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(II) Two blocks, with masses $m_{A}$ and $m_{B},$ are connected toeach other and to a central post by cords as shown inFig. $46 .$ They rotate about the post at frequency $f$(revolutions per second) on a frictionless horizontal surfaceat distances $r_{A}$ and $r_{B}$ from the post. Derive an algebraicexpression for the tension in each segment of the cord(assumed massless).

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$$4 \pi^{2} f^{2}\left(m_{\Lambda} r_{\Lambda}+m_{\mathrm{B}} r_{\mathrm{B}}\right)$$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

University of Sheffield

McMaster University

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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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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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so consider a free body diagram for both masses from a side view at the instant that there to the left off the post. So that means if the post is on the left then we have Sorry If the masses are on the left of the post then we have ah detention force. Then we applied on m A. Let's mark that as if off t a. So here on the right we have the post then. Since both of the masses have the same frequency over patient, that means they will always stay in the line and ah, from mutants. Third Law, We know that since they're connected, the tension between M a n N b will be equal. Let's call that force f off TB because as we mentioned that the tension must be equal. So if we call ftv for Massa be then that must be equal to mass in a So let's call that ftv then for M V. We have m B G or the weight off weight which is acting vertically downwards to counterbalance that we have the normal force since there's no motion in the vertical direction. Let's call that the again similarly in M A. Will have to force is one is M A G and the other is f off an eight. Now we can solve Newton's second law, uh, for each masses individually. So for mass m A. We have use a different color, So yeah, for mass m A. We have f off our A, which is basically the force on the radial direction. So if here is the Post, then this is this FDA is acting radial Ian Ward's so that's gonna be f off t a minus riff off TV. And that's responsible for the centripetal accelerations. Ex elation times the mass where by definition, we know that centripetal acceleration is basically velocity over our velocity squared over r Also notice that we took from left to right as positive x direction now, um far and B if off r b must be equal to just ftv, which is again equal toe m b times a b on by definition, that's gonna be envy. B B squared over R now. Ah, the speed can be expressed in terms of frequency as follows. So speed is, uh, frequency times, ref bar second, which is nothing works then we multiply that with two pi ours by one Rev. So from here we see that velocity is equal to pi are times if so Now we know the connection between frequency and ah, Velocity says it will be easy to solve for FT. ES and F TV. So we want to solve FDA and F TV individually. So we know that F t B is M v V squared over R. So let's solve that first in terms of frequency, so F t B is equal. D'oh and B B B squared over r. Let's call it R B because no, the radius has changed. So let's do the same thing here. It's R B. This is our A um Now from here, this is gonna be m times two pi r b f. But a square They're divided by R V is equal to for pi squared and B R B excluded now for FT. A. We know that that's equal to F off TV plus M v. A squared over R A, which is equal to four bys cleared M b R B f squared. Plus I am off to buy R a X squared divided by our eight combining everything together. We have four pi squared, X squared times, M A r A plus and B are big Thank you.

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