Answer:
0.407 rev/s
Explanation:
The centripetal force on the passenger equals the normal force acting on the passenger due to the hollow steel cylinder.
So mrω² = N
Also, the frictional force equals the passenger's weight
F = μN = mg
So. μmrω² = mg ⇒ ω = √g/μr.
where ω is the angular frequency,
ω = √(g/μr)
Given that r = radius of cylinder = 5.0/2 = 2.5 m and μ = 0.6 which is the minimum value of the coefficient of static friction given,
ω = √(g/μr) = √(9.8/(0.6 × 2.5)) = √(9.8/1.5) = 2.556 rad/s
ω = 2.556 rad/s ÷ 2π = 0.407 rev/s
Answer:
Minimum rotational frequency for which the ride is safe is 24.402 rpm
Explanation:
For the rotational frequency to be minimum, the frictional force acting on the passengers by the wall should be maximum
[tex]f_{r} = \mu R\\R = mr\omega^{2} \\f_{r} = mg[/tex]
[tex]mg =\mu mr \omega^{2} \\g = \mu r \omega^{2} \\g = 9.8 m/s^{2} \\diameter, d= 5.0 m\\radius, r = d/2 = 5/2\\r = 2.5 m\\\mu = 0.6\\9.8 = 0.6 * 2.5 * \omega^{2} \\\omega^{2} =9.8/1.5\\\omega^{2} =6.53\\\omega = \sqrt{6.53} \\\omega = 2.56 rad/sec[/tex]
Multiply by 60/2π to convert to rev/min
[tex]\omega = 2.56 * (60/2\pi )\\\omega = 24.402 rpm[/tex]
