Question

A torque of \(0.97 \mathrm{~N} \cdot \mathrm{m}\) is applied to a bicycle wheel of radius \(35 \mathrm{~cm}\) and mass \(0.75 \mathrm{~kg}\). Treating the wheel as a hoop, find its angular acceleration.

Short answer

The angular acceleration of the wheel is approximately 10.56 rad/s².

Step-by-step solution

Determine the Moment of Inertia

For a hoop, the moment of inertia \( I \) is calculated using the formula \( I = m r^2 \), where \( m = 0.75 \mathrm{~kg} \) is the mass and \( r = 0.35 \mathrm{~m} \) is the radius of the wheel after converting from centimeters to meters. Substitute the values: \( I = 0.75 \times (0.35)^2 = 0.091875 \mathrm{~kg} \cdot \mathrm{m}^2 \).

Calculate Angular Acceleration

Angular acceleration \( \alpha \) can be found using the equation \( \tau = I \alpha \), where \( \tau = 0.97 \mathrm{~N} \cdot \mathrm{m} \) is the torque and \( I \) is the moment of inertia. Rearrange the equation to solve for \( \alpha \): \( \alpha = \frac{\tau}{I} = \frac{0.97}{0.091875} \approx 10.56 \mathrm{~rad/s^2} \).

Key concepts

Torque

Torque is a concept that revolves around the rotational equivalent of linear force. It tells us how effectively a force can cause an object to rotate about an axis. When you apply torque to something like a bicycle wheel, you essentially twist it. This is much like using a wrench to tighten a bolt.

The formula to find torque \( \tau \) is \( \tau = F r \sin(\theta) \), where \( F \) is the applied force, \( r \) is the distance from the pivot point to where the force is applied (also known as the lever arm), and \( \theta \) is the angle between\ the force vector and the lever arm. However, when the force is perpendicular to the lever arm, \( \sin(\theta) \) becomes 1, simplifying the formula to \( \tau = F r \).

In our context, you might imagine pushing the edge of a wheel with a certain force at some distance from the center, effectively bending the wheel into motion. It's key to remember that the greater the distance or the stronger the force, the larger the torque.

Moment of Inertia

The moment of inertia, often denoted as \( I \), plays a similar role in rotational motion to what mass does in linear motion. It measures an object's resistance to any change in its rotational speed. For various shapes, the calculation of this value differs based on their mass distribution.

For a hoop, the moment of inertia is straightforward. The formula \( I = m r^2 \) tells us how the mass \( m \), and radius \( r \) of the hoop influence its inertial resistance. Here, all the mass is at a certain radius from the center, making this calculation very convenient. This distribution is a crucial factor because if the mass were spread differently, like in a solid disk, the equation would differ.

In our exercise, this calculation helps in understanding how the hoop's mass and radius affect its behavior when torque is applied. The calculated value of \( 0.091875 \mathrm{~kg} \cdot \mathrm{m}^2 \) tells us just how much the bicycle wheel resists the change in motion.

Hoop

A hoop is a simple physical shape, often visualized as a circular ring. In physics problems, it's important because it represents a common type of rigid body where mass is evenly distributed along a circular path. This characteristic distinctly affects its moment of inertia, making it different from other shapes like spheres or solids.

When you assume a wheel is like a hoop, you're implying that almost all the mass is concentrated at the circumference at a uniform radius from the center. This assumption simplifies calculations for rotational motion. For a hoop, this assumption means the formula for the moment of inertia is just \( I = m r^2 \), showcasing how its mass and radius shape its rotational characteristics.

This is especially useful when dealing with real-world objects like bicycle wheels, which, for small mass thicknesses, behave quite similarly to theoretical hoops in terms of their rotational dynamics.