91Ó°ÊÓ

The given data listed below as,

• The diameter of horizontal rotating platform is, $d=0.520\text{m}$.
• The revolution speed of button is, $\mathrm{ν}=40.0\text{rev/min}$.
• The distance of button from the axis is, $r=0.220\text{m}$.

If on application of a force the body does not move, this means that the sliding tendency in the body is cancelled by a force acting between the surface of body and the base. This force is known as static friction.

Part (a)

the angular velocity is given as-

$\mathrm{Ó\neg }=2\mathrm{Ï€}\mathrm{ν}$

Here, $\mathrm{ν}$is the revolution speed per minute.

for $\mathrm{ν}=40.0\text{rev/min}$, the above equation becomes-

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}(40\text{rev/min})\\ & =& \frac{2\mathrm{Ï€}40}{60}\text{rad/s}\\ \mathrm{Ó\neg }& =& 4.19\text{rad/s}\end{array}$$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}(40\text{rev/min})\\ & =& \frac{2\mathrm{Ï€}40}{60}\text{rad/s}\\ \mathrm{Ó\neg }& =& 4.19\text{rad/s}\backslash \end{array}$

Hence, the angular velocity is, $4.19\text{rad/s}$.

Centripetal force is given as-

$F=m{\mathrm{Ó\neg }}^{2}r$

Here $\mathrm{Ó\neg }$is the angular velocity, and r is the distance of button from axis, m is the mass of the object.

The expression for the frictional force can be expressed as,

$F=\mathrm{μ}mg$

Here, $\mathrm{μ}$is the coefficient of static friction, g is the acceleration due to gravity.

for $F=m{\mathrm{Ó\neg }}^{2}r$, the above equation becomes-

$$\begin{gathered} m{\mathrm{Ó\neg }}^{2}r=\mathrm{μ}mg \\ \mathrm{μ}g={\mathrm{Ó\neg }}^{2}r \end{gathered}$$

for $g=9.81{\text{m/s}}^{\text{2}}$, $\mathrm{Ó\neg }=4.19\text{rad/s}$and data-custom-editor="chemistry" $r=0.220\text{m}$the above equation becomes-

$$\begin{gathered} \mathrm{μ}×9.81{\text{m/s}}^{\text{2}}=(4.19{\text{rad/s)}}^2(0.220\text{m}) \\ \mathrm{μ}=0.40 \end{gathered}$$

Hence, required coefficient of static friction is, 0.40.

Part (b)

the angular velocity is given as,

$\mathrm{Ó\neg }=2\mathrm{Ï€}\mathrm{ν}$

For $\mathrm{ν}=60.0\text{rev/min}$the above equation becomes-

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}(60\text{rev/min})\\ & =& \frac{2\mathrm{Ï€}60}{60}\text{rad/s}\\ \mathrm{Ó\neg }& =& 6.28\text{rad/s}\end{array}$

Hence, the angular velocity is, $6.28\text{rad/s}$.

Centripetal force is given as-

$F=m{\mathrm{Ó\neg }}^{2}r$

The expression for the frictional force can be expressed as,

$F=\mathrm{μ}mg$

for $F=m{\mathrm{Ó\neg }}^{2}r$, the above equation becomes-

$$\begin{gathered} m{\mathrm{Ó\neg }}^{2}r=\mathrm{μ}mg \\ \mathrm{μ}g={\mathrm{Ó\neg }}^{2}r \end{gathered}$$

For $g=9.81{\text{m/s}}^{\text{2}}$, $\mathrm{Ó\neg }=6.28\text{rad/s}$and $\mathrm{μ}=0.40$the above equation becomes-

$$\begin{gathered} (0.40)(9.81{\text{m/s}}^{\text{2}})=(6.28{\text{rad/s)}}^{2}r \\ r=0.099\text{m} \end{gathered}$$

Hence, required distance is, 0.099 m.