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(III) An Atwood's machine consists of two masses, $m_{\mathrm{A}}$ and $m_{\mathrm{B}}$ which are connected by a massless inelastic cord that passes over a pulley, Fig. $57 .$ If the pulley has radius $R$ and moment of inertia $I$ about its axle, determine the acceleration of the masses$m_{\mathrm{A}}$ and $m_{\mathrm{B}},$ and compare to the situation in which the moment of inertia of the pulley is ignored. [Hint: The tensions $F_{\mathrm{TA}}$ and $F_{\mathrm{TB}}$ are not equal. With the Atwood machine, assume $I=0$ for the pulley.]

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$a=\frac{\left(m_{\mathrm{B}}-m_{\mathrm{A}}\right)}{\left(m_{\mathrm{A}}+m_{\mathrm{B}}+I / R^{2}\right)} g$$a_{I=0}=\frac{\left(m_{\mathrm{B}}-m_{\mathrm{A}}\right)}{\left(m_{\mathrm{A}}+m_{\mathrm{B}}\right)} g \mid$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Washington

University of Winnipeg

McMaster University

Lectures

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.

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.

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An Atwood's machine c…

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(III) An Atwood's mac…

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(III) An Atwood machine co…

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An Atwood Machine consists…

An Atwood machine with pul…

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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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(III) The double Atwood ma…

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Double Atwood's Machi…

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Double Atwood's Mach…

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Two masses are suspended f…

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06:59

An Atwood machine is shown…

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(III) Two blocks are conne…

here is the diagram of the system. We're going to assume that the Massa be is greater than the mass of A and we know that this system will essentially have ah, static equilibrium as well as translational equilibrium in both the Y and ex directions. So we can first say that here amass sub they will accelerate up and Massa be will accelerate down. We can also then say that the ex angular acceleration of the pulley would be equaling a the acceleration of the linear acceleration of the system multiple, rather divided by are the radius of the pulley. And so we could apply Newton's second law some of forces in the wider action for Block A. This would be equal. It is to be equal to the force tension someday minus M sub a. G and this would be equal to the mass sub eight times the acceleration of the system. We know that then the sum of forces in the UAE direction for Block B. This would be equaling m sub b g minus the force tension so be and this would be equal to the massive block B times the XL linear acceleration of the system. We can then solve and say forced tensions of a will be equaling M sub a times G plus I'm sub eight times a and forced tension said B will be equaling. Um, sub b times G minus m should be times, eh? And so now we can apply the new apply the sum of works knowing that this is this has, um, static equilibrium. So this would be the force Tension said The times are minus force. Tensions of eight times are equaling the moment of inertia of the pulley multiplied by the angular acceleration of the pulley. And this is equaling the moment of inertia of the pulley times. The linear acceleration of the system divided by the radius are of the pulley. And so we can say that we're going to substitute the tension expressions into the torque equation. And so we can say that forced tension B R minus force attention sub a r. Equaling the moment of inertia times the linear acceleration of the system divided by our and so we have that and sub B G minus times B times eight times are minus. I'm sub eight times G plus temps of eight times a multiplied by again are this is equaling. I multiplied by a divided by our and here we know that we're simply going to solve for a so the acceleration of the system would be equally the difference of the masses. M sub d minus m step a time The acceleration G and then divided by the mass of a plus the mass of B plus the moment of inertia of the police divided by the radius of the pulley squared. Ah, weakened then say that if the moment of inertia is ignored than from the torque equation, we see that the force tension said B equals the force tensions of a And so this would be the answer if the moment of inertia cannot be ignored. But if it can be ignored, we can then say that the acceleration of the system with the moment of inertia equaling zero. This would simply be the difference of the masses and sub d minus. M. C. A times again g divided by the sum of the masses and said a plus m sub B and so this would be your answer. If the moment of inertia can be ignored, that is the end of the solution. Thank you for watching

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