Problem 78 - University Physics with Modern Physics

Question

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.45. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 m and moment of inertia for rotation about the axle of \(I =\) 0.480 kg \(\cdot\) m\(^2\). A small block with mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-\)9.00 J of work as the block descends 3.00 m. What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 m?

Step-by-Step Solution

Answer

The angular velocity of the wheel is approximately 3.12 rad/s.

Step 1: Understand the Given Values and Problem

We have a wheel with radius \( r = 0.180 \) m and moment of inertia \( I = 0.480 \) kg \( \cdot \) m\(^2\). A block of mass \( m = 0.340 \) kg descends 3.00 m, and friction does \(-9.00 \) J of work. We need the angular velocity of the wheel after the block descends.

Step 2: Calculate the Potential Energy Change

The change in potential energy \( \Delta U \) when the block descends is given by \( \Delta U = m g h \), where \( g = 9.81 \) m/s\(^2\) and \( h = 3.00 \) m. Hence, \( \Delta U = 0.340 \times 9.81 \times 3.00 \).

Step 3: Apply the Work-Energy Principle

The initial potential energy is converted into kinetic energy of the block and wheel, minus work done by friction. The relation is \( \Delta U = \Delta K_{block} + \Delta K_{wheel} + \text{work by friction} \).

Step 4: Write the Equations for Kinetic Energies

The kinetic energy of the block \( \Delta K_{block} = \frac{1}{2} m v^2 \) and for the wheel \( \Delta K_{wheel} = \frac{1}{2} I \omega^2 \). The velocities are related by \( v = r \omega \).

Step 5: Solve for Block's Linear Velocity

Substitute \( v = r \omega \) into \( \Delta K_{block} \) and combine with previous equations to write \( m g h - 9.00 = \frac{1}{2} m (r \omega)^2 + \frac{1}{2} I \omega^2 \).

Step 6: Isolate \( \omega \) and Solve

Rearrange to \( m g h - 9.00 = \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} I \omega^2 \). Combine terms: \( \omega^2 = \frac{2(m g h - 9.00)}{m r^2 + I} \).

Step 7: Calculate Angular Velocity

Substitute values: \( \omega = \sqrt{ \frac{2(0.340 \times 9.81 \times 3.00 - 9.00)}{0.340 \times 0.180^2 + 0.480} } \). Compute \( \omega \).

Step 8: Conclude with \( \omega \) Value

After calculations, \( \omega \approx 3.12 \) rad/s. The magnitude of the angular velocity of the wheel is \( 3.12 \) rad/s.

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept in rotational dynamics. It is analogous to mass in linear motion, indicating how much torque is needed for a desired angular acceleration. Moment of inertia ( I ) depends on the mass distribution relative to the axis of rotation. In this exercise, we have a wheel with a specified moment of inertia of 0.480 kg · m ² . This means that the resistance of the wheel to any change in its rotational motion is determined by this value. The greater the moment of inertia, the more difficult it is to change the object's rotational state. In practical terms, if two wheels have the same size but different mass distributions, the one with more mass spread farther from the axis will have a higher moment of inertia. This concept helps in calculating the angular velocity by integrating with other principles like energy transformation.

Work-Energy Principle

The work-energy principle is essential in understanding how energy changes forms but is conserved within a system. It states that the work done on a system is equal to the change in its kinetic energy. In this problem, the wheel's friction does -9.00 J of work, which affects the energy conservation during the block’s descent. Using the principle, we equate the change in potential energy to the work done by friction and the change in kinetic energy of both the wheel and block. This allows us to understand how the system's energy evolves and aids in solving for unknowns, like angular velocity, by balancing all energy changes. It assures us that the total mechanical energy, minus work done by friction, gets distributed between the kinetic energies of the wheel and the block.

Potential Energy

Potential energy is stored energy due to an object's position, generally in a gravitational field. In this case, when the block hangs from the wheel, it possesses gravitational potential energy determined by its height above the ground level. The potential energy ( U ) is quantified by the equation U = mgh , where m is mass, g is gravitational acceleration, and h is the height. As the block descends 3.00 meters, its potential energy decreases, converting this energy into other forms. This decrease in potential energy drives the motion of the wheel, culminating in mechanical work, specifically rotational kinetic energy, while considering the energy lost due to friction, which is how the work-energy principle is applied.

Kinetic Energy

Kinetic energy is the energy of motion, applicable to both linear and rotational movements. For linear motion, like that of the falling block, kinetic energy is defined as \( rac{1}{2} mv^2 \), where m is mass and v is velocity.For rotational motion, shown by the wheel, it's described by \( rac{1}{2} I \omega^2 \), indicating the influence of both the angular velocity \( \omega \)and the moment of inertia \( I \).In this exercise, when the block descends, the system’s potential energy shifts substantially to these kinetic energy forms. The relationship between linear and angular velocities is given by v = r\omega, ensuring interconnected transformations.Solving the exercise involves equating the potential energy change to combined kinetic energies, showing how various energies interplay under the influence of the work-energy principle.