Question

A string is wrapped many times around a pulley and is connected to a block that is hanging vertically. The pulley consists of a wheel of radius \(48.77 \mathrm{~cm}\) and mass \(m_{\mathrm{p}}=5.991 \mathrm{~kg}\), with spokes that have negligible mass. The block accelerates downward at \(4.539 \mathrm{~m} / \mathrm{s}^{2}\). What is the mass of the block, \(m_{\mathrm{b}} ?\)

Short answer

Answer: The mass of the hanging block is approximately 2.646 kg.

Step-by-step solution

1. Identify the forces and moment of inertia for the system

The forces involved are the tension in the string acting upward and the gravitational force acting downward on the block. The moment of inertia of the pulley, assuming it is a solid disk, can be calculated using the equation \(I = \frac{1}{2}m_{\mathrm{p}}R^2\), where \(I\) is the moment of inertia, \(m_{\mathrm{p}}\) is the mass of the pulley, and \(R\) is the radius.

2. Calculate the gravitational force on the block

To calculate the gravitational force acting on the block (\(F_{\mathrm{g}}\)), we can use the equation \(F_{\mathrm{g}} = m_{\mathrm{b}}g\), where \(m_{\mathrm{b}}\) is the mass of the block (our unknown) and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)).

3. Create an equation for the tension in the string

The tension in the string (\(T\)) is the force acting upward on the hanging block. Using Newton's second law, we can create an equation for the tension as \(T = m_{\mathrm{b}}(g - a)\), where \(a\) is the downward acceleration of the block.

4. Apply the torque equation to the pulley

The torque (\(\tau\)) acting on the pulley is equal to the moment of inertia multiplied by the angular acceleration (\(\alpha\)). We can write this equation as \(\tau = I \alpha\). Since the torque is also equal to the tension in the string multiplied by the radius of the pulley (\(\tau = T R\)), we can substitute this into our torque equation: \(T R = I \alpha\).

5. Relate angular acceleration to linear acceleration

To connect the angular acceleration of the pulley to the linear acceleration of the block, we can use the relationship \(a = R \alpha\). Rearranging for \(\alpha\), we get \(\alpha = \frac{a}{R}\).

6. Substitute equations and solve for the mass of the block

Substitute the tension and angular acceleration equations from step 3 and 5 into the torque equation from step 4: \(m_{\mathrm{b}}(g - a)R = I \frac{a}{R}\). Next, substitute the moment of inertia equation from step 1: \(m_{\mathrm{b}}(g - a)R = \frac{1}{2}m_{\mathrm{p}}R^2 \frac{a}{R}\). Now, solve for the mass of the block (\(m_{\mathrm{b}}\)): \(m_{\mathrm{b}}(g - a)R = \frac{1}{2}m_{\mathrm{p}}R a\) \(m_{\mathrm{b}} = \frac{\frac{1}{2}m_{\mathrm{p}}R a}{(g - a)R}\) [| R Cancel out] \(m_{\mathrm{b}} = \frac{\frac{1}{2}(5.991 \mathrm{~kg})(4.539 \mathrm{~m/s^2})}{(9.81 \mathrm{~m/s^2} - 4.539 \mathrm{~m/s^2})}\) \(m_{\mathrm{b}} \approx 2.646 \mathrm{~kg}\) Therefore, the mass of the block is approximately \(2.646 \mathrm{~kg}\).

Key concepts

Moment of Inertia

Moment of inertia, often represented by the symbol 'I', is a measure of an object's resistance to changes in its rotational motion. It is the rotational equivalent to mass in linear motion. The higher the moment of inertia, the more torque you need to start or stop spinning an object.

For simple shapes, the moment of inertia can be calculated using basic formulas. For instance, the moment of inertia of a solid disc, which is often seen in pulleys, is given as \( I = \frac{1}{2}m_{\text{p}}R^2 \), where 'm' represents mass and 'R' is the radius of the disc.

Understanding the concept of moment of inertia is crucial for solving physics problems related to rotational motion because it directly affects the angular acceleration of an object when a torque is applied.

Rotational Motion

Rotational motion refers to the movement of an object in a circular path around a central point or axis. This type of motion is everywhere in the physical world, from the spinning of a bicycle wheel to the orbits of planets.

In the context of the problem, the pulley's rotational motion is triggered by the unwrapping of the string as the block falls. The key to solving rotational motion problems is correlating the linear motion of the falling block (linear acceleration 'a') to the angular motion of the pulley (angular acceleration '\(\alpha\)'). This relationship is described by the equation \( a = R\alpha \).

Additionally, understanding the effects of the moment of inertia and the applied torque (force times the radius) will help in quantifying this motion. Only by accounting for both the rotational attributes and the linear forces at play can one accurately resolve such physics problems.

Newton's Second Law

Newton's second law explains how the velocity of an object changes when it is subjected to an external force. The law states that the acceleration (a) of an object is directly proportional to the net external force (F) acting on the object and inversely proportional to its mass (m), i.e., \( F = ma \).

In our pulley system, Newton's second law applies to both the linear dynamics of the falling block and the rotational dynamics of the pulley. For the block, the law takes the familiar form \( F_g = m_{\text{b}}g \), whereas for the pulley, it's expressed in terms of torque (\(\tau \)) and angular acceleration (\(\alpha \)), as \( \tau = I\alpha \).

When solving for the mass of the block, it’s essential to consider the force of gravity acting on it and the tension in the string, which provides the torque leading to the pulley's rotation. Linking these linear and rotational expressions of Newton's second law enables the solution of complex motion problems by breaking them down into more manageable components.