Problem 72
Question
A CD has a mass of \(15.0 \mathrm{~g}\), an inner diameter of \(1.5 \mathrm{~cm},\) and an outer diameter of \(11.9 \mathrm{~cm} .\) Suppose you toss it, causing it to spin at a rate of 4.3 revolutions per second. a) Determine the moment of inertia of the \(\mathrm{CD}\), approximating its density as uniform. b) If your fingers were in contact with the CD for 0.25 revolutions while it was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque?
Step-by-Step Solution
Verified Answer
\(I = \frac{1}{2}(15)(0.00005625 + 0.00354025)\)
\(I = 7.5(0.0035965)\)
\(I = 0.027\mathrm{~kg\cdot m^2}\)
The moment of inertia of the CD is approximately \(0.027\mathrm{~kg\cdot m^2}\).
1Step 1: Identify given information
From the problem: A CD has a mass of \(15.0 \mathrm{~g}\), an inner diameter of \(1.5 \mathrm{~cm},\) and an outer diameter of \(11.9 \mathrm{~cm} .\) Suppose you toss it, causing it to spin at a rate of 4.3 revolutions per second. a) Determine the moment of inertia of the \(\mathrm{CD}\), approximating its density as uniform. b) If your fingers were in contact with
2Step 2: Determine what to find
We identify the unknown quantity in this Physics problem and select the appropriate approach.
3Step 3: Select the appropriate formula
Based on the given data and unknown, we choose the relevant equation from Physics.
4Step 4: Substitute values and solve
Substituting known values into the equation, we perform the calculation.
5Step 5: State the result
The result is: \(I = \frac{1}{2}(15)(0.00005625 + 0.00354025)\) \(I = 7.5(0.0035965)\) \(I = 0.027\mathrm{~kg\cdot m^2}\) The moment of inertia of the CD is approximately \(0.027\mathrm{~kg\cdot m^2}\).
Key Concepts
Understanding Angular VelocityExploring TorquePrinciples of Circular MotionUniform Density and Its Implications
Understanding Angular Velocity
Angular velocity measures how quickly an object rotates or spins around an axis. Imagine a CD spinning like a top. Its angular velocity tells us how many rotations it completes in a given time, typically measured in radians per second. In the problem, the CD spins at 4.3 revolutions per second, which you can convert to radians per second by knowing there are \(2\pi\) radians in one revolution.
Angular velocity is essential because it helps us understand the kinetic energy and momentum of rotating objects. When you toss a CD and make it spin, you're imparting energy that translates into this spinning motion.
By grasping angular velocity, you can better understand how different speeds of rotation affect an object's dynamic behavior and how applied forces or torques change this velocity.
Angular velocity is essential because it helps us understand the kinetic energy and momentum of rotating objects. When you toss a CD and make it spin, you're imparting energy that translates into this spinning motion.
- Angular velocity (\(\omega\)) = \(\frac{\text{Angle in radians}}{\text{Time}}\)
- 1 revolution = \(2\pi\) radians
By grasping angular velocity, you can better understand how different speeds of rotation affect an object's dynamic behavior and how applied forces or torques change this velocity.
Exploring Torque
Torque is like the rotational equivalent of a force. It causes an object to start rotating, stop, or change its rotational speed. When you apply a torque to an object, such as the CD in this exercise, you're exerting a twisting force. This is similar to opening a jar lid—a force applied at a distance from the axis turns it.
Torque is measured in Newton-meters (Nm) and is calculated by multiplying the force applied by the distance from the force's line of action to the axis of rotation. In the given problem, your fingers supply the torque necessary for the CD to reach its angular velocity.
Torque is measured in Newton-meters (Nm) and is calculated by multiplying the force applied by the distance from the force's line of action to the axis of rotation. In the given problem, your fingers supply the torque necessary for the CD to reach its angular velocity.
- Torque (\(\tau\)) = Force (\(F\)) \(\times\) Lever arm (\(r\))
- The longer the lever arm, the more torque generated.
Principles of Circular Motion
Circular motion occurs when an object moves along a circular path. This motion is characterized by parameters such as angular velocity and centripetal force. For example, when a CD spins, it showcases circular motion, where every point on the CD follows a circular trajectory at any point in time.
In circular motion, each part of the object is at the same distance from a fixed point, commonly called the center of rotation. This consistent motion is possible due to centripetal force, which pulls the object toward the center and keeps it from flying off the circular path.
In circular motion, each part of the object is at the same distance from a fixed point, commonly called the center of rotation. This consistent motion is possible due to centripetal force, which pulls the object toward the center and keeps it from flying off the circular path.
- Angular Velocity \(\omega = \frac{\theta}{t}\), with \(\theta\) being angular displacement in radians.
- Centripetal Force \(F_c = \frac{mv^2}{r}\), where \(m\) is mass, \(v\) is velocity, and \(r\) is radius.
Uniform Density and Its Implications
When an object like the CD has uniform density, its mass is evenly distributed throughout its volume. This type of density matters when calculating properties such as the moment of inertia, which depends on the mass distribution relative to the rotation axis.
For the CD, assuming uniform density simplifies calculations. We use formulas that require mass as an input, assuming it's spread out evenly without any variations in different sections. The even distribution assists in determining the exact inertia accurately.
For the CD, assuming uniform density simplifies calculations. We use formulas that require mass as an input, assuming it's spread out evenly without any variations in different sections. The even distribution assists in determining the exact inertia accurately.
- Uniform Density \(\rho = \frac{m}{V}\), with \(m\) as mass and \(V\) as volume.
- This assumption makes mathematical modeling manageable and often more accurate for calculating rotational dynamics.
Other exercises in this chapter
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