Respuesta :
Answer:
The coefficient of static friction is 0.578.
Explanation:
Given that,
Mass of coin = 54 gm
Revolution r = 1.0
Distance = 14.4 cm
We need to calculate the force when the coin located more than 14.4 cm from the axis of rotation
Using formula of centripetal force
[tex]F= mR\omega^2[/tex]
Where, m = mass
R = radius
Put the value into the formula
[tex]F=0.054\times10\times14.4^{-2}\times(2\pi)^2[/tex]
[tex]F=0.306\ N[/tex]
We need to calculate the coefficient of static friction
Using formula of friction
[tex]F=\mu mg[/tex]
[tex]\mu=\dfrac{F}{mg}[/tex]
[tex]\mu=\dfrac{0.306}{0.054\times9.8}[/tex]
[tex]\mu=0.578[/tex]
Hence, The coefficient of static friction is 0.578.
Answer:
[tex]\mu = 0.578 [/tex]
Explanation:
given data:
revolution per second [tex]\omega = 1 rev/s = 1 \times 2\pi = 6.28 rad/s [/tex]
for sliding off
maximum static friction [tex] = m \omega^2 \times r[/tex]
maximum static friction is frictional force, hence we have
[tex]\mu mg = m \omega^2 r[/tex]
solving for \mu
[tex]\mu = \frac{\omega^2 r}{g}[/tex]
[tex]\mu = \frac{6.28^2 \times .144}{9.81}[/tex]
[tex]\mu = 0.578 [/tex]
