Most Important Previous Year Questions
Rotational Dynamics
Question 1: Angular Momentum Conservation (5 marks)
A platform of radius 2m and mass 100kg is rotating with angular velocity of 10 rad/s. A person of mass 60kg standing at the center walks to the edge of the platform. Calculate:
- Initial angular momentum of the system
- Final angular velocity of the platform-person system
- Change in rotational kinetic energy
Solution:
Let's solve this step by step:
1. Initial angular momentum (Li):
Li = I₁ω₁ = (½MR²)ω₁
Li = ½(100)(2²)(10) = 2000 kg⋅m²/s
2. Final moment of inertia:
If = ½(100)(2²) + (60)(2²) = 440 kg⋅m²
3. Final angular velocity (by conservation of angular momentum):
2000 = 440ωf
ωf = 4.55 rad/s
4. Change in rotational KE = ½If(ωf²) - ½Ii(ωi²)
Question 2: Moment of Inertia (4 marks)
Derive the moment of inertia of a solid cylinder about its axis using integration. How does the moment of inertia change when the cylinder is hollow?
Solution:
For solid cylinder:
1. Consider a cylindrical shell of radius r and thickness dr
2. dI = r²dm where dm = ρ(2πr)(h)dr
3. I = ∫dI = ∫r²ρ(2πr)(h)dr from 0 to R
4. I = ½MR² where M is total mass
For hollow cylinder:
I = ½M(R₁² + R₂²) where R₁ and R₂ are outer and inner radii
Question 3: Torque and Angular Acceleration (5 marks)
A wheel of moment of inertia 0.5 kg⋅m² is acted upon by a torque that varies with time according to the relation τ = 3t² - 4t + 2 (where τ is in N⋅m and t in seconds). Find the angular acceleration at t = 2s and the angular velocity at t = 2s if the wheel starts from rest.
Solution:
1. Torque equation: τ = 3t² - 4t + 2
2. Angular acceleration: α = τ/I
3. At t = 2s:
τ = 3(2)² - 4(2) + 2 = 12 - 8 + 2 = 6 N⋅m
α = 6/0.5 = 12 rad/s²
4. For angular velocity:
ω = ∫α dt = (1/I)∫(3t² - 4t + 2)dt
ω = (1/0.5)(t³ - 2t² + 2t) from 0 to 2
Final ω = 8 rad/s
