A salt shaker sits 0.102 m from
the center of a rotating Lazy
Susan with us = 0.307. At what
maximum speed can the shaker
move without sliding off?
(Unit = m/s)​

Respuesta :

The maximum speed is 0.55 m/s

Explanation:

For an object in uniform circular motion, the force of friction between the object and the ground provides the centripetal force required to keep the body in motion. Therefore we can write:

[tex]\mu_s mg = m\frac{v^2}{r}[/tex]

where the term on the left is the frictional force and the term on the right is the centripetal force, and where

[tex]\mu_s[/tex] is the coefficient of static friction

m is the mass of the body

g is the gravitational acceleration

v is the speed of the body

r is the radius of the circular path

In this problem, we have:

[tex]\mu_s = 0.307[/tex]

r = 0.102 m

[tex]g=9.8 m/s^2[/tex]

Substituting and re-arranging, we find the maximum speed v at which the salt shaker can rotate:

[tex]v=\sqrt{\mu gr}=\sqrt{(0.307)(9.8)(0.102)}=0.55 m/s[/tex]

Learn more about circular motion:

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.55 works for acellus

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