Q 52E - University Physics with Modern Physics

Question

Question: A small button placed on a horizontal rotating platform with diameter 0.520 m will revolve with the platform when it is brought up to a speed of $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{rev/min}}$, provided the button is no more than 0.220 m from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{rev/min}}$?

Step-by-Step Solution

Answer

a) The coefficient of static friction is, 0.40.

b) The distance from the axis is, 0.099 m.

Step 1: Identification of given data

The given data listed below as,

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

Step 2: Significance static friction

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.

Step 3: Determination of coefficient of static friction.

Part (a)

the angular velocity is given as-

$\omega =2 \pi \nu$

Here, $\nu$ is the revolution speed per minute.

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

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

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

Centripetal force is given as-

$F=m {\omega }^2 r$

Here $\omega$ 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,

 $F=\mu m g$

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

 

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

 $$\begin{gathered} m {\omega }^2 r=\mu m g \\ \mu g={\omega }^2 r \end{gathered}$$

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

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

Hence, required coefficient of static friction is, 0.40.

Step 4: Determination of distance from the axis.

Part (b)

the angular velocity is given as,

$\omega =2 \pi \nu$

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

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

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

Centripetal force is given as-

$F=m {\omega }^2 r$ 

The expression for the frictional force can be expressed as,

 $F=\mu m g$

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

$$\begin{gathered} m {\omega }^2 r=\mu m g \\ \mu g={\omega }^2 r \end{gathered}$$

 

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