A ball of mass 2 kg is moving with a speed of +6 m/s directly towards

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0

2 years ago

(c) If η = 60% and TC = 40°F, what is TH, in °F?

arlik [135]

2b2t hope that helps

5 0

3 years ago

Explain the difference between balanced forces and action and reaction forces.

Monica [59]

Action-reaction forces<span> act on different objects; </span>balanced forces<span> act on the same object. </span>Balanced forces<span> can result in acceleration, </span>action-reaction forces<span> cannot. ... Newton's Third Law of Motion does not apply to </span>balanced forces<span>.</span>

5 0

3 years ago

As gravity continues to crush the inner core, the fusion of carbon and oxygen begins. What temperature is needed for the fusion

Olin [163]

Answer:

600,000,000 degree C

Explanation:

This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.

The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.

8 0

3 years ago

A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling

Thepotemich [5.8K]

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

7 0

3 years ago