/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 127 Member \(A B C\) has a mass of \... [FREE SOLUTION] | 91Ó°ÊÓ

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Member \(A B C\) has a mass of \(2.4 \mathrm{kg}\) and is attached to a pin support at \(B\). An \(800-\mathrm{g}\) sphere \(D\) strikes the end of member \(A B C\) with a vertical velocity \(\mathrm{v}_{1}\) of \(3 \mathrm{m} / \mathrm{s}\). Knowing that \(L=750 \mathrm{mm}\) and that the coefficient of restitution between the sphere and member \(A B C\) is \(0.5,\) determine immediately after the impact \((a)\) the angular velocity of member \(A B C,(b)\) the velocity of the sphere.

Short Answer

Expert verified
The angular velocity of member ABC is [calculated value] rad/s, and the velocity of the sphere is [calculated value] m/s.

Step by step solution

01

Convert Masses

Convert the masses from grams and kilograms to consistent SI units. The mass of the sphere is given as 800 g, which converts to 0.8 kg. The mass of member ABC is already given as 2.4 kg.
02

Compute Initial Momentum

Write the equation for conservation of linear momentum for the system immediately before and after the collision.\[mv_1 + I\omega_1 = mv_2 + I\omega_2\]Here, initial angular velocity \(\omega_1\) is zero because the member is initially at rest.
03

Use Coefficient of Restitution

The coefficient of restitution equation relates the relative velocities before and after the collision:\[e = \frac{v_2 + r\omega_2}{v_1}\]Given that the coefficient of restitution, \(e = 0.5\), solve for one of the final unknowns. Since it's easier, solve for \(v_2\).
04

Express Final Velocities

Use the equations from steps 2 and 3 to express the final velocities:- Substitute \(v_2\) from Step 3 into the momentum equation from Step 2 to calculate \(\omega_2\) (angular velocity of ABC).- Substitute \(\omega_2\) into Step 3 as needed to calculate \(v_2\), the final velocity of the sphere.
05

Calculate Angular Velocity

Using the equations from previous steps, compute the angular velocity of member ABC:\[\omega_2 = \frac{v_2}{L}\]Substitute known values for \(v_2\) and \(L = 0.75 \ m\) to find \(\omega_2\).
06

Calculate Sphere's Final Velocity

Substitute \(\omega_2\) back into the equation in Step 3 to determine the sphere's velocity \(v_2\) after impact.

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Key Concepts

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

Angular Velocity Calculation
Understanding angular velocity is key in analyzing dynamic systems with rotation, such as the member ABC in our problem. Angular velocity, denoted typically as \( \omega \), describes how fast an object rotates or spins around a particular axis. It is measured in radians per second (rad/s).

When a force is applied to an object at a distance from a point of rotation, it can create rotational motion. In the case of member ABC, the impact from sphere D causes the member to rotate about its fixed pin support at B.

To calculate the angular velocity immediately after impact, we use the relationship between linear and angular quantities. Specifically, you will encounter the equation \( \omega_2 = \frac{v_2}{L} \), where \( v_2 \) is the linear velocity at the end of the member and \( L \) is the distance from the point of rotation to where the sphere strikes the member. This formula emerges from recognizing that the linear velocity is equivalent to the tangential component of the angular velocity.
Conservation of Linear Momentum
The principle of conservation of linear momentum is vital for solving collision problems. This principle states that the total momentum of a closed system remains constant provided no external forces act. In simpler terms, the momentum before the collision is equal to the momentum after the collision.

For our system comprising member ABC and sphere D, we express this as:

\[ mv_1 = mv_2 + I\omega_2 \]

Here, \( m \) is the mass, \( v_1 \) and \( v_2 \) are the initial and final velocities of the sphere, \( I \) is the moment of inertia of member ABC, and \( \omega_2 \) is its angular velocity after the impact. Before the impact, member ABC is stationary, implying its initial angular velocity is zero.

Using this principle allows us to relate complex interactions in the collision, setting the groundwork to find unknowns such as \( \omega_2 \) and \( v_2 \). This connection between linear and rotational components is crucial in dynamics and offers profound insight into the post-collision state.
Coefficient of Restitution
The coefficient of restitution is a measure that describes how 'bouncy' a collision is. It is a dimensionless number ranging from 0 to 1, representing the ratio of relative velocities after and before an impact.

In our equation, it is expressed as:

\[ e = \frac{v_2 + r\omega_2}{v_1} \]

Where \( e \) is the coefficient of restitution, \( v_2 \) is the final velocity of the sphere, \( r \) is the radius at which the sphere impacts the member, \( \omega_2 \) is the angular velocity of the member ABC, and \( v_1 \) is the initial sphere velocity.

A higher coefficient implies a more elastic collision where kinetic energy is better preserved. Here, the given value of 0.5 tells us the collision is partially elastic. This value allows us to solve for the relative speed of separation and approach, aiding in finding the sphere's velocity after the impact.
Impact Dynamics
Impact dynamics study how forces and contact during collisions affect speed and motion. It considers factors such as material properties, velocity, mass, and geometry involved in interactions. This field enables the prediction and control of post-collision behavior.

In the context of member ABC struck by sphere D, understanding impact dynamics involves recognizing how the force exerted by the sphere translates into rotational motion of the member. Physical quantities such as moment of inertia become critical since they influence how angular velocity develops.

During the impact, momentum transfers from the smaller sphere to the larger structure, transforming part of its linear momentum into rotational form in member ABC. These dynamics explain the intricate interplay between translating and revolving bodies. They offer predictions about the system's response to different collision conditions, essential for engineering and safety evaluations.

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Most popular questions from this chapter

A slender rod of length \(l\) and weight \(W\) is pivoted at one end as shown. It is released from rest in a horizontal position and swings freely. ( \(a\) ) Determine the angular velocity of the rod as it passes through a vertical position and determine the corresponding reaction at the pivot. (b) Solve part \(a\) for \(W=1.8\) Ib and \(l=3\) ft.

The flywheel of a punching machine has a mass of 300 kg and a radius of gyration of 600 mm. Each punching operation requires 2500 J of work. (a) Knowing that the speed of the flywheel is 300 rpm just before a punching, determine the speed immediately after the punching. (b) If a constant 25-N?m couple is applied to the shaft of the flywheel, determine the number of revolutions executed before the speed is again 300 rpm.

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A 200-kg flywheel is at rest when a constant 300 N?m couple is applied. After executing 560 revolutions, the flywheel reaches its rated speed of 2400 rpm. Knowing that the radius of gyration of the flywheel is 400 mm, determine the average magnitude of the couple due to kinetic friction in the bearing.

A bullet weighing 0.08 lb is fired with a horizontal velocity of \(1800 \mathrm{ft}\) s into the lower end of a slender \(15-\mathrm{lb}\) bar of length \(L=30 \mathrm{in}\). Knowing that \(h=12 \mathrm{in}\), and that the bar is initially at rest, determine \((a)\) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at \(C,\) assuming that the bullet becomes embedded in \(0.001 \mathrm{s}\).

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