Problem 38 - College Physics

Question

A grinding wheel in the shape of a solid disk is 0.200 m in diameter and has a mass of 3.00 kg. The wheel is rotating at 2200 rpm about an axis through its center. (a) What is its kinetic energy? (b) How far would it have to drop in free fall to acquire the same amount of kinetic energy?

Step-by-Step Solution

Answer

(a) 398.43 J; (b) 13.55 m.

Step 1: Calculate the Moment of Inertia

The moment of inertia \(I\) for a solid disk rotating about its center is given by the formula \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius. First, convert the diameter to radius: \( r = \frac{0.200\, \text{m}}{2} = 0.100\, \text{m} \). Now, substitute in the values: \[ I = \frac{1}{2} \times 3.00\, \text{kg} \times (0.100\, \text{m})^2 = 0.015\, \text{kg}\,\text{m}^2. \]

Step 2: Convert Rotational Speed to Radians per Second

The rotational speed \( \omega \) must be in radians per second. Convert \( \text{rpm} \) to \( \text{radians per second} \) using the conversion factor \( \frac{2\pi}{60} \). So, \( \omega = 2200\, \text{rpm} \times \frac{2\pi}{60} = \frac{2200 \times 2\pi}{60}\, \text{rad/s}. \) Simplifying gives \( \omega \approx 230.4\, \text{rad/s}. \)

Step 3: Calculate Rotational Kinetic Energy

The rotational kinetic energy \( K \) is given by \( K = \frac{1}{2} I \omega^2 \). Substitute the previously calculated values into this equation: \[ K = \frac{1}{2} \times 0.015\, \text{kg}\,\text{m}^2 \times (230.4\, \text{rad/s})^2. \] This gives \( K \approx 398.43\, \text{J} \).

Step 4: Calculate Equivalent Free Fall Height

The potential energy \( U \) in free fall is given by \( U = mgh \), where \( g = 9.81\, \text{m/s}^2 \) is the gravitational acceleration and \( h \) is the height. Set this equal to the kinetic energy: \( 398.43\, \text{J} = 3.00\, \text{kg} \times 9.81\, \text{m/s}^2 \times h \). Solve for \( h \): \[ h = \frac{398.43\, \text{J}}{3.00\, \text{kg} \times 9.81\, \text{m/s}^2} \approx 13.55\, \text{m}. \]

Key Concepts

Moment of Inertia

The moment of inertia is an essential concept in rotational dynamics. Think of it as the rotational analog of mass in linear motion. It helps determine how difficult it is to change the rotational motion of an object. For a solid disk, like our grinding wheel, the moment of inertia is calculated using the formula: \[ I = \frac{1}{2} m r^2 \] where

• \( m \) represents the mass of the object,
• \( r \) is the radius of the disk.

It's crucial to convert the diameter of the disk into the radius by dividing by two. In this example: \[ r = \frac{0.200\, \text{m}}{2} = 0.100\, \text{m} \]The moment of inertia gives insight into the object's resistance to change in its angular speed, and it's necessary for calculating rotational kinetic energy.

Angular Velocity

Angular velocity is a measure of how fast an object rotates or spins. It indicates the angle an object sweeps out per unit of time, much like speed in linear motion. However, in rotational dynamics, we express this in radians per second.To convert from revolutions per minute (rpm) to radians per second, use the formula:\[ \omega = \text{rpm} \times \frac{2\pi}{60} \]For the grinding wheel, with a speed of 2200 rpm, the conversion looks like this:\[ \omega = 2200 \times \frac{2\pi}{60} = 230.4\, \text{rad/s} \]Understanding angular velocity helps in determining the rotational kinetic energy since it reflects the speed at which the object is spinning.

Potential Energy Gravitational

Gravitational potential energy is the energy stored due to an object's position relative to Earth. It's the energy an object possesses because of its height above the ground. The formula used is:\[ U = mgh \]where

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity (9.81 m/s²),
• \( h \) is the height above the ground.

In the problem, you equate this potential energy to the rotation's kinetic energy to find how far the grinding wheel would need to drop to gain the same energy. Knowing gravitational potential energy helps connect concepts of mechanics, showing how energy transforms from one type to another while remaining conserved.