Problem 40 - Fundamentals of Physics

Question

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a time-dependent torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m} .\) At time \(t=1.00 \mathrm{~s},\) its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

The angular momentum at \(t=3.00\,\text{s}\) is \(23.00\,\text{kg m}^2/\text{s}\).

Step 1: Understand the Problem

We need to find the angular momentum of a disk at a certain time given a time-dependent torque and initial angular momentum. The disk's rotational inertia is also given.

Step 2: Use the Relationship between Torque and Angular Momentum

The relationship between torque \(\tau(t)\) and angular momentum \(L(t)\) is given by the equation \(\tau(t) = \frac{dL(t)}{dt}\). We can integrate this to find the change in angular momentum over time.

Step 3: Integrate the Torque to Find Change in Angular Momentum

Integrate the torque function \(\tau(t) = (5.00 + 2.00t)\) with respect to time from \(t_1 = 1.00\, s\) to \(t_2 = 3.00\, s\):\[\Delta L = \int_{1}^{3} (5.00 + 2.00t) \, dt.\]

Step 4: Calculate the Integral

Compute the integral:\[ \Delta L = \left[ 5.00t + \frac{2.00}{2} t^2 \right]_{1}^{3} = \left[ 5.00(3) + 1(3)^2 \right] - \left[ 5.00(1) + 1(1)^2 \right]= (15.00 + 9.00) - (5.00 + 1.00)= 24.00 - 6.00 = 18.00 \, \text{kg m}^2/\text{s}.\]

Step 5: Apply Initial Angular Momentum

Add the change in angular momentum to the initial angular momentum to find \(L(3)\):\[L(3) = L(1) + \Delta L = 5.00 \, \text{kg m}^2/\text{s} + 18.00 \, \text{kg m}^2/\text{s} = 23.00 \, \text{kg m}^2/\text{s}.\]

Step 6: Confirm the Units

Ensure that all calculations are consistent in terms of units, using \(\text{kg m}^2/\text{s}\) for angular momentum throughout the steps.

Key Concepts

Rotational Inertia

Rotational inertia, also known as the moment of inertia, is a fundamental concept in rotational dynamics. It is the rotational analog of mass for linear motion. This quantity determines how much torque is needed for a desired angular acceleration, analogous to how mass affects the force needed for linear acceleration. The rotational inertia of an object depends on its mass distribution. For example:

• A solid disk has a different rotational inertia compared to a ring of the same mass and radius.
• Rotational inertia increases if the mass is distributed further from the axis of rotation.

In the original exercise, we deal with a disk that has a rotational inertia of 7.00 kg·m². This directly affects how the disk responds to applied torque, dictating the changes in its angular speed.

Torque

Torque is the rotational equivalent of linear force. It is what causes an object to rotate and is expressed in units of Newton meters (N·m). Torque takes into account not just the force applied, but also the distance from the pivot point where this force is applied. This is crucial in determining how effective a force is at causing rotational motion.

For instance:

• A small force at a larger radius can generate the same torque as a larger force closer to the axis.
• The direction of the force also matters, as it needs to be perpendicular to the radius to produce optimal torque.

In our exercise, the disk experiences a time-dependent torque, \[tau(t) = (5.00 + 2.00t) \]which changes as time varies. Understanding torque helps us predict how the angular momentum of an object like the disk evolves over time.

Integration

Integration in physics is often used to find a quantity that accumulates over time, such as distance, area, or, as in our case, angular momentum. When we encounter a situation where the rate of change of a quantity is known, as given by a function of time, integration allows us to find the actual change or total value of that quantity over a specified period.

In the original problem:

• We use integration to find how angular momentum changes with a time-dependent torque.
• The torque function \( \tau(t) = (5.00 + 2.00t) \) is integrated over the time interval from 1.00 s to 3.00 s.
• This integration provides the change in angular momentum, \( \Delta L \), during this period.

Calculating this integral allows us to accurately determine the total angular momentum at any given time by adding this change to the initial known value.