More Problems

Problem 1:

In a demonstration, a bicycle wheel with moment of inertia I_wheel = 0.37 kg*m² is spun up to rotate about a vertical axis with ω = 14 rad/s k.  A wheel is handed to a student on a rotatable platform by a person standing directly in front of the student.  The student and the platform are initially stationary and have a moment of inertia equal to 3.6 kg*m².  The student reaches up and very quickly stops the rotation of the wheel.  He holds on to the rim with his hand.  How long will it take before the student faces forward again?

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  After the wheel has been handed to the student, the system consists of the student and the wheel.  No external torques act on the system about the vertical axis, so the total angular momentum of the system is about the vertical axis conserved.
• Details of the calculation:
  The wheel receives an angular impulse, so the student and the platform receive an angular impulse of equal magnitude and opposite direction.
  L = Iω.
  L_before = (0.37 kg*m²)(14/s) = 5.18 kgm²/s.
  Since the student holds on to the rim with his hand, the student, the platform, and the wheel all rotate together.
  L_after = (0.37 kg*m² + 3.6 kg*m²)ω = 5.18 kgm²/s.
  ω = 1.3 rad/s k.
  ω = dθ/dt.  t = (2π/1.3)s = 4.8 s.
  After 4.8 s the student will face forward again.

Problem 2:

A truck is traveling on a level road.  The driver suddenly applies the brakes, causing the truck to decelerate by an amount 3m/s².  There is a 90 kg box in the back of the truck.  The coefficient of sliding friction between the truck and the box is 0.2.  Find the acceleration of the box relative to
(a)  the truck and
(b)  the road.

Solution:

• Concepts:
  Motion in an accelerating frame.
• Reasoning:
  In an accelerating frame fictitious forces appear.  The net force in such a frame is
  F = F_inertial - ma, where a is the acceleration of the frame.
• Details of the calculation:
  This is a one-dimensional problem.  Let the positive direction be the direction of the trucks initial velocity.
  (a)  F = -0.2m(9.8 m/s²) + m(3m/s²) = +m(1.04 m/s²).  The acceleration of the box relative to the truck is 1.04 m/s². 
  (b)  F = -0.2m(9.8 m/s²).  The acceleration of the box relative to the road is -1.96 m/s².

Problem 3:

A sphere has a mass m with uniform density ρ and radius a.  Calculate the moment of inertia about an axis through its center.  Use spherical coordinates and do not just write down the answer, show your calculation.

Solution:

• Concepts:
  The moment of inertia
• Reasoning:
  The moments of inertia about all axes through the center are the same.
  The moment of inertia about the z-axis is
  I_z = ∫_VdV ρ(x² + y²) = ρ∫_Vdx ∫dy ∫dz (x² + y²).
• Details of the calculation:
  We are integrating over the volume of a sphere.
  Change variables to spherical coordinates.
  x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ.
  I = ρ∫₀^ar²dr ∫₀^πsinθdθ ∫₀^2πdφ (r²sin²θcos²φ + r²sin²θsin²φ)
  = ρ∫r⁴dr ∫sin³θdθ ∫dφ (cos²φ + sin²φ)
  = ρ (a⁵/5)   (4/3)      2π.
  For a sphere we have ρ = M/[(4π/3)a³].  Therefore I = 2Ma²/5.

Problem 4:

A small airplane (M = 990 kg) comes in for a landing at a speed of v₀ = 35 m/s.  Before touching the ground, the wheels are not rotating.  How long a skid mark do the two wing wheels leave if the mass of each wheel is m = 25 kg distributed uniformly, the diameter of each wheel is d = 0.5 m, and the coefficient of friction with the ground is μ = 0.5?

Solution:

• Concepts:
  Rolling
• Reasoning:
  The tires will leave a skid mark if ωR < v. Here ω is the angular speed of the tires, R is the radius of the tires, and v is the linear speed of the plane.  The tires will start rolling when ω = v/R.  Until rolling sets in, the force of kinetic friction decelerates the plane and produces a torque which increases ω.
• Details of the calculation:
  
  
  Before the wheels start rolling:
  The normal force on each wheel is equal to the weight each wheel has to support,
  N = W_w = 0.5 Mg = 4851 N.
  The magnitude of the force of kinetic friction on each wheel is f = μN = μW_w = 2425.5 N.
  (Assume μ_k = μ.)
  The magnitude of the acceleration of the plane is a = 2f/M = 2μW_w/M = 0.5 g = 4.9 m/s².
  Its speed as a function of time is v(t) = v₀ - at.
  The magnitude of the torque on each wheel is τ = f*d/2 = μW_wd/2 = 606.375 Nm.
  The magnitude of the angular acceleration of each wheel is
  α = τ/I = τ/(0.5mR²) = 8τ/(md²) = 776.16/s².
  The angular speed of each wheel as a function of time is ω(t) = αt.
  
  When  v₀ - at = α(d/2)t, or t = 2v₀/(αd + 2a) = 0.176 s the wheels will start rolling.
  In the time interval between 0 and t the plane skids for a distance s = v₀t - 0.5at².
  The length of the skid marks is 6.1 m.

Problem 5:

A uniform rod of length b and mass m stands vertically upright on a rough floor and tips over.  What is the rod's angular speed just before it hits the floor?

Solution:

• Concepts:
  Energy conservation
• Reasoning:
  Gravity is a conservative force.  We have pure rotation about a point.
• Details of the calculation:
  ½Iω² = mgb/2.  I = mb²/3.
  ω = (3g/b)^½.