Question

In an Atwood's machine one block has a mass of \(512 \mathrm{~g}\) and the other a mass of \(463 \mathrm{~g}\). The pulley, which is mounted in horizontal frictionless bearings, has a radius of \(4.90 \mathrm{~cm}\). When released from rest, the heavier block is observed to fall \(76.5 \mathrm{~cm}\) in \(5.11\) s. Calculate the rotational inertia of the pulley.

Short answer

The rotational inertia of the pulley is \(0.0198 \mathrm{~kg \cdot m^2}\).

Step-by-step solution

Calculate the Acceleration of the System

Given that the heavier block falls \(76.5 \mathrm{~cm}\) in \(5.11\) s from rest, the acceleration \(a\) of the system can be calculated using the equation of motion \(s = ut + \frac{1}{2} a t^2\), where \(s\) is the displacement (\(76.5 \mathrm{~cm} = 0.765 \mathrm{~m}\)), \(u\) is the initial velocity (which is zero, as the blocks are released from rest), \(a\) is the acceleration, and \(t\) is the time interval (\(5.11 \mathrm{~s}\)). Solving for \(a\) gives \(a = \frac{2s}{t^2} = \frac{2 \times 0.765 \mathrm{~m}}{(5.11 \mathrm{~s})^2}= 0.0583 \mathrm{~m/s^2}\).

Calculate the Net Force Acting on the System

The net force \(F_{net}\) acting on the system is the difference of the forces due to gravity on the two blocks, which is given by \(F_{net} = (m1 - m2)g\), where \(m1\) and \(m2\) are the masses, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). Substituting the given values (\(m1 = 512 \mathrm{~g} = 0.512 \mathrm{~kg}\), \(m2 = 463 \mathrm{~g} = 0.463 \mathrm{~kg}\)) yields \(F_{net} = (0.512 \mathrm{~kg} - 0.463 \mathrm{~kg}) \times 9.81 \mathrm{~m/s^2} = 0.480 \mathrm{~N}\).

Equate Net Torque to the Product of Rotational Inertia and Angular Acceleration

The net torque \(\tau\) on the pulley due to the net force is \(F_{net} r\), where \(r\) is the radius of the pulley (\(4.90 \mathrm{~cm} = 0.049 \mathrm{~m}\)). Therefore, \(\tau = 0.480 \mathrm{~N} \times 0.049 \mathrm{~m}= 0.02352 \mathrm{~Nm}\). By Newton's second law for rotation, \(\tau = I \alpha\), where \(I\) is the rotational inertia and \(\alpha\) is the angular acceleration. The angular acceleration can be calculated from the linear acceleration as \(\alpha = \frac{a}{r} = \frac{0.0583 \mathrm{~m/s^2}}{0.049 \mathrm{~m}} = 1.1898 \mathrm{~rad/s^2}\). Now, it's possible to solve for \(I\) by rearranging the equation to \(I = \frac{\tau}{\alpha}= \frac{0.02352 \mathrm{~Nm}}{1.1898 \mathrm{~rad/s^2}}= 0.0198 \mathrm{~kg \cdot m^2}\).

Conclusion

The rotational inertia of the pulley, given the parameters of the Atwood's machine setup and the observed motion, is thus \(0.0198 \mathrm{~kg \cdot m^2}\).

Key concepts

Atwood's Machine

An Atwood's machine is a device composed of two masses connected by a string that passes over a pulley. When the masses are different, gravity causes the heavier mass to fall and the lighter mass to rise, converting potential energy into motion. This apparatus is commonly used in physics to study the principles of mechanics and to measure the acceleration due to gravity effectively. The pulley in an Atwood's machine is ideally frictionless and massless; however, when considering real-life applications or more advanced problems, the rotational inertia of the pulley needs to be considered, as it affects the motion of the system. In the provided exercise, the rotational inertia is sought while the system's masses, pulley radius, and distance covered by the heavier block are given.

Angular Acceleration

Angular acceleration is a measure of how quickly the angular velocity of an object changes with time. In the case of rotational motion, such as a pulley in an Atwood's machine, angular acceleration is incredibly important, as it connects the linear acceleration of the blocks to the rotation of the pulley. It’s calculated by the equation \( \alpha = \frac{a}{r} \) where \( a \) is the linear acceleration and \( r \) is the radius of the pulley. This quantity not only enables us to delve into rotational dynamics but also is crucial for calculating the rotational inertia of the pulley when we apply the concepts of torque and Newton's second law for rotation.

Torque

Torque is a measure of the force that causes an object to rotate about an axis. It's an essential part of rotational dynamics and is inherently present in systems like the Atwood's machine. Torque is calculated by multiplying the force applied perpendicularly to the lever arm from the axis of rotation by the length of the lever arm (\( \tau = rF \sin(\theta) \) where \( r \) is the lever arm’s length and \( F \) is the force). In simpler cases where the force is applied at a 90-degree angle to the lever arm, the sine term becomes 1, and the formula simplifies to \( \tau = rF \) - just as shown in the calculation of the net torque acting on the pulley due to the net force of the Atwood's machine.

Equations of Motion

Equations of motion are fundamental tools in classical mechanics that describe the relationship between the motion of an object and the forces acting upon it. They allow physicists and engineers to predict future movement and analyze past motion. For linear motion, equations such as \( s = ut + \frac{1}{2} at^2 \) link displacement (\( s \) with initial velocity (\( u \) with acceleration (\( a \) and time (\( t \) - as demonstrated in the step-by-step solution wherein the equation enabled us to find the linear acceleration of the heavier block in the Atwood's machine. Understanding these fundamental concepts allows us to bridge the gap between linear and rotational dynamics, thereby enabling students to solve complex mechanics problems.