Problem 11 - College Physics

Question

A \(\mathrm{A} 2.00 \mathrm{kg}\) stone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0 \(\mathrm{kg}\) cylindrical pulley shown in Figure \(10.47 .\) The inner diameter of the pulley is \(60.0 \mathrm{cm},\) while the outer diameter is 1.00 \(\mathrm{m}\) . The system is released from rest, and there is no friction at the axle of the pulley. Find (a) the acceleration of the stone, (b) the tension in the wire, and (c) the angular acceleration of the pulley.

Step-by-Step Solution

Answer

The stone's acceleration is 3.24 m/s², the tension is 16.12 N, and the angular acceleration is 3.24 rad/s².

Step 1: Understand the Problem

We need to calculate (a) the acceleration of a 2.00 kg stone connected to a cylindrical pulley, (b) the tension in the wire connecting the stone to the pulley, and (c) the angular acceleration of the pulley. Key parameters include the mass of the pulley (10.0 kg), the outer radius (0.50 m), and the inner radius (0.30 m).

Step 2: Apply Newton's Second Law for Linear Motion

For the stone, we apply Newton's second law: \[ T - mg = ma \]where \( T \) is the tension in the wire, \( m = 2.00 \) kg is the mass of the stone, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( a \) is the linear acceleration of the stone.

Step 3: Use Newton's Second Law for Rotational Motion

For the pulley, apply Newton's second law for rotation:\[ I \alpha = au \]where \( I \) is the moment of inertia of the pulley, \( \alpha \) is the angular acceleration, and \( \tau \) is the torque produced by the tension. Calculate \( I \) for a cylindrical pulley with both a solid and hollow part:\[ I = \frac{1}{2} M (R_o^2 + R_i^2) \]where \( M = 10.0 \, \text{kg} \), \( R_o = 0.50 \, \text{m} \), \( R_i = 0.30 \, \text{m} \).

Step 4: Solve for Linear and Angular Variables

From relation between linear and angular acceleration, \( a = R_o \alpha \). Combine with torque equation:\[ \tau = T \cdot R_o = I \alpha \]Replace \( \alpha = \frac{a}{R_o} \) to find\[ T R_o = I \frac{a}{R_o} \rightarrow T = \frac{I a}{R_o^2} \]

Step 5: Solve the System of Equations

Use the equations from steps 2 and 4 to solve for acceleration:\[ T - mg = ma \] \[ T = \frac{I a}{R_o^2} \] Substitute for \( I \) and solve for \( a \). With known values:\[ a = \frac{mg}{m + \frac{I}{R_o^2}} \]

Step 6: Calculate Tension and Angular Acceleration

With \( a \) known, use it to find tension \( T \) using:\[ T = m(g - a) \]Finally, calculate angular acceleration \( \alpha \) using:\[ \alpha = \frac{a}{R_o} \]

Step 7: Numerical Calculations

Compute the actual values:- Use \( m = 2.00 \) kg, \( g = 9.81 \, \text{m/s}^2 \), \( I = 1.09 \text{kg} \cdot \text{m}^2 \) (from substitution), and \( R_o = 0.50 \, \text{m} \).- Find linear acceleration \( a \).- Calculate tension \( T \) as \( 16.12 \, \text{N} \).- Find angular acceleration \( \alpha \) as \( 3.24 \, \text{rad/s}^2 \).

Key Concepts

Linear Acceleration

Linear acceleration is the change in velocity of an object per unit time along a straight path. In the context of a mass attached to a pulley system, like the 2.00 kg stone in our exercise, it's crucial to understand how forces affect its motion.
The linear acceleration of the stone can be derived from Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. Using the equation:\[\ T - mg = ma \]Where:

• \(T\) is the tension in the wire
• \(m\) is the mass of the stone (2.00 kg)
• \(g\) is the acceleration due to gravity (9.81 \(m/s^2\))
• \(a\) is the linear acceleration of the stone

To solve for \(a\), you rearrange the equation to find:\[ a = \frac{T - mg}{m} \]This gives you the stone's linear acceleration once \(T\) is known. It's connected to the tension in the wire and the gravitational force pulling the stone down. By understanding how to manipulate these forces, you can predict how fast the stone will accelerate when the system is released.

Angular Acceleration

Angular acceleration refers to how quickly an object changes its rotational velocity. In simple terms, it’s the rate at which the spinning speed of an object increases or decreases. This concept is particularly vital when dealing with objects in rotational motion, like our cylindrical pulley.
The angular acceleration \(\alpha\) of the pulley is related to the linear acceleration \(a\) of the stone. They are connected through the radius of the pulley by the formula:\[ \alpha = \frac{a}{R_o} \]Where:

• \(R_o\) is the outer radius of the pulley (0.50 m)
• \(a\) is the linear acceleration of the stone

The moment of inertia \(I\) and the torque \(\tau\) produced by tension in the wire contribute to angular acceleration. Using:\[ \tau = T \cdot R_o = I \alpha \]It helps explain how force applied at a distance (like the tension in the wire) causes the pulley to spin. Pulley systems rely on angular acceleration to determine when and how quickly they start rotating. The balance between the moment of inertia, tension, and radius determines the angular speed, just as weight and tension influence linear speed.

Newton's Second Law

Newton's second law is fundamental in mechanics, linking force, mass, and acceleration. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be expressed as:\[ F = ma \]Where:

• \(F\) is the net force
• \(m\) is mass
• \(a\) is acceleration

For linear motion, this law is directly used in the stone's movement in the exercise, helping us find its acceleration when only tension and gravitational forces are at play. For rotational dynamics, a similar principle applies:\[ \tau = I \alpha \]Where:

• \(\tau\) is the torque
• \(I\) is the moment of inertia
• \(\alpha\) is the angular acceleration

Both equations show how mass or inertia influences acceleration, either linearly or rotationally. By using these principles, we can solve complex motion problems like our pulley system, identifying how forces and masses interact both in straight lines and rotational movements.