91影视

The given data listed below as,

• The diameter of horizontal rotating platform is, $d=0.520\text{m}$.
• The revolution speed of button is, $谓=40.0\text{rev/min}$.
• 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-

$蝇=2蟺谓$

Here, $谓$is the revolution speed per minute.

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

$\begin{array}{rcl}蝇& =& 2蟺(40\text{rev/min})\\ & =& \frac{2蟺40}{60}\text{rad/s}\\ 蝇& =& 4.19\text{rad/s}\end{array}$$\begin{array}{rcl}蝇& =& 2蟺(40\text{rev/min})\\ & =& \frac{2蟺40}{60}\text{rad/s}\\ 蝇& =& 4.19\text{rad/s}\backslash \end{array}$

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

Centripetal force is given as-

$F=m{蝇}^{2}r$

Here $蝇$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,

$F=渭mg$

Here, $渭$is the coefficient of static friction, g is the acceleration due to gravity.

for $F=m{蝇}^{2}r$, the above equation becomes-

$$\begin{gathered} m{蝇}^{2}r=渭mg \\ 渭g={蝇}^{2}r \end{gathered}$$

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

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

Hence, required coefficient of static friction is, 0.40.

Part (b)

the angular velocity is given as,

$蝇=2蟺谓$

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

$\begin{array}{rcl}蝇& =& 2蟺(60\text{rev/min})\\ & =& \frac{2蟺60}{60}\text{rad/s}\\ 蝇& =& 6.28\text{rad/s}\end{array}$

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

Centripetal force is given as-

$F=m{蝇}^{2}r$

The expression for the frictional force can be expressed as,

$F=渭mg$

for $F=m{蝇}^{2}r$, the above equation becomes-

$$\begin{gathered} m{蝇}^{2}r=渭mg \\ 渭g={蝇}^{2}r \end{gathered}$$

For $g=9.81{\text{m/s}}^{\text{2}}$, $蝇=6.28\text{rad/s}$and $渭=0.40$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.