University:
Massachusetts Institute of Technology
Course:
8.01X | Physics I: Classical Mechanics with an Experimental Focus
Academic year:

2020
Views:

455

Pages:

25
Author:

Adrian Davila

Practice Final Exam SOLUTIONS Problem 1: (Experiment AM) In experiment AM, a circular steel washer, of mass m₁ and moment of inertia Iw = 4.0 x 10⁻⁵ kg-m² is mounted on the shaft of a small motor. Assume that the inertia of the motor is negligible. The washer is set into motion. When it reaches 200 rad/sec, at t=0 the power to the motor is shut off, and the washer slows down until it reaches an angular velocity of 140 rad/sec at t=6 sec. At that moment a second identical circular steel washer of mass m₂ = m₁ is dropped on the first. The collision takes place over a time 0.2 sec. a) What is the initial angular momentum of the washer at t=0? b) What is the initial kinetic energy of the washer at t=0? c) What is the magnitude of the frictional torque acting from t=0 to 6 sec? d) What is the angular velocity of both washers at t=6.2 sec? e) What is the average torque which the dropped washer exerts on the spinning washer during the time between 6.0 sec and 6.2 sec? a) L₀ = Icmω₀ = (4.0 x 10⁻⁵ kg-m²)(200 rad/sec) = 8.0 x 10⁻³ kg-m²/sec b) K₀ = ½Icmω₀² = ½(4.0 x 10⁻⁵ kg-m²)(200 rad/sec)² = 8.0 x 10⁻¹ J c) τ = Icm(Δω/Δt) = (4.0 x 10⁻⁵ kg-m²)(140 rad/sec - 200 rad/sec) / 6 sec = 4 x 10⁻⁴ J e) -I total Δω = ΔL = 2Icmωf - Icmωb ωf = -τΔt + Icmωb / 2Icm ωf = (-4 x 10⁻⁴ N-m)(0.2 sec) + (4.0 x 10⁻⁵ kg-m²)(140 rad/sec) / (2)(4.0 x 10⁻⁵ kg-m²) ωf = 69 rad/sec -τ total Δt = Icm(ωf - ωb) τ total = Icm(ωf - ωb) / Δt = (4 x 10⁻⁵ kg-m²)(69 rad/sec - 140 rad/sec) / 0.2 sec = -1.42 x 10⁻² J τ total = τ friction + τ dropped washer τ dropped washer = τ total - τ friction = -1.42 x 10⁻² J - 4 x 10⁻⁴ J = -1.38 x 10⁻² J Problem 2: (Bernoulli's Principle) 2 A water bottle of area A = 3.8 x 10⁻³ m² containing water at atmospheric pressure is sealed shut with an initial volume of air V₀ = 1.5 x 10⁻⁴ m³ and brought to the top of Mount Washington where the pressure is 0.88 atm. A hole is punched in the side of the bottle a distance h₀ = 10.0 cm below the water level. Assume the flow is steady out of the hole, and ignore any effects due to the hole. Note that 1 atm = 1.013 x 10⁵ N/m². a) What is the magnitude of the velocity of the efflux through the hole immediately after the hole is punched? b) Assume the air is an ideal gas, what is the air pressure in the bottle as a function of the height h of the water level above the hole? c) What is the magnitude of the velocity of the efflux through the hole as a function of the height h of the water level above the hole? 1 [Diagram of the water bottle with labels: Patm, h₀, h, P, hole] the pressure inside the bottle is atmosphere pressure P₀ = 1 atm = 1.013 x 10⁵ N/m² while the pressure outside the hole is the pressure atop Mt. Washington P₁ = 0.88 atm = 0.89 x 10⁵ N/m² Bernoulli's Principle states that P₀ + ρgh₀ + ½ρv₀² = P₁ + ρgh₁ + ½ρv₁² when the hole is first punched, v₀ ≈ 0, h₁ = 0 so P₀ + ρgh₀ = P₁ + ½ρv₁² v₁ = (2(P₀ - P₁) + 2gh₀)^(½) / ρ = ((2)(12 atm)(1.013x10⁵ Pa/atm) + (2)(9.8 m/s²)(0.1 m))^(½) / 10³ kg/m³ ≈ 5.1 m/s Once the hole is punched and the water level starts dropping, the pressure inside drops.

8.01X Practice Final Exam (Solutions)

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