In a carnival ride,passengers stand with their backs against the wall of a cylinder.The cylinder is set into rotation and the floor is lowered away from the passengers,but they remain stuck against the wall of the cylinder.For a cylinder with a 2.0-m radius,what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25?
A)8.9 m/s
B)2.3 m/s
C)3.0 m/s
D)4.9 m/s
E)It depends on the mass of the passengers.

Question

Quizplus · Accepted Answer

The minimum speed can be found using the formula for centripetal force and the force of friction. The force keeping the passengers from falling is the force of friction, which is $f = \mu N$, where $\mu$ is the coefficient of static friction and $N$ is the normal force, which in this case is provided by the centripetal force required to keep the passengers moving in a circle, $F_c = \frac{mv^2}{r}$. Setting the force of friction equal to the gravitational force (since this is what needs to be overcome to prevent falling), we get $\mu \frac{mv^2}{r} = mg$, where $m$ is the mass of the passenger, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $v$ is the velocity. Solving for $v$, we find $v = \sqrt{\frac{gr}{\mu}}$. Substituting $g = 9.8$ m/s$^2$, $r = 2.0$ m, and $\mu = 0.25$, we get $v = \sqrt{\frac{(9.8)(2.0)}{0.25}} = \sqrt{78.4} \approx 8.9$ m/s. This calculation shows that the mass of the passengers cancels out, indicating that the minimum speed does not depend on the mass of the passengers.

In a Carnival Ride,passengers Stand with Their Backs Against the Wall