Question

A generator's flywheel, which is a homogeneous cylinder of radius \(R\) and mass \(M\), rotates about its longitudinal axis. The linear velocity of a point on the rim (side) of the flywheel is \(v\). What is the kinetic energy of the flywheel? a) \(K=\frac{1}{2} M v^{2}\) b) \(K=\frac{1}{4} M v^{2}\) c) \(K=\frac{1}{2} M v^{2} / R\) d) \(K=\frac{1}{2} M v^{2} R\) e) not given enough information to answer

Short answer

Answer: b) \(K=\frac{1}{4} M v^{2}\)

Step-by-step solution

Determine the moment of inertia for the cylinder

First, let's find the moment of inertia for the homogeneous cylinder. For a solid cylinder rotating about its longitudinal axis, the moment of inertia (I) can be calculated as follows: \(I = \frac{1}{2}MR^2\)

Convert linear velocity to angular velocity

Next, we need to convert the given linear velocity (v) of a point on the rim of the flywheel to its angular velocity (ω). This can be done by using the relation between linear and angular velocity: \(v = Rω\) Now, we can solve for angular velocity: \(ω = \frac{v}{R}\)

Calculate the kinetic energy using moment of inertia and angular velocity

Now, we can calculate the kinetic energy (K) of the flywheel using the following expression: \(K = \frac{1}{2}Iω^2\) Plug in the moment of inertia and angular velocity expressions into the equation: \(K = \frac{1}{2}\cdot\frac{1}{2}MR^2 \cdot \left(\frac{v}{R}\right)^2\) After simplifying the expression, we get the formula for the kinetic energy: \(K = \frac{1}{4}Mv^2\) Therefore, the correct answer is: b) \(K=\frac{1}{4} M v^{2}\)

Key concepts

Moment of Inertia

When evaluating the rotational motion of an object, the moment of inertia plays a crucial role, analogous to mass in linear motion. It is a measure of an object's resistance to changes in its rotational motion. The higher the moment of inertia, the harder it is to spin the object. The formula for the moment of inertia depends on the object's shape and the axis about which it is rotating. For common shapes, such as the homogeneous cylinder in the given exercise, the moment of inertia about its longitudinal axis is \(I = \frac{1}{2}MR^2\).

This expression indicates that the cylinder’s mass (M) and the square of its radius (R) proportionately affect its resistance to rotational acceleration. The moment of inertia is a scalar quantity and has units of kilograms times meters squared (kg\(\cdot\)m^2). In practice, knowing the moment of inertia allows us to predict how an object will behave when rotational forces are applied, which is crucial for understanding machines like the generator’s flywheel from our problem.

Angular Velocity

Angular velocity is a measure of the rate of rotation, describing how fast an object spins around a fixed axis. It tells us how many radians (the standard unit for measuring angles) the object rotates through per second. In the context of our exercise, angular velocity (\(\omega\)) is related to the linear velocity (v) of a point on the flywheel's rim by the simple relationship \(v = R\omega\).

This equation implies that the linear velocity at the rim of the flywheel is equal to the product of the radius and the angular velocity. By rearranging the terms, we can solve for angular velocity: \(\omega = \frac{v}{R}\), allowing us to transition from linear motion to rotational dynamics. It is vital to understand angular velocity when calculating rotational kinetic energy or examining the dynamics of rotating systems.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by a rotating object due to its motion. It is directly related to both the moment of inertia and the angular velocity of the object. In our textbook's example of a generator's flywheel, the rotational kinetic energy (K) is calculated using the formula \(K = \frac{1}{2}I\omega^2\).

The formula encapsulates how rotational kinetic energy increases with both a greater moment of inertia and a higher angular velocity. Similar to kinetic energy in linear motion, which depends on mass and velocity squared, rotational kinetic energy depends on the moment of inertia and the square of angular velocity. When we substitute the given values of moment of inertia and angular velocity of the flywheel into the kinetic energy formula, we find that \(K = \frac{1}{4}Mv^2\), indicating that the kinetic energy of the flywheel is proportional to the mass and the square of the linear velocity at its rim. This explanation helps us understand why option (b) provides the correct evaluation of the kinetic energy for the flywheel problem.