Question

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (\(\textbf{Fig. E10.40}\)). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. (a) Is the angular momentum of the block conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

Short answer

(a) Yes, angular momentum is conserved. (b) New angular speed: 11.4 rad/s. (c) Change in KE: 0.06125 J. (d) Work done: 0.06125 J.

Step-by-step solution

Analyze Angular Momentum Conservation

Angular momentum (\(L\)) is given by \(L = I \cdot \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular speed. For a point mass, \(I = mr^2\). Because there is no external torque acting on the system, the angular momentum is conserved.Thus, \(L_{initial} = L_{final}\).

Calculate the Initial Angular Momentum

Using \(L_{initial} = mr^2 \omega\), with \(m = 0.0250\, \text{kg}\), \(r = 0.300\, \text{m}\), and \(\omega = 2.85\, \text{rad/s}\), we find:\[L_{initial} = 0.0250 \times (0.300)^2 \times 2.85 = 0.0064275\, \text{kg } \text{m}^2/\text{s}\]

Calculate the Final Angular Speed

Using \(L_{initial} = L_{final}\) and \(L_{final} = mr'^2 \omega'\), where \(r' = 0.150\, \text{m}\), solve for \(\omega'\):\[0.0064275 = 0.0250 \times (0.150)^2 \times \omega'\]\[\omega' = \frac{0.0064275}{0.0250 \times 0.0225} = 11.4\, \text{rad/s}\]

Compute Change in Kinetic Energy

Kinetic energy is given by \(K = \frac{1}{2} I \omega^2\). Calculating the initial and final kinetic energies:\[K_{initial} = \frac{1}{2} \times 0.0250 \times (0.300)^2 \times (2.85)^2 = 0.03081\, \text{J}\]\[K_{final} = \frac{1}{2} \times 0.0250 \times (0.150)^2 \times (11.4)^2 = 0.09206\, \text{J}\]The change in kinetic energy is:\[\Delta K = K_{final} - K_{initial} = 0.09206 - 0.03081 = 0.06125\, \text{J}\]

Determine Work Done on the System

The work-energy theorem states that the work done is equal to the change in kinetic energy. Therefore:\[W = \Delta K = 0.06125\, \text{J}\]

Key concepts

Understanding Kinetic Energy

Kinetic energy represents the energy that an object possesses due to its motion. Whenever an object is moving, it has kinetic energy, and this energy is calculated based on the object's mass and speed. The formula for calculating kinetic energy is:

• \(K = \frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the velocity of the object.

Kinetic energy is always a positive value, since it is a scalar, and it indicates the energy due to the motion of the object. In the context of circular motion, even though the object moves in a circle, its speed remains constant due to the conservation of angular momentum, causing variations in kinetic energy. By looking at the example of the block moving on a cord, we calculate its kinetic energy at different distances from the center to understand how changes in speed affect energy.
In the exercise, initially, the block revolves with a kinetic energy determined by its angular speed (\(2.85\, \text{rad/s}\)) and initial radius (\(0.300\, \text{m}\)). When the radius shortens to \(0.150\, \text{m}\), the block's speed increases to \(11.4\, \text{rad/s}\), leading to an increase in kinetic energy from \(0.03081\, \text{J}\) to \(0.09206\, \text{J}\). This change quantifies how much more energy the object has due to its increased speed after pulling the cord.

Exploring the Work-Energy Theorem

The work-energy theorem is a key concept in mechanics that states the net work done by forces on an object equals the change in its kinetic energy. Simply put, any work done on an object will result in a change in its kinetic energy. This can be expressed as:

• \(W = \Delta K\), where \(W\) is the work done and \(\Delta K\) is the change in kinetic energy.

In the exercise, as the cord pulls the block closer to the hole, work is done on the block. This directly relates to the amount of energy needed to change the block's speed as it moves in an increasingly tighter circle.
The change in kinetic energy, calculated earlier as \(0.06125\, \text{J}\), is exactly the amount of work done to pull the cord. This is because the work-energy relationship indicates that the work done should equal this change in energy, connecting motion to energy perspectives in a seamless manner. It highlights how energy is transferred into the system by an external force, in this case, pulling the cord.

The Role of Moment of Inertia

Moment of inertia, often termed the rotational inertia, extends the concept of mass to rotational systems. It determines how much resistance an object has to changing its rotational state. For a point mass moving in a circle, like the block in our exercise, the moment of inertia is given by:

• \(I = mr^2\), where \(m\) is the mass and \(r\) is the radius of the circle.

The farther the mass is from the axis of rotation, the greater the moment of inertia.
In actions like pulling the cord to change the radius of rotation, the moment of inertia decreases. This is because the block's radius is reduced, and since moment of inertia is directly proportional to the square of the radius, it plays a major role in the result that the angular speed increases to keep the product of \(I\) and the angular speed constant (conservation of angular momentum).
Understanding the concept of moment of inertia helps to see why changes in the radius lead to changes in angular speed and, consequently, kinetic energy. It showcases how rotational dynamics mirror linear dynamics, enhancing comprehension of circular motion phenomena.