Problem 6 - College Physics

Question

\(\cdot\) A cord is wrapped around the rim of a wheel 0.250 \(\mathrm{m}\) in radius, and a steady pull of 40.0 \(\mathrm{N}\) is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 5.00 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Compute the angular acceleration of the wheel.

Step-by-Step Solution

Answer

The angular acceleration of the wheel is 2.0 rad/s².

Step 1: Identify the given values

The radius of the wheel is 0.250 m, the force applied is 40.0 N, and the moment of inertia is 5.00 kg \(\cdot\) m\(^2\). We need to find the angular acceleration \(\alpha\).

Step 2: Relate torque to force and radius

Calculate the torque \(\tau\) exerted by the force using the formula \(\tau = F \cdot r\), where \(F\) is the force and \(r\) is the radius. Substitute \(F = 40.0\, \mathrm{N}\) and \(r = 0.250\, \mathrm{m}\) to find \(\tau = 40.0 \times 0.250 = 10.0\, \mathrm{N} \cdot \mathrm{m}\).

Step 3: Use the rotational dynamics equation

The relationship between torque and angular acceleration is given by \(\tau = I \cdot \alpha\), where \(I\) is the moment of inertia. We can rearrange this equation to solve for angular acceleration: \(\alpha = \frac{\tau}{I}\).

Step 4: Calculate the angular acceleration

Substitute the values for torque \(\tau = 10.0\, \mathrm{N} \cdot \mathrm{m}\) and moment of inertia \(I = 5.00\, \mathrm{kg} \cdot \mathrm{m}^{2}\) into the equation \(\alpha = \frac{\tau}{I}\) to find \(\alpha = \frac{10.0}{5.00} = 2.0\, \mathrm{rad/s}^{2}\).

Key Concepts

Torque

Torque is a key concept in rotational dynamics. It can be thought of as the 'twisting force' that causes an object to rotate. In simpler terms, torque depends on how much force you apply to an object and where you apply that force relative to its pivot point. Let's break this down:

• The formula for torque is given by \( \tau = F \cdot r \), where \( F \) is the force applied, and \( r \) is the distance from the pivot point or axis of rotation.
• This means the further you apply the force from the axis, the more torque you generate.
• In the context of our problem, the force applied on the cord creates a torque around the wheel's center, which makes it rotate.

Understanding how torque works helps in solving problems related to how things spin, like wheels, gears, or even planets!

Moment of Inertia

The Moment of Inertia is sometimes called the rotational inertia. It is a measure of an object's resistance to changes in its rotational motion.Just like mass is inertia in linear motion, the moment of inertia plays a similar role in rotational settings.

• It's denoted by \( I \) and has units of kg \( \cdot \) m².
• It depends on how an object's mass is distributed relative to its axis of rotation. More mass further from the axis means a larger moment of inertia.
• In our example, the wheel has a moment of inertia of \( 5.00 \ \text{kg} \cdot \text{m}^2 \).

The concept helps us understand how easy or hard it is to change an object's rotation speed. It is crucial in analyzing everything from flywheels in engines to toys your child might play with.

Rotational Dynamics

Rotational Dynamics deals with forces and motions revolving around rotational movement. At its core, it revolves around using Newton's laws of motion in a rotational context.For an object to change how fast it spins, it needs an angular acceleration and a torque acting on it.

• Angular acceleration (\( \alpha \)) is the rate of change of rotational velocity. It's how quickly something speeds up or slows down its spinning.
• In our problem, the relationship between torque and angular acceleration is given by \( \tau = I \cdot \alpha \). This is similar to \( F = m \cdot a \) in linear dynamics.
• The solution involves calculating the angular acceleration using the known torque and moment of inertia.
• This gives us \( \alpha = \frac{\tau}{I} = \frac{10.0}{5.00} = 2.0 \ \text{rad/s}^2 \).

Understanding rotational dynamics helps us solve practical problems involving anything that spins—from simple wheels to more complicated machinery.