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Chapter 11: Problem 38

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Short Answer

Expert verified
Angular momentum is 0.528 N路m路s, and angular velocity is 440 rad/s.

Step by step solution

01

Identify Given Values

First, collect all the relevant information from the problem statement:- Rotational inertia (I): \(1.2 \times 10^{-3} \; \text{kg} \cdot \text{m}^2\)- Torque (\tau): \(16 \; \text{N} \cdot \text{m}\)- Time (\Delta t): \(33 \times 10^{-3} \; \text{s}\) (converted from milliseconds)
02

Calculate Angular Momentum

Angular momentum can be calculated using the formula:\[ L = \tau \Delta t \]Substitute the values for torque and time:\[ L = 16 \; \text{N} \cdot \text{m} \times 33 \times 10^{-3} \; \text{s} \]Calculate \( L \):\[ L = 0.528 \; \text{N} \cdot \text{m} \cdot \text{s} \]
03

Calculate Angular Velocity

Angular velocity is calculated using the relationship between angular momentum, rotational inertia, and angular velocity:\[ L = I \omega \]Here \( \omega \) is the angular velocity. Rearrange the formula to solve for \( \omega \):\[ \omega = \frac{L}{I} \]Substitute \( L = 0.528 \; \text{N} \cdot \text{m} \cdot \text{s} \) and \( I = 1.2 \times 10^{-3} \; \text{kg} \cdot \text{m}^2 \):\[ \omega = \frac{0.528}{1.2 \times 10^{-3}} \]Calculate \( \omega \):\[ \omega = 440 \; \text{rad/s} \]

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Key Concepts

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

Angular Momentum
Angular momentum is a concept in physics that describes the rotational analogy to linear momentum. It's a measure of an object's tendency to continue rotating. When thinking about angular momentum, it鈥檚 essential to consider the "twistiness" or "spin" of an object in motion. Angular momentum (L) is determined by the formula:\[ L = I \omega \] In this formula, I represents the rotational inertia, which depends on how mass is distributed relative to an axis of rotation, and \omega is the angular velocity, or how quickly something is rotating.For the sanding disk, we can also compute angular momentum using torque when the torque (\tau) acts over a period, \Delta t:\[ L = \tau \Delta t \] This is what happens when an external force is applied, like using the electric drill to spin the disk. The torque here acts over a certain time, changing the angular momentum of the disk.
Torque
Torque is like a rotational force and it plays a crucial role in rotational dynamics. It鈥檚 the measure of the force that causes an object to rotate around an axis. Think of torque as the "twisting strength" that makes an object spin. The formula for torque (\tau) is:\[ \tau = F \cdot r \cdot \sin(\theta) \] where F is the force applied, r is the distance from the pivot point (or axis of rotation) to where the force is applied, and \theta is the angle between the force and the lever arm.In the context of the sanding disk:- The electric drill provides a torque of 16 N路m.- This torque influences how quickly the disk begins to spin, depending on how long the torque is applied. - Applying torque over time causes a change in the angular momentum of the disk.The idea is that more force or a longer lever arm (more distance from the axis) increases torque, thus influencing how rapidly an object's rotational motion changes.
Angular Velocity
Angular velocity is the rate at which an object rotates or spins around a central axis. It's measured in radians per second (rad/s). Imagine how fast a DVD spins inside a player, which is an example of angular velocity.The relationship between angular momentum, rotational inertia, and angular velocity is:\[ L = I \omega \] To determine the angular velocity (\omega) of the disk, rearrange the formula:\[ \omega = \frac{L}{I} \] This shows that for a given angular momentum, the rotational inertia will affect how fast something spins. In another manner, if you have a small rotational inertia, the angular velocity can be quite high for the same angular momentum. It鈥檚 essential to know that high angular velocity points out rapid spinning.For the sanding disk in the exercise, the outcomes demonstrate how a torque applied over a brief duration can result in a sizable angular velocity, making the disk spin swiftly.

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Most popular questions from this chapter

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m}\). Just then, in unitvector notation, what is its angular momentum relative to (a) the origin and (b) the point located at \((-2.0,-2.0) \mathrm{m} ?\)

In unit-vector notation, what is the net torque about the origin on a flea located at coordinates \((0,-4.0 \mathrm{~m}, 5.0 \mathrm{~m})\) when forces \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{j}}\) act on the flea?

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11- 18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Aparticle moves through an \(x y z\) coordinate system while a force acts on the particle. When the particle has the position vector \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}+(2.00 \mathrm{~m}) \hat{\mathrm{k}}\), the force is given by \(\vec{F}=F_{x} \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}-(6.00 \mathrm{~N}) \hat{\mathrm{k}}\) and the corresponding torque about the origin is \(\vec{\tau}=(4.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{i}}+(2.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{j}}-(1.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{k}}\). Determine \(F_{x^{*}}\)

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm}\). Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the \((\mathrm{g})\) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, \((\mathrm{k})\) top, and (1) bottom of each tire?
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