Question

A string is wrapped many times around a pulley and is connected to a block of mass \(m_{\mathrm{b}}=4.701 \mathrm{~kg},\) which is hanging vertically. The pulley consists of a wheel of radius \(47.49 \mathrm{~cm},\) with spokes that have negligible mass. The block accelerates downward at \(4.330 \mathrm{~m} / \mathrm{s}^{2}\) What is the mass of the pulley, \(m_{\mathrm{p}}\) ?

Short answer

Based on the given parameters and the equations set up for the linear motion of the block and the rotational motion of the pulley, the mass of the pulley was found to be approximately 4.96 kg.

Step-by-step solution

Write down the given parameters

The given parameters are: - Mass of the block: \(m_b = 4.701\ kg\) - Radius of the pulley: \(r = 47.49\ cm = 0.4749\ m\) - Acceleration of the block: \(a = 4.330\ m/s^2\) Our goal is to find the mass of the pulley, \(m_p\).

Set up the linear motion equation for the block

Newton's second law states that the net force on an object is equal to its mass times its acceleration: \(F_{net} = m_b \cdot a\). For the block, the net force is the difference between tension force \(T\) and gravitational force \(m_b \cdot g\) (where \(g = 9.81\ m/s^2\) is the acceleration due to gravity). Thus, our equation for the linear motion of the block is: $$m_b \cdot g - T = m_b \cdot a$$

Set up the rotational motion equation for the pulley

For the rotational motion of the pulley, we will use the equation \(I \cdot \alpha = \tau\), where \(I\) is the moment of inertia, \(\alpha\) is the angular acceleration, and \(\tau\) is the net torque. Since the spokes have negligible mass, we will treat the pulley as a solid disk, so its moment of inertia is \(I = \frac{1}{2} m_p \cdot r^2\). The net torque on the pulley is due to the tension force on the string, and it can be calculated as \(\tau = T \cdot r\). Angular acceleration \(\alpha\) can be related to the linear acceleration of the block by the equation \(a = \alpha \cdot r\). Now our equation for the rotational motion of the pulley is: $$\frac{1}{2} m_p \cdot r^2 \cdot \frac{a}{r} = T \cdot r$$

Solve for the tension force T

We can solve for the tension force \(T\) from the rotational motion equation and then plug it into the linear motion equation: $$T = \frac{1}{2} m_p \cdot r \cdot a$$

Substitute this expression for T and solve for mass of the pulley

Substitute the expression for \(T\) into the linear motion equation: $$m_b \cdot g - \frac{1}{2} m_p \cdot r \cdot a = m_b \cdot a$$ Now we can solve for the mass of the pulley \(m_p\): $$m_p = \frac{2 \cdot m_b \cdot (g - a)}{r \cdot a}$$ Plug in the given values and calculate \(m_p\): $$m_p = \frac{2 \cdot 4.701\ kg \cdot (9.81\ m/s^2 - 4.330\ m/s^2)}{0.4749\ m \cdot 4.330\ m/s^2}$$ $$m_p \approx 4.96\ kg$$ The mass of the pulley is approximately \(4.96\ kg\).

Key concepts

Moment of Inertia

The moment of inertia is a measure of an object’s resistance to changes in its rotation rate. It plays a similar role in rotational motion as mass does in linear motion. Essentially, it is the rotational mass for a spinning object, reflecting how the mass is distributed relative to the axis of rotation.

Each shape has a unique moment of inertia equation that depends on its geometry and mass distribution. For a solid disk, like the pulley in our exercise, the moment of inertia is given by the formula \(I = \frac{1}{2} m_p \cdot r^2\), where \(m_p\) is the mass of the pulley and \(r\) is its radius. Because of the way the moment of inertia is defined, if more mass is concentrated farther from the axis, the object is harder to rotate and thus has a higher moment of inertia.

Understanding the moment of inertia is crucial when we deal with rotational dynamics because it fundamentally affects how an object spins and accelerates around an axis.

Angular Acceleration

Angular acceleration, often denoted as \(\alpha\), is the rate at which the angular velocity of an object changes with respect to time. It's the rotational analogue to linear acceleration in translational motion and can be expressed in radians per second squared (\(\text{rad/s}^2\)).

For a pulley system like the one in our exercise, the angular acceleration can be directly related to the linear acceleration of the hanging mass. If the string is not slipping, we can derive the relation \(\alpha = \frac{a}{r}\), which allows us to link the rotational motion of the pulley to the linear motion of the block.

This concept is important for solving problems in rotational motion as it connects the rotational effects to the more intuitive linear perspective, allowing us to use the familiar principles of linear dynamics to solve rotational problems.

Newton's Second Law

Newton's second law for rotation is the backbone of rotational dynamics. Just as Newton's second law in linear form involves force, mass, and acceleration (\(F=ma\)), the rotational form involves torque (\(\tau\)), moment of inertia (\(I\)), and angular acceleration (\(\alpha\)). The rotational form of Newton's second law can be written as \(\tau = I \cdot \alpha\), expressing that the net torque on an object is equal to the moment of inertia times the angular acceleration.

The torque, representing a rotational force, causes the object to change its rotational motion. In the case of the pulley system, the tension in the string exerts a torque on the pulley, which leads to its angular acceleration. When the mass and radius of the pulley are known, the pulley's moment of inertia can be calculated, and with the angular acceleration, we can apply Newton's second law to find related forces and ultimately solve for unknown variables, like the pulley mass in the exercise provided.