Problem 73 - University Physics with Modern Physics

Question

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 15.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During, the 15.0 -rad angular displacement of part (b), the kinetic energy \(y\) y of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J}\) . What is the moment of inertia of the wheel about the rotation axis?

Step-by-Step Solution

Answer

Tangential acceleration is 2.0 m/s² and moment of inertia is 0.208 kg·m².

Step 1: Analyze Radial Acceleration Change

The radial acceleration ar of a point on the wheel at a distance r from the axis is given by ar = ω²r, where ω is the angular velocity. If the angular velocity changes by Δω due to an angular acceleration α over a time interval, then the change in radial acceleration ∆a_r can be expressed as ∆a_r = (ω + ∆ω)²r - ω²r. Simplifying, we find that ∆a_r = (2ωΔω + (Δω)²)r. If ω changes over an angular displacement θ with constant angular acceleration α, Δω = αθ and thus ∆a_r = 2αθr.

Step 2: Calculate Tangential Acceleration

From part (a), we have ∆a_r = 2αθr. Given that r = 0.250 m, initial a_r is 25.0 m/s², final a_r is 85.0 m/s², and θ = 15.0 rad, substitute ∆a_r = 85.0 - 25.0 = 60.0 m/s² into 60.0 = 2α(15.0)(0.250). Solving for α gives α = 8.0 rad/s² and then using a_t = αr gives a_t = 8.0(0.250) = 2.0 m/s².

Step 3: Relate Kinetic Energy Change to Moment of Inertia

The change in kinetic energy ΔKE of the wheel is given by ΔKE = 0.5I(ω_f^2 - ω_i^2), where I is the moment of inertia and ω_i, ω_f are the initial and final angular velocities. Using constant angular acceleration, ω_f^2 - ω_i^2 = 2αθ. Therefore, ΔKE = Iαθ.

Step 4: Calculate Moment of Inertia

Using part (c), the moment of inertia I is found from ΔKE = Iαθ. Given that ΔKE = 45.0 - 20.0 = 25.0 J and using θ = 15.0 rad and α = 8.0 rad/s² from part (b), substitute into 25 = I(8.0)(15.0) to find I = 0.208 kg·m².

Key Concepts

Angular Acceleration

Angular acceleration is a key concept in rotational motion, similar to how linear acceleration applies to linear motion. It measures how quickly or slowly the angular velocity of an object is changing over time. Angular acceleration is denoted by the symbol \( \alpha \), and is usually measured in radians per second squared (\( \text{rad/s}^2 \)). For a wheel or object rotating around an axis, it’s defined as the rate of change of angular velocity. In our exercise, the angular acceleration indicates how the wheel's speed of rotation is increasing from one moment to the next. The formula is expressed as \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the change in time.

In cases of constant angular acceleration, predicting other motion parameters like angular displacement or final velocity becomes straightforward. For example, if you know an object’s angular acceleration and the time for which it accelerates, you can determine the change in angular velocity and eventual angular displacement.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, occurs when an object moves along a curved path, even if its speed remains constant. This acceleration is always directed towards the center of the path around which the object is rotating. In the exercise, the radial acceleration of a point on a wheel changes as the wheel spins. Radial acceleration's formula is \( a_r = \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the radius or the distance from the axis of rotation.

One important detail is that changes in angular velocity lead to changes in radial acceleration. The exercise shows us that if the angular velocity alters due to an angular acceleration over a displacement, the change in radial acceleration can be calculated as \( \Delta a_r = 2\alpha \theta r \). This formula helps us understand how radial acceleration varies with both angular acceleration and displacement.

Kinetic Energy

Kinetic energy in the context of rotational motion refers to the energy possessed by an object due to its motion around a point. For rotating objects, the kinetic energy is different from linear kinetic energy and depends on both the angular velocity and the object's moment of inertia. The formula for rotational kinetic energy is \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity of the rotating object.

The change in kinetic energy for a rotating object can be described as a function of the moment of inertia, angular acceleration, and the angular displacement, such that \( \Delta KE = I\alpha\theta \). The step-by-step solution illustrates this clearly by relating the changes in kinetic energy with the rotational properties of the wheel, providing insight into how these rotational variables interact during motion.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. It is the rotational equivalent of mass in linear motion. The greater the moment of inertia, the harder it is to change the object's rotational speed. Moment of inertia depends on the object's mass distribution relative to the axis of rotation and is denoted by \( I \).

The calculation of the moment of inertia is specific to the object's shape and the axis about which it rotates. For a point mass, it is calculated as \( I = mr^2 \), where \( m \) is mass and \( r \) is the distance from the axis. In our exercise, the moment of inertia is calculated as \( I = \frac{\Delta KE}{\alpha \theta} \). This allows us to understand how the energy stored in the system during rotation changes based on how the mass is distributed around the axis. Understanding it helps in solving problems involving rotational motion and energy changes.