Question

A $62kg$ woman is a passenger in a "rotor ride" at an amusement park. A d$5.0m$ is spun with an angular velocity of $25rpm$ . The woman is pressed against the wall of the rotating drum. While the drum rotates, the floor is lowered. A vertical static friction force supports the woman's weight. What must the coefficient of friction be to support her weight?

• A. $0.412$
• B. $0.292$
• C. $0.135$
• D. $0.5$

Answer 1

Answer: (B)

$0.292$

The speed is given as,

$\omega =25rpm$

$\omega =25\times \frac{2\pi }{60}rps$

The force is given as,

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

$F=62\times {\left(25\times \frac{2\pi }{60}\right)}^2\times 5$

$F=2124.71N$

The coefficient of friction is given as,

$\mu F=mg$

$\mu \times 2124.71=62\times 10$

$\mu =0.291$