The person must stand at a radius of 0.99 m
Explanation:
In order for the person to stand on the merry go round, the force of friction acting on the person must provide the centripetal force necessary to keep the person in uniform circular motion.
Therefore, we can write:
[tex]\mu mg = m\omega^2 r[/tex]
where:
- the term on the left is the force of friction, and the term on the right is the centripetal force
[tex]\mu = 0.45[/tex] is the coefficient of friction
m is the mass of the person
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\omega = 20 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=2.09 rad/s[/tex] is the angular velocity
r is the radius of the circular path
Solving the equation for r, we find the radius at which the person must be standing:
[tex]r=\frac{\omega^2}{\mu g}=\frac{(2.09)^2}{(0.45)(9.8)}=0.99 m[/tex]
Learn more about circular motion:
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