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A solid cylindrical disk has a radius of 0.15 m. It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a 45-N force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of 120 \(\mathrm{rad} / \mathrm{s}^{2} .\) What is the mass of the disk?

Short Answer

Expert verified
The mass of the disk is 5 kg.

Step by step solution

01

Identify the Given Information

We are given the following values: the radius of the disk \( r = 0.15 \) m, the applied tangential force \( F = 45 \) N, and the angular acceleration \( \alpha = 120 \; \mathrm{rad/s^2} \). We need to determine the mass of the disk.
02

Use the Formula for Torque

The torque \( \tau \) applied on the disk is given by the formula \( \tau = F \cdot r \). Substituting the given values, we find the torque: \( \tau = 45 \; \mathrm{N} \times 0.15 \; \mathrm{m} = 6.75 \; \mathrm{Nm} \).
03

Relate Torque to Angular Acceleration

The torque is also related to the angular acceleration through the formula \( \tau = I \cdot \alpha \), where \( I \) is the moment of inertia. For a solid cylinder, the moment of inertia \( I \) is \( \frac{1}{2} m r^2 \).
04

Express the Moment of Inertia

Substitute \( I = \frac{1}{2} m r^2 \) into the torque formula: \( 6.75 = \frac{1}{2} m (0.15)^2 \cdot 120 \). Simplify and solve for \( m \).
05

Solve for Mass

Rearrange the equation: \( m = \frac{6.75}{0.5 \times 120 \times (0.15)^2} \). Calculate \( (0.15)^2 = 0.0225 \). Then, \( m = \frac{6.75}{0.5 \times 120 \times 0.0225} = \frac{6.75}{1.35} = 5 \; \mathrm{kg} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It measures how difficult it is to change the rotational motion of an object. This is similar to mass in linear motion.
For a solid body, this property is calculated based on how the mass is distributed around the axis of rotation. In the case of a cylindrical disk, the moment of inertia, denoted as \( I \), is given by:
  • \( I = \frac{1}{2}mr^2 \)
Here, \( m \) is the mass of the disk, and \( r \) is its radius. This formula specifically applies to solid cylinders, showing that not only the mass, but also how far it is from the center, affects its inertia.
The larger the moment of inertia, the more torque is required to achieve the same angular acceleration.
Torque
Torque is essentially the rotational equivalent of force in linear motion. It is what causes an object to rotate. A good way to think about torque is to imagine opening a door. The force applied at a certain distance from the hinge creates torque.In the context of our problem, the torque \( \tau \) is calculated by multiplying the force \( F \) applied to the disk by the radius \( r \):
  • \( \tau = F \cdot r \)
For the disk, given \( 45 \text{ N} \) force and \( 0.15 \text{ m} \) radius, the torque is \( 6.75 \text{ Nm} \).
This relationship shows how even a small force can cause a substantial rotation if applied far from the axis.
Torque also links directly with angular acceleration \( \alpha \) and the moment of inertia:\( \tau = I \cdot \alpha \). This equation illustrates how the angular speed of an object can be manipulated by adjusting the torque applied.
Cylindrical Disk
A cylindrical disk is a simple yet essential shape in mechanics. It's essentially a 3D object with two identical flat circular ends and a specific height or thickness. When dealing with physical properties such as rotation, the disk's dimensions become vital.
  • The radius affects the moment of inertia.
  • The mass distribution around the axle determines how it spins.
In exercises like ours, knowing the radius helps compute the torque and moment of inertia, both of which play key roles in solving for unknowns like mass.
A careful analysis of these properties allows you to understand the motion dynamics of everyday objects like wheels or disks. By mastering concepts related to a cylindrical disk, you gain insights into broader rotational dynamics applicable in engineering and physics.

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Most popular questions from this chapter

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