A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis through speed $2rad{s^{ - 1}}.$ the radius of the disc is$50\,{\text{cm}}$. Find the minimum coefficient of friction between disc and coin so that the coin does not slip ($g = 10m{s^{ - 2}}$)
$
A.{\text{ 0}}{\text{.1}} \\
{\text{B}}{\text{. 0}}{\text{.2}} \\
{\text{C}}{\text{. 0}}{\text{.3}} \\
{\text{D}}{\text{. 0}}{\text{.4}} \\
$
Answer
635.7k+ views
Hint: Friction is the resisting force between the two surfaces that are sliding or slipping across each other which always work in the opposite direction in the object is moving. Use the given terms and its correlation to find out the minimum coefficient of the friction.
Complete step by step answer:
Angular Speed, $\omega = 2{\text{ radse}}{{\text{c}}^{ - 1}}$
Radius, $r = 50{\text{ cm = }}\dfrac{{50}}{{100}}m$
$g = 10m{s^{ - 2}}$
Mass is $ = m$
The coefficient of the friction is $\mu $
Also Friction,
$
f = m{\omega ^2}{\text{r }}.......{\text{(1)}} \\
f = \mu N \\
f = \mu mg\,{\text{ }}........{\text{(2)}} \\
$
Equate the right hand side of the equations $(1){\text{ and (2)}}$
$m{\omega ^2}{\text{r}} = \mu mg$
Take mass “$m$” common from both the sides of the equation and remove it.
${\omega ^2}{\text{r}} = \mu g$
Now, place the values of the known terms.
${(2)^2} \times \dfrac{{50}}{{100}} = \mu \times 10$
Make the coefficient of the friction, $\mu $ the subject –
$
\mu = \dfrac{{4 \times 50}}{{100 \times 10}} \\
\mu = \dfrac{{200}}{{1000}} \\
\mu = 0.2 \\
$
Therefore, the required answer is - the minimum coefficient of friction between disc and coin so that the coin does not slip is $0.2$.
Hence, from the given multiple choices – the option B is the correct answer.
Note:Since the friction and load are measured in units of force, they cancel each other as a result the unit of coefficient of friction ($\mu = \dfrac{F}{L}$ ) is dimensionless. Basically there are two types of coefficient of friction.
-Static coefficient friction – It is applied to the objects which are motionless.
-Kinetic coefficient friction – It is applied to the objects which are in motion.
Complete step by step answer:
Angular Speed, $\omega = 2{\text{ radse}}{{\text{c}}^{ - 1}}$
Radius, $r = 50{\text{ cm = }}\dfrac{{50}}{{100}}m$
$g = 10m{s^{ - 2}}$
Mass is $ = m$
The coefficient of the friction is $\mu $
Also Friction,
$
f = m{\omega ^2}{\text{r }}.......{\text{(1)}} \\
f = \mu N \\
f = \mu mg\,{\text{ }}........{\text{(2)}} \\
$
Equate the right hand side of the equations $(1){\text{ and (2)}}$
$m{\omega ^2}{\text{r}} = \mu mg$
Take mass “$m$” common from both the sides of the equation and remove it.
${\omega ^2}{\text{r}} = \mu g$
Now, place the values of the known terms.
${(2)^2} \times \dfrac{{50}}{{100}} = \mu \times 10$
Make the coefficient of the friction, $\mu $ the subject –
$
\mu = \dfrac{{4 \times 50}}{{100 \times 10}} \\
\mu = \dfrac{{200}}{{1000}} \\
\mu = 0.2 \\
$
Therefore, the required answer is - the minimum coefficient of friction between disc and coin so that the coin does not slip is $0.2$.
Hence, from the given multiple choices – the option B is the correct answer.
Note:Since the friction and load are measured in units of force, they cancel each other as a result the unit of coefficient of friction ($\mu = \dfrac{F}{L}$ ) is dimensionless. Basically there are two types of coefficient of friction.
-Static coefficient friction – It is applied to the objects which are motionless.
-Kinetic coefficient friction – It is applied to the objects which are in motion.
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