Math Problem Statement

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.3 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.5 m/s in the opposite direction. Part A What is the angular speed of the gate just after it is struck by the unfortunate raven? Express your answer in radians per second.

Solution

To solve for the angular speed of the gate (ω\omega) just after it is struck by the raven, we will apply the principle of conservation of angular momentum.


Step 1: Determine the moment of inertia of the gate

The gate is a square solid object rotating about a horizontal axis through its upper edge. The moment of inertia for a square solid object rotating about this edge is given by:

I=13ML2I = \frac{1}{3} M L^2

where:

  • M=4.5kgM = 4.5 \, \mathrm{kg} (mass of the gate),
  • L=2.0mL = 2.0 \, \mathrm{m} (length of the side of the square).

Substituting:

I=13(4.5)(2.0)2=134.54.0=6.0kgm2I = \frac{1}{3} (4.5) (2.0)^2 = \frac{1}{3} \cdot 4.5 \cdot 4.0 = 6.0 \, \mathrm{kg \, m^2}


Step 2: Compute the angular momentum of the raven before and after collision

The angular momentum of the raven is calculated about the pivot point of the gate.

The angular momentum of a particle is:

L=mvdL = m v d

where:

  • m=1.3kgm = 1.3 \, \mathrm{kg} (mass of the raven),
  • vv is the velocity of the raven,
  • d=1.0md = 1.0 \, \mathrm{m} (distance from the pivot to the point of collision, which is the center of the gate).

Before collision:

The velocity of the raven is v=5.0m/sv = 5.0 \, \mathrm{m/s}, so the angular momentum is:

Lbefore=(1.3)(5.0)(1.0)=6.5kgm2/sL_{\text{before}} = (1.3)(5.0)(1.0) = 6.5 \, \mathrm{kg \, m^2/s}

After collision:

The velocity of the raven after collision is v=2.5m/sv = -2.5 \, \mathrm{m/s} (negative because it bounces back), so the angular momentum is:

Lafter=(1.3)(2.5)(1.0)=3.25kgm2/sL_{\text{after}} = (1.3)(-2.5)(1.0) = -3.25 \, \mathrm{kg \, m^2/s}

Net angular momentum change due to the raven:

The change in angular momentum imparted to the gate is:

ΔL=LbeforeLafter=6.5(3.25)=6.5+3.25=9.75kgm2/s\Delta L = L_{\text{before}} - L_{\text{after}} = 6.5 - (-3.25) = 6.5 + 3.25 = 9.75 \, \mathrm{kg \, m^2/s}


Step 3: Relating angular momentum to angular velocity

The angular momentum imparted to the gate is:

ΔL=Iω\Delta L = I \omega

Solving for ω\omega:

ω=ΔLI=9.756.0=1.625rad/s\omega = \frac{\Delta L}{I} = \frac{9.75}{6.0} = 1.625 \, \mathrm{rad/s}


Final Answer:

The angular speed of the gate just after it is struck is:

1.63rad/s\boxed{1.63 \, \mathrm{rad/s}}


Would you like further explanation or help with related concepts? Here are 5 related questions:

  1. How does the conservation of angular momentum apply to this system?
  2. What would happen if the gate's pivot had friction?
  3. How would the calculation change if the raven stuck to the gate instead of rebounding?
  4. What is the kinetic energy of the gate after the collision?
  5. How would the angular speed change if the gate were rectangular instead of square?

Tip: Always confirm whether external forces like friction or additional torques act on the system before assuming conservation laws.

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Math Problem Analysis

Mathematical Concepts

Angular Momentum
Rotational Motion
Moment of Inertia

Formulas

Moment of Inertia for a square solid object: I = (1/3)ML^2
Angular Momentum: L = mvd
Conservation of Angular Momentum: ΔL = Iω

Theorems

Conservation of Angular Momentum

Suitable Grade Level

Grades 11-12 or introductory college-level physics