Angular Motion objective question answer quiz

We know that two equal and unlike parallel forces acting on a body form a couple. And moment of the couple is equal to the product of the magnitude of either force and perpendicular distance between the lines of action of the forces.
This couple tends the body to have purely rotational motion.

Given: Angular velocity of wheel (ω) = 70 rad-s^-1 and radius of wheel (r) = 0.5m.
We know that linear velocity of the wheel (v) = ω r = 70 × 0.5 = 35 m-s^-1

Given: Radius of dics (r) = 1m; Angular velocity of dics (ω) = 10 rad-s^-1 and distance of the point from its centre (d) = 0.5m.
We know that linear velocity of dice on its extreme point (v) = d ω = 0.5 × 10 = 5 m-s^-1.

Given that diameter of wheel (d) = 400cm = 4m or Radius (r) = 2m and velocity of the point on the rim (v) = 16 m-s^-1.
We know that angular velocity of wheel (ω) = v/r = 16/2 = 8 rad-s^-1.

Given that angular frequency of flywheel (N) = 120 rev-min^-1 = 2 rev-s^-1.
We know that angular speed of the flywheel (ω) = 2 π N = 2 π × 2 = 4 π rad-s^-1.

Given that diameter of flywheel (d) = 1m or Radius (r) = 0.5m and angular frequency of flywheel (N) = 600 rev-min^-1 = 10 rev-s^-1.
We know that angular speed of flywheel (ω) = 2 π N = 2 π × 10 = 20 π m-s^-1.
Therefore linear velocity of a particle on the periphery of the wheel (v) = ωr
= 0.5 × 20 π = 10 π m-s^-1.

Given that initially angular frequency (N₁) = 600 rev-m^-1 = 10 rev-s^-1 and final angular frequency (N₂) = 1200 rev-min^-1 = 20 rev-s^-1.
We know that increase in angular velocity (∆ω)          = ω₂ – ω₁ = 2 π N₂ – 2 π N₁  = (2π × 20) – (2π × 10)
= (40 π × 20 π) = 20 π rad-s^-1.

Given that initially angular frequency (N₁) = 0 (because its starts from rest); Final angular frequency (N₂) = 540 rev-min^-1 = 9 rev-s^-1 and time taken (t) = 6s.
We know that angular acceleration (α) = [(ω₂ – ω₁) / t] = [(2 π N₂ – 2 π N₁) / t]  = [2π (N₂ – N₁)/ 6]
= (2 π ( 9 – 0) / 6 = 3 π rad-s^-2.

Given that initial angular frequency (N₁) = 900 rev-min^-1 = 15 rev-s^-1; Final angular frequency (N₂) = 0 (because it comes to rest) and time taken (t) = 1 min = 60s.
We know that angular acceleration (α) = [(ω₁ – ω₂) / t] = [(2 π N₁ – 2 π N₂) / t]  = [2π (N₁ – N₂)/ 60]
= (2 π (15 – 0) / 60 = 30 π / 60 = π / 2 rad-s^-2.

Given that Mass of the body = m; Spring constant = k; length of the spring = l and the angular velocity of the system = ω.
We know that when the system is rotated, it will cause some centrifugal force, which will increase the length of the spring.
We also know that centrifugal force (F) = mω² (l + x). [where x = Extension in the spring]
Therefore increase in the length of spring (x) = Centrifugal force / Spring constant  =  F / k
or, F = kx.
Equating two values of F, we get kx = mω² (l + x) = mω²l + m²x
or, mω²l = kx – mω²x = x (k – mω²)
or, x = (mω²l / k) – mω².