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A machine part consists of a thin, uniform $4.00-\mathrm{kg}$ bar that is 1.50 $\mathrm{m}$ long, hinged perpendicular to a similar vertical bar of mass 3.00 $\mathrm{kg}$ and length 1.80 $\mathrm{m}$ . The longer bar has a small but dense $2.00-\mathrm{kg}$ ball at one end (Fig. 8.40$) .$ By what distance will the center of mass of this part move horizontally and vertically if the vertical bar ispivoted through $90^{\circ}$ to make the entire part horizontal?

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$x=0.70 \mathrm{m} \quad$ and $\quad y=0.70 \mathrm{m}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

09:31

A machine part consists of…

07:00

03:54

A uniform metal bar is 5.0…

03:15

Two bowling balls are at r…

01:55

04:02

00:45

04:57

A $5.00$ -kg ball is dropp…

01:11

A $240-\mathrm{kg}, 50.0 \…

03:20

A uniform bar has two smal…

A thin uniform bar has two…

04:21

03:16

A uniform plank with a len…

12:11

A thin, uniform $3.80-\mat…

in this question, we have a mechanism that is composed off two arms. One arm has a mass of 4 kg and the length off 1.5 m and another arm has the mass off 3 kg and the length off 1.80 m. At the end of the second arm there is a ball with a mass off 2 kg. Now these two arms are connected by a Hinch which is considered to be massless. Then, under this configuration, the center of mass is at some position. Now let us pivot counterclockwise this arm so that the mechanism now looks like this. When the mechanism looks like this, the center of mass, we'll move somewhere else. So say before it was here, now it is. Maybe here. Then we have to calculate what is the change in the position off the center of mass in both directions, vertical on a horizontal. We begin this question by calculating what is the center of mass in the mechanism before people 10 counterclockwise. For that we have to choose a reference frame. I would choose the hinge as my reference frame. So this is why you call zero X equals zero. Additionally, I would choose my access like that. So everything that is pointing upwards is positive in the right direction, and everything that is pointing to the right is positive in the X direction. This is also true for these other situation. Everything that is pointing upwards. It's positive and everything that is pointing to the right. It's positive now, before performing any other calculation. Let's locate one of the center of mass is off this arm, this arm on the ball. So the center of mass off this arm is located exactly in the middle of this arm. And why is that? Because the arm is considered to be very, very thin and also uniforms. The same thinking is true for the second arm, so the center off mass off the second arm will be exactly in the middle off the second arm just right here. Now, the center of mass off the ball is the easiest one, because the ball is point like to reform the center of mass is the position off the ball, drawing the center of mass on the second situation. We have something like that and we already know the positions off the centers off mass. So here we have 0.75 m to the left, 0.90 m downwards on 1.8 m downwards. On the second situation, we have 0.75 m to the left. Here we have 0.90 m to the right and here we have 1.18 m to the right. Now we can calculate where is the center off mass off the world mechanism before and after pivoting So before pivoting. The exposition off the center of mass is given by the mass off this arm, which is 4 kg times the exposition off its center of mass, which in our reference frame his minus 0.75 plus the mass off the second arm, which is this one and that mass is 3 kg times the exposition off the center of mass Off that arm, the exposition off the center of mass off that arm is the cost of zero because that arm is oriented along the y axis, and finally we have the mask off the ball times the exposition off the center off mass off the ball again. It's just zero because the ball is oriented along the Y axis. So here we have zero again. Then we divide this by the total mass off the mechanism which is four plus three plus two. These results in the following the exposition off the center of mass is equals to minus four times 0.75 which is equals to treat plus zero plus zero divided by nine. Then the exposition off the center of mass off the mechanism before pivoting is given by minus one third, which is approximately minus 0.33 m to the left, doing the same for the white position off the center of mass, we get the following the white position off the center of mass is given by four times the white position off the center of mass off this arm, which now is equals to zero because this arm is oriented along the X axis. So here we have zero plus the mass off the second car, which is 3 kg times the white position off its center off mass, which is minus 0.90 m. So here we have 0.90 with a negative sign in front of it, plus the mass off the ball times the white position off the center of mass off the ball, which is minus 1.8. And this is divided by the total mass which you already know that is equal to nine. Then the white position off the center off mass is given by zero minus 2 70 minus 3 60. This is divided by nine on these results in minus 6.3, divided by nine, which is equal to minus 0.70 m. Now we do the same calculation, but for the mechanism after pivoting counterclockwise, the exposition off the center of Mass is given by four times minus 0.75 which is the position off the center off mass off this piece after people 10 counterclockwise plus the mass off the second arm, which is 3 kg times the exposition off its center of Mass, which is now 0.90 plus the mass off the ball times. The exposition off its center off mass, which is 1 80 on this wall thing, is divided by the total mass, which is nine then the exposition off the center off Mass is given by minus three plus 2.7 plus 3.6 divided by nine and this is equals to 3.3, divided by nine, which results in approximately 0.37 m. Now the white position off the center of mass is a straightforwardly equals to zero because all pieces off the mechanism are oriented along the X axis. So the exposition off the center of mass off all of the pieces is equals to zero. The reform the center of mass off the world. Peace must be equal to zero. Now we can give our answer. So what is the change in the position off the center of mass? The change in the exposition off the center of mass is given by the final exposition 0.37 minus the initial exposition which waas minus 0.33. These results in 0.37 plus 0.33 on this is E goes to 0.70 m. So the center off mass moved 0.7 m to the right. Now for the vertical axis, we have the final position minus the initial position which is minus 0.70. Therefore, the change in the vertical position off the center of Mass waas 0.70 m to and this is the answer to this question

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