Question 55
Minimum angular speed at which the platform starts sliding
Given:
r = 1 m, mass of platform + pole + motor = ηm, η = 4, μ = 1/√3.
r = 1 m, mass of platform + pole + motor = ηm, η = 4, μ = 1/√3.
Key idea: Take the platform, pole, motor, rod and ball together as one system. Forces between these parts are internal. The only external horizontal force is friction from the ground.
1. Horizontal force required
Let the rod make an angle θ with the downward vertical. The ball has centripetal acceleration ω2r towards the motor.
ax = ω2r sinθ
Hence the required external horizontal force is
F = mω2r sinθ
2. Normal reaction
The total mass of the system is (η + 1)m. The upward component of acceleration of the ball is ω2r cosθ.
N − (η + 1)mg = mω2r cosθ
N = (η + 1)mg + mω2r cosθ
3. Condition for impending sliding
At limiting friction, F = μN.
mω2r sinθ = μ[(η + 1)mg + mω2r cosθ]
ω2r(sinθ − μcosθ) = μ(η + 1)g
4. Minimum angular speed
For the minimum ω, the factor (sinθ − μcosθ) must be maximum.
maximum value = √1 + μ2
ω02 = μ(η + 1)g / [r√1 + μ2]
5. Substitute the values
μ = 1/√3, η = 4, r = 1 m, g = 10 m s−2
ω02 = 25
ω0 = 5 rad s−1
ω0 = 5 rad s−1