}{3}ma鈥�鈥�鈥�鈰�鈰�\left(4\right) \end{gathered}$$

Using this in $\left(2\right)$, we get,

$$\begin{gathered} ma=mg\sin 胃-\frac{2}{3}ma \\ 鈬�a=g\sin 胃-\frac{2}{3}a \\ 鈬�\left(1+\frac{2}{3}\right)a=g\sin 胃 \end{gathered}$$

$鈬�a=\frac{3}{5}g\sin 胃$

$$\begin{gathered} 鈬�\mathrm{a}=\frac{3}{5}\left(9.80\right)\sin 38掳 \\ 鈬�\mathrm{a}=\frac{3}{5}\left(9.80\right)\left(0.62\right) \\ 鈬�\mathrm{a}=3.65,\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the acceleration is,$\mathrm{a}=3.65\mathrm{m}/{\mathrm{s}}^2$.

Substituting this value in $\left(4\right)$, we get,

$$\begin{gathered} F=\frac{2}{3}\left(2.00\right)\left(3.65\right) \\ F=4.86鈥�N \end{gathered}$$

Hence, the friction force is, $F=4.86鈥�N$.

Since this is without slipping, we have to justify this. In order to find the minimum coefficient of friction that is need to prevent slipping, we have,

$$\begin{gathered} F=渭N \\ F=渭mg\cos 胃 \\ 鈬�渭=\frac{F}{mg\cos 胃} \\ 鈬�渭=\frac{4.86}{2\left(9.80\right)\left(0.79\right)} \\ 鈬�渭=0.31 \end{gathered}$$

The coefficient of frictions is, $渭=0.31$.

If the mass would be doubled, the acceleration a wouldn鈥檛 be affected since from part (a) it doesn鈥檛 depends on mass. But the friction force will be doubled as it depends on mass directly. Doubling of mass wouldn鈥檛 change the minimum coefficient of friction because it doesn鈥檛 depend on mass.