Problem 43 - University Physics with Modern Physics

Question

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley (\(\textbf{Fig. E9.43}\)), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Step-by-Step Solution

Answer

(a) The stone must fall 0.167 m. (b) The pulley has 45.6% of the total kinetic energy.

Step 1: Identify Known Quantities

The mass of the pulley, \(m_p = 2.50 \text{ kg}\). The radius of the pulley, \(r = 0.20 \text{ m}\). The mass of the stone, \(m_s = 1.50 \text{ kg}\). The initial kinetic energy is \(0\) J because the system starts from rest.

Step 2: Understand Total Kinetic Energy of the System

The total kinetic energy of the system is given as \(4.50 \text{ J}\). This combines the translational kinetic energy of the stone and the rotational kinetic energy of the pulley.

Step 3: Expression for Kinetic Energy

The total kinetic energy \(K\) for the system is the sum of the translational kinetic energy of the stone and the rotational kinetic energy of the pulley:\[ K = \frac{1}{2} m_s v^2 + \frac{1}{2} I \omega^2 \] where \(v\) is the speed of the stone and \(\omega\) is the angular velocity of the pulley.

Step 4: Relate Angular and Linear Velocity

The angular velocity \(\omega\) is related to the linear velocity \(v\) by the equation \(\omega = \frac{v}{r}\). Substitute \(\omega\) into the expression for rotational kinetic energy:\[ \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{2} m_p r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{4} m_p v^2 \]

Step 5: Substitute Back to Total Kinetic Energy Equation

Substitute \(\frac{1}{4} m_p v^2\) for the rotational kinetic energy into the total kinetic energy equation:\[ 4.50 = \frac{1}{2} m_s v^2 + \frac{1}{4} m_p v^2 \]Combine terms:\[ = \left( \frac{1}{2} m_s + \frac{1}{4} m_p \right) v^2 \]

Step 6: Solve for Velocity \(v\)

Solve for \(v\):\[ 4.50 = \left( \frac{1}{2} \times 1.50 + \frac{1}{4} \times 2.50 \right) v^2 \]\[ 4.50 = (0.75 + 0.625) v^2 \]\[ 4.50 = 1.375 v^2 \]\[ v^2 = \frac{4.50}{1.375} \]\[ v = \sqrt{\frac{4.50}{1.375}} \approx 1.81 \text{ m/s}\]

Step 7: Calculate Distance Fallen

Use the kinematic equation relating velocity, acceleration due to gravity \(g\), and distance \(d\):\[ v^2 = 2gd \]Solving for \(d\):\[ d = \frac{v^2}{2g} = \frac{1.81^2}{2 \times 9.8} \approx 0.167 \text{ m}\]

Step 8: Calculate Percentage of Total Kinetic Energy for the Pulley

The kinetic energy of the pulley \(K_p\) is given by:\[ K_p = \frac{1}{4} m_p v^2 \]Substitute the known values:\[ K_p = \frac{1}{4} \times 2.50 \times 1.81^2 \approx 2.05 \text{ J}\]The percentage of the pulley's energy to total energy:\[ \text{Percentage} = \left( \frac{2.05}{4.50} \right) \times 100\% \approx 45.6\% \]

Key Concepts

Rotational Motion

Rotational motion is a concept you encounter in systems where objects spin around an axis. In this exercise, the pulley represents a body that undergoes rotational motion. Unlike linear motion, rotational motion concerns quantities like angular velocity \( \omega \), which describes how fast something spins.

• **Angular velocity** \( \omega \) is related to linear velocity \( v \) through the radius \( r \) of the pulley: \( \omega = \frac{v}{r} \).
• The pulley's mass and radius are crucial to determining its ability to rotate since they influence its moment of inertia \( I \).

Understanding rotational motion helps in analyzing systems with rotating parts, like this one where a pulley determines how a stone moves downwards. This kind of motion is crucial in real-world applications ranging from engines to amusement park rides.

Kinetic Energy

Kinetic energy is the energy an object has due to its motion. In this exercise, we focus on two forms: translational kinetic energy and rotational kinetic energy.

• **Translational kinetic energy** for the falling stone is given by: \[ KE_{trans} = \frac{1}{2} m_s v^2 \]
• **Rotational kinetic energy** for the pulley is expressed as: \[ KE_{rot} = \frac{1}{2} I \omega^2 \]

For the pulley, we convert \( \omega \) using \( \omega = \frac{v}{r} \), resulting in a term where the rotational energy depends on \( v^2 \) and other parameters of the pulley, like mass \( m_p \) and radius \( r \). Understanding how these two energies combine is critical; the total kinetic energy of this system is the sum: \[ K = \frac{1}{2} m_s v^2 + \frac{1}{4} m_p v^2 \].This equation allows us to solve for the speed \( v \) and analyze the stone's motion as well as the energy dynamics in a frictionless system.

Frictionless Systems

In a frictionless system, energy transformations occur without the loss of energy due to heat or resistance. Theoretically, these systems allow us to focus solely on conversion between potential and kinetic energy.

• Without friction, the only forces acting are due to gravity, causing objects like the stone to fall.
• This exercise has no energy loss due to friction, which simplifies calculations and ensures that all potential energy converts to kinetic energy.

The benefit of considering such idealized systems is the ability to predict outcomes with greater accuracy in an educational sense. These principles apply, for example, in systems designed to minimize energy loss in engineering contexts, enhancing efficiency.