Math Problem Statement

A force F is applied over time t to a rotating rigid body with rotational inertia I_0. This force is applied at a fixed moment arm of distance R from an axis of rotation about the rotating object’s center of mass. The applied force as a function of time can be expressed as F(t) = βt^2 + γ, where β and γ are constants with appropriate units. You may assume negligible due to the rotational axis and non-slipping conditions. Derive expressions for the object’s angular impulse, angular momentum after 2 seconds, and various rotational and translational quantities after 3 seconds.

Solution

Let's go through each part of this problem.

  1. Problem Summary:
    • A force F(t)=βt2+γF(t) = \beta t^2 + \gamma is applied to a rotating rigid body with rotational inertia I0I_0, at a fixed moment arm RR from the axis of rotation.
    • We need to derive expressions for the angular impulse, determine the angular momentum after a specific time, and calculate quantities like angular speed, translational speed, and instantaneous power given certain initial conditions.

Part (a): Deriving Angular Impulse

  • Angular Impulse JrotJ_{\text{rot}} over a time interval tt is given by: Jrot=0tτ(t)dtJ_{\text{rot}} = \int_0^t \tau(t) \, dt where τ(t)\tau(t) is the torque applied by the force.
  • Torque τ(t)\tau(t) can be calculated as: τ(t)=RF(t)=R(βt2+γ)\tau(t) = R \cdot F(t) = R (\beta t^2 + \gamma)
  • Substituting τ(t)\tau(t) into the integral: Jrot=0tR(βt2+γ)dtJ_{\text{rot}} = \int_0^t R (\beta t^2 + \gamma) \, dt
  • Solving this integral: Jrot=R(0tβt2dt+0tγdt)J_{\text{rot}} = R \left( \int_0^t \beta t^2 \, dt + \int_0^t \gamma \, dt \right) After evaluating, this results in: Jrot=R(βt33+γt)J_{\text{rot}} = R \left( \frac{\beta t^3}{3} + \gamma t \right)

Part (b): Determining Angular Momentum at t=2t = 2 seconds

  • Angular momentum LL can be found using the relationship L=JrotL = J_{\text{rot}} (if the object starts from rest).
  • Substitute t=2t = 2 into the expression derived in Part (a): L=R(β(2)33+γ2)L = R \left( \frac{\beta (2)^3}{3} + \gamma \cdot 2 \right)

Part (c): Calculations at t=3t = 3 seconds

  1. Angular Speed:

    • Given L=IωL = I \omega, solve for ω\omega: ω=LI\omega = \frac{L}{I}
    • Use JrotJ_{\text{rot}} at t=3t = 3 for LL.
  2. Translational Speed at Location of Applied Force:

    • Translational speed v=Rωv = R \cdot \omega.
  3. Instantaneous Power after t=3t = 3 seconds:

    • Power P=τωP = \tau \cdot \omega.
    • Substitute τ(t)=R(βt2+γ)\tau(t) = R(\beta t^2 + \gamma) and ω\omega from previous calculations.

Would you like me to proceed with detailed calculations for each part?

Here are 5 questions for deeper understanding:

  1. How is angular impulse related to torque in rotational dynamics?
  2. Why is the moment arm important when calculating torque?
  3. How does starting from rest affect angular momentum calculations?
  4. What are the steps to compute instantaneous power in rotational systems?
  5. How does rotational inertia impact angular speed?

Tip: Always keep track of units, especially with time-dependent forces and torques, to ensure consistent results.

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Math Problem Analysis

Mathematical Concepts

Rotational dynamics
Angular impulse
Torque
Angular momentum
Translational speed
Instantaneous power

Formulas

Torque τ = R * F(t)
Angular impulse J_rot = ∫ τ(t) dt
Angular momentum L = J_rot if initial angular momentum is zero
Angular speed ω = L / I
Translational speed v = R * ω
Instantaneous power P = τ * ω

Theorems

Angular impulse-momentum theorem
Newton’s second law for rotation

Suitable Grade Level

College Level (Physics)