A merry go round rotates from rest with an constant angular acceleration ratio of time
Answers
Given that merry go round starts from rest. So ω0 = 0 rad/s.
For first two revolutions, θ = 2×2π rad
θ = ω.t1 + 1/2 α.t1^2
2×2π = 0×t1 + 1/2 × α × t1^2
t1^2 = 8π/α
t1 = √(8π/α)
For first four revolutions, θ = 4×2π rad
θ = ω.t2 + 1/2 α.t2^2
4×2π = 0×t2 + 1/2 × α × t2^2
t2^2 = 16π/α
t2 = √(16π/α)
Time taken for 3rd and 4th revolution is -
t3 = t2 - t1
t3 = √(16π/α) - √(8π/α)
t3 = (√2-1) √(8π/α)
Now, ratio of time to rotate first 2 revolutions and next 2 revolutions is -
t1/t3 = √(8π/α) / [(√2-1)√(8π/α)]
t1/t3 = 1/(√2-1) × (√2+1)/(√2+1)
t1/t3 = √2+1
Therefore, ratio of time to rotate first 2 revolutions & next 2 revolutions is √2+1.
Explanation:
let at any time t, q = EC (1 – e^-t/CR) E = Energy stored = q^2/2c = E^2C^2/2c (1 – e^-t/CR)^2 = E^2C/2 (1 – e^-t/CR)^2 R = rate of energy stored = dE/dt = -E^2C/2 (-1/RC)^2 (1 – e^-t/RC)e^-t/RC = E^2/CR . e^-t/RC (1 – e^-t/CR) dR/dt = E^2/2R [-1/RC e^-t/CR . (1 – e^-t/CR)+(-) . e^-t/CR(1-/ RC) . e^-t/CR] E^2/2R = (-e^-t/CR/RC + e^-2t/CR/RC + 1/RC. E^-2t/CR) = E^2/2R(2/RC . e^-2t/CR – e^-t/CR/RC) ….(1) For R base max dR/dt = 0 ⇒ 2.e^-t/RC -1 = 0 ⇒ e^-t/CR = ½ ⇒ -t/RC = -In^2 ⇒ t = RC In 2 ∴ Putting t = RC In 2 in equation (1) We get dR/dt = E^2/4R .