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Atwood's machine. Figure 10.73 illustrates an Atwood's machine. Find the linear accelerations of blocks $A$ and $B,$ the angular acceleration of the wheel $C,$ and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks $A$ and $B$ be 4.00 $\mathrm{kg}$ and 2.00 $\mathrm{kg}$ , respectively, the moment of inertia of the wheel about its axis be $0.300 \mathrm{kg} \cdot \mathrm{m}^{2},$ and the radius of the wheel be 0.120 $\mathrm{m} .$

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0.73 $\mathrm{m} / \mathrm{s}^{2}$6.09 $\mathrm{rad} / \mathrm{s}^{2}$36.3 $\mathrm{N}$$21.1 \mathrm{N}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

University of Michigan - Ann Arbor

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02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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Atwood's Machine. Fig…

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Atwood's machine. Fig…

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The Atwood's Machine.…

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$\textbf{Figure P10.59}$ i…

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EXERCISE [010.59 The A…

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The figure below illustrat…

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An Atwood's machine c…

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(III) An Atwood's mac…

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An Atwood machine is shown…

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An object of mass $M=$ 12.…

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A $12.0-\mathrm{kg}$ objec…

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A $12 \mathrm{kg}$ mass is…

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An Atwood machine (Fig. $1…

Hello Problems 63. That's about an Atwood machine that has a mass be here and a mass A on the right side. Now our masses. We won't look at the individual to three bodies. Nasty is two kilograms, which means it's wait. It is 19.6 now, thinking from our mass. We know it is held up by tension in this string here, which is what's pulling on our pulley on our A. This is four kilograms, which is 39. What to Newton's That wait is held up by a string of tension A, which is the other side of our Atwood machine. Atwood machine pulley is 0.12 meter radius with a moment of inertia 0.3. So we will look at this and look at our net Force is on both of our objects and then we'll look at our net force on our polling. So one must be. We know our net force is in the wind direction are causing the acceleration and same thing goes here for a mess. Hey, differences. Since Macey's larger, our acceleration is going to be clockwise about this pulleys. So our way will be greater than our attention here that attention will be greater than our weight in our mass. Here is to We're here. Are Mass is for we actually saw both of these for tensions. We get 19.6 plus 2 a.m. We get 39.2 minus for it now. Taking a look at our torques. We know that our networks o r I maker now r t a torque will be larger than our TB Twerk Both have the same mom lever arm the radius. But our ta is larger because that tension would be larger. You know, we remember our moment of inertia is three mother saying we need to look at is how is it angular in translational acceleration related in this case? Aye, over on ours 0.12 so we can put in a over 0.12 here, please. 2.5. Hey, what gonna help us out here? Factor that out. A little simpler equation. However, we still have three variables in it. Going back here. We can take what we know already for TV and t A and sub in. Over here, we end up with just being able to solve for acceleration no, Uh, ta we know is 39.2 minus for a and minus TV, which is 19.6 plus to it that is equal to 2.5. Careful with distributing that negative people your positives and negatives in the right spot. Let's go through the algebra. Here you will get an acceleration of 0.73 meters per second squared. Also, we confined our angular acceleration by divided by the radius and that will give us an angular acceleration of 6.9 radiance per second squared that gets both of our accelerations. Other part of our question asked us for what is ta in? What his TV. Since we know now, our acceleration is 0.73 We can sub in both those here and we would get TV to be 21.1 new and t a 36.3. That makes sense. Since we're moving here, our attention we're moving down. Distension should be greater than that Tension. Thank you for lunch.

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