Review problem. A 30.0-kg hammer strikes a steel spike 2.30 cm in diameter while moving with spe

step 1 Chapter 12, problem 32. In this problem, we're going to find the average strain when a spike is

Review problem. A 30.0-kg hammer strikes a steel spike 2.30...

Review problem. A $30.0-\mathrm{kg}$ hammer strikes a steel spike $2.30 \mathrm{cm}$ in diameter while moving with speed $20.0 \mathrm{m} / \mathrm{s}$ The hammer rebounds with speed $10.0 \mathrm{m} / \mathrm{s}$ after $0.110 \mathrm{s} .$ What is the average strain in the spike during the impact?

Key Concepts

Impulse-Momentum Theorem

This theorem states that the change in momentum of an object is equal to the impulse applied to it, where impulse is the product of average force and the time interval over which the force acts. It is used to calculate the average force during an impact by relating the momentum change to the time duration of the collision.

Average Force Calculation

Average force is determined by dividing the computed impulse (change in momentum) by the duration of the impact. This concept is crucial in dynamics when analyzing collisions and impact events.

Stress and Strain Relationship

Stress is defined as the force applied per unit area, and strain is the measure of deformation experienced by a material (change in length relative to original length). Their relationship, often linear within the elastic limit as given by Hooke's law, helps in understanding how materials deform under applied forces.

Cross-Sectional Area Determination

Calculating the cross-sectional area of a component, such as a steel spike with a circular cross-section, is critical because stress is computed by dividing the applied force by this area. Geometric formulas, like using the diameter or radius of a circle, are employed to find the area.

Transcript

00:02 Chapter 12, problem 32.

00:04 In this problem, we're going to find the average strain when a spike is hit by a big steel hammer at high speed.

00:14 We're told that it is indeed a steel hammer.

00:18 It's given a mass of 30 kilos.

00:21 The diameter of the spike is 2 .3 centimeters.

00:25 The starting velocity of the hammer is 20 meters per second.

00:29 The velocity after rebounding from the spike is negative 10 meters per second, and the time between the two is 0 .110 second.

00:41 And then because we're going to need it here in a moment, we're going to use from table 12 .1, the young's modulus of steel, which is 20 times 10 to the 10th newtons per meter squared.

00:56 So we know from section 124, or actually from equation 126, that the young's, sorry, should be in pen mode, young's modulus is actually defined.

01:15 All right, i'm on a roll writing wise here, is actually defined as the stress, the tensile stress divided by the tensile strain.

01:25 The book actually spells it out.

01:28 I use the little sigma and little epsilon from when i was learning this stuff.

01:35 And from that, we know that the stress, or from that chapter, we know the stress is the force times the cross -sectional, or force divided by the cross -sectional area.

01:51 And from much earlier in the book, we know that force is equal to mass, times acceleration.

02:02 That's newton's second law, quick review.

02:06 And then in this instance, we're looking for the average acceleration, which is going to be the change in speed.

02:25 That looks like a five, doesn't it? that's supposed to be an s.

02:27 Let's make it a little clearer.

02:30 Change in speed.

02:31 Yeah, it's not much better, is it? divided by the time that that change in speed took place.

02:38 So that'll give us our average acceleration...