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A solid cylindrical disk has a radius of \(0.15 \mathrm{~m} .\) It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a \(45-\mathrm{N}\) 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 approximately 5 kg.

Step by step solution

01

Understand the Problem

We need to find the mass of a solid cylindrical disk given the force applied tangentially, the radius, and the angular acceleration. According to Newton's second law for rotation, torque results in angular acceleration.
02

Identify Relevant Equations

We use the formula for torque, \( \tau = I \alpha \), where \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. The formula for torque can also be \( \tau = F \, r \), where \( F \) is the force, and \( r \) is the radius of the disk.
03

Calculate Torque

The torque exerted on the disk by the force is \( \tau = F \times r = 45 \, \text{N} \times 0.15 \, \text{m} = 6.75 \, \text{Nm} \).
04

Express Moment of Inertia

For a solid cylinder, the moment of inertia \( I = \frac{1}{2} m r^2 \). We need to find the value of mass \( m \), given \( r \) and \( \alpha \).
05

Relate Torque and Known Quantities

Set the expressions for torque equal: \( 6.75 = \frac{1}{2} m (0.15)^2 \times 120 \).
06

Solve for Mass

Simplify the equation: \[ 6.75 = 0.5 \times 120 \times 0.0225 \times m \] \[ 6.75 = 1.35 m \] Divide both sides by 1.35 to find \( m \): \[ m = \frac{6.75}{1.35} \approx 5 \]
07

Conclusion

The mass of the disk is approximately \(5\) 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, often symbolized as \( I \), is a measure of an object's resistance to change in its rotational motion. It's like mass in linear motion, but for spinning bodies. It's crucial in understanding how different distribution of mass affects the rotational inertia.

For simple shapes, the moment of inertia can be calculated using predefined formulas. In the case of a solid cylindrical disk like the one in our problem, it is given by:
  • \( I = \frac{1}{2} m r^2 \)
where \( m \) represents the mass of the disk, and \( r \) is the radius.

This formula shows us how both the radius and the mass of the object affect its rotation. As the radius increases or as the mass is distributed further from the axis, the moment of inertia also increases, making it harder for the object to accelerate rotationally.
Angular Acceleration
Angular acceleration refers to how quickly an object's rotational speed changes, denoted as \( \alpha \). It's analogous to linear acceleration but in a rotational context.

Angular acceleration is measured in radians per second squared (\( \mathrm{rad}/\mathrm{s}^2 \)), which tells us how much the angular velocity changes every second. In our exercise, the disk experiences an angular acceleration of \( 120 \mathrm{rad}/\mathrm{s}^2 \), which means every second, its angular velocity increases by 120 radians per second. This is caused due to the applied torque.

Understanding angular acceleration helps in analyzing how soon an object will reach its desired rotational speed and is influenced by both the applied force and the object's moment of inertia.
Newton's Second Law for Rotation
Newton's Second Law for rotation connects torque and angular acceleration, showing how much torque is needed to cause a certain angular acceleration. The law states that:
  • \( \tau = I \alpha \)
Here, \( \tau \) denotes torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.

Torque can be imagined as a rotational force that causes an object to rotate around an axis. Just as force produces linear acceleration, torque produces angular acceleration.

In our problem, this principle helped set up the relationship between the known force, the moment of inertia (dependent on mass and radius), and the resulting angular acceleration. The torque calculated using \( \tau = F \cdot r \) equals the product of moment of inertia and angular acceleration, providing a way to solve for unknown variables, such as mass in this case. The connection between these elements illustrates how rotational dynamics operates seamlessly with linear principles in physics.

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Most popular questions from this chapter

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of \(2.00 \mathrm{~N}\) and is applied perpendicular to the length of the stick at the free end. The other force has a magnitude of \(6.00 \mathrm{~N}\) and acts at a \(30.0^{\circ}\) angle with respect to the length of the stick. Where along the stick is the 6.00 -N force applied? Express this distance with respect to the end that is pinned.

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

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