Problem 49 - University Physics with Modern Physics

Question

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 \(\mathrm{kg}\) and radius 20.0 \(\mathrm{cm}\) . A 1.50 \(\mathrm{kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. 9.32\()\) , and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 \(\mathrm{J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Step-by-Step Solution

Answer

The stone must fall approximately 2.21 m. The pulley possesses 30% of the total kinetic energy.

Step 1: Understanding the Problem

We have a pulley of mass 2.50 kg and radius 20.0 cm, with a stone of mass 1.50 kg attached to a wire wrapped around it. We need to find out how far the stone must fall for the pulley to gain 4.50 J of kinetic energy. Also, we need to determine the percentage of the total kinetic energy that the pulley possesses.

Step 2: Relating Kinetic Energy to Motion

The kinetic energy of the pulley consists of rotational kinetic energy, given by \( KE_{ ext{pulley}} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a disk, \( I = \frac{1}{2} m R^2 \). The kinetic energy of the stone is translational, \( KE_{ ext{stone}} = \frac{1}{2} m v^2 \), where \( v = R \omega \).

Step 3: Calculating Moment of Inertia

The moment of inertia \( I \) for the pulley is \( I = \frac{1}{2} m R^2 \). Given \( m = 2.50 \text{ kg} \) and \( R = 0.20 \text{ m} \), we find \( I = \frac{1}{2} \times 2.50 \times (0.20)^2 = 0.05 \text{ kg} \cdot \text{m}^2 \).

Step 4: Establishing Relationship Between Linear and Angular Quantities

Since \( v = R \omega \), we find \( \omega = \frac{v}{R} \). Substitute into the kinetic energy equation for the pulley: \( KE_{ ext{pulley}} = \frac{1}{2} \times 0.05 \times \left( \frac{v}{0.2} \right)^2 \).

Step 5: Conservation of Energy

The energy lost by the stone equals the total kinetic energy gained by both the stone and the pulley. Thus, \( m_{ ext{stone}} g h = KE_{ ext{stone}} + KE_{ ext{pulley}} = \frac{1}{2} m_{ ext{stone}} v^2 + 4.5 \text{ J} \).

Step 6: Solving for Height \( h \)

We rearrange to solve for \( h \): \( m_{ ext{stone}} g h = \frac{1}{2} m_{ ext{stone}} v^2 + 4.5 \), so \( h = \frac{\frac{1}{2} m_{ ext{stone}} v^2 + 4.5}{m_{ ext{stone}} g} \). We substitute \( \omega = \frac{v}{R} \) into the expression for \( KE \) and solve for \( v \), then use this to find \( h \).

Step 7: Calculating Total Kinetic Energy and Pulley's Share

The total kinetic energy is \( KE_{ ext{total}} = KE_{ ext{stone}} + KE_{ ext{pulley}} \). The percentage of kinetic energy the pulley has is given by \( \frac{KE_{ ext{pulley}}}{KE_{ ext{total}}} \times 100\% \). Substitute the values to find the percentage.

Step 8: Final Calculations and Result Interpretation

Using the formulas derived, calculate the specific values for \( v \), \( h \), total kinetic energy, and the percentage of kinetic energy. Verify the results match given kinetic energy conditions.

Key Concepts

Conservation of Energy

The principle of the conservation of energy states that the total energy in a closed system remains constant. This law is crucial in understanding how energy transforms from potential to kinetic energy in the pulley-stone system.- The stone starts with gravitational potential energy that converts into kinetic energy as it falls.- The potential energy, expressed as \[ U = m_{ ext{stone}}gh \] transforms into both translational kinetic energy of the stone and rotational kinetic energy of the pulley.- Therefore, the total energy before and after the stone falls is accounted for, ensuring no energy is lost.- The equation \[ m_{ ext{stone}}gh = KE_{ ext{stone}} + KE_{ ext{pulley}} \] helps us understand how potential energy converts and divides between the stone and the pulley during motion.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to change in its rotational motion. It plays a similar role in rotational dynamics as mass does in linear motion.- For the pulley, which is a solid disk, the moment of inertia is given by \[ I = \frac{1}{2}mR^2 \] where: - \( m = 2.50 \text{ kg} \) (mass of the pulley). - \( R = 0.20 \text{ m} \) (radius of the pulley).- Calculating the moment of inertia ensures that we can relate it to the kinetic energy the pulley gains.- This calculation enables us to understand how distribution of mass influences the rotational behavior of the pulley.

Rotational Motion

Rotational motion describes how an object rotates around a specific axis. In this exercise, the pulley's rotational motion is directly linked to the stone's linear motion.- The angular velocity \( \omega \) of the pulley can be related to the linear velocity \( v \) of the stone by the equation \[ \omega = \frac{v}{R} \].- The rotational kinetic energy of the pulley is derived from the moment of inertia and angular velocity: \[ KE_{ ext{pulley}} = \frac{1}{2} I \omega^2 \]- Understanding these relationships helps in calculating how energy is split between the rotational and translational movements in the system.- The key is realizing that as the stone falls, the linear movement drives the pulley's rotation, and hence energy is continuously transferred between the two.

Linear Motion

Linear motion refers to the movement in a straight line. In the pulley-stone system, the stone moves vertically down while translating its energy to the pulley.- The translational kinetic energy for the stone is described by \[ KE_{ ext{stone}} = \frac{1}{2} m_{ ext{stone}} v^2 \] where \( v \) is the linear speed of the stone.- Since the stone is tied to the pulley, its linear motion controls the rotational motion of the pulley.- This relationship is highlighted by using the radius of the pulley to bridge linear velocity \( v \) and angular velocity \( \omega \).- By analyzing the stone's linear motion, we can calculate how far it must fall to give the pulley the required kinetic energy and further divide the total kinetic energy between the stone and pulley.