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If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$ \mathbf{F}=m\cdot \mathbf{a} \tag{01} $$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...


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Note that you have to swing the pail with a certain minimum speed for the water to stay in. That minimum speed is such that when the pail is at the top of the arc, the rope accelerates the pail downward faster than gravity accelerates the water downward. Otherwise, the water falls out.


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If water particles move in a circular path its because of some net force towards the centre. This net force is usually called "centripetal force". Don't ever put centrifugal force into the description. It is not a force, but just a name of the "feeling" that your body (or in this case water particles) want to move out of the circular motion but can't.


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The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...


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First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...


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Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...


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So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis". That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and ...


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I think the error occurs where you state : Now because we found that both disks have the same amount of kinetic energy (and the same mass), that means that they have the same translational speed. (In fact, my professor also did a demo of this in class and we observed that they did have the same speed). The two discs have the same total KE but ...


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The angular velocity is related to the tangential speed $v_t$ by: $$ \omega = \frac{v_t}{r} $$ However the velocity $v$ being used here is not the tangential speed. It is the linear velocity $v_l$ of the centre of mass of the disk relative to the floor. The question has chosen a value for the angular velocity of the disk that appears similar to the ...


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$\omega=\frac{v}{r}$ is used for rotating without slipping. Your disc is rotating and slipping. So, it is possible that $\omega=\frac{v}{3r}$.


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I am not sure what you actually want to ask. So I would recommend that you put more effort into your question. Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange: Say the toque applied to the joint of the scissors is $\tau$. What is this force $F$, then? The distance of the rope contact point to ...


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Rolling of a circular body, on a flat surface, and without sliding results in $v_\text{tan}=v_\text{cm}$ where $v_\text{tan}=ωr$ is the tangential speed of any point on the rim of the body in the center-of-mass frame of reference. This is understood by studying the motion in the c.m. frame: there, the flat surface has velocity $v_\text{cm}$ (backwards). The ...


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Are you confused about how you get into this? $$\begin{align} v & = \omega\,r \\ ({\rm m/s}) & = ({\rm rad/s})\,({\rm m}) = ({\rm m\,rad/s}) \end{align} $$ Radians are not units with dimensions. They can be seen as $({\rm rad}) = ({\rm m/m})$ like arc length to radius. This makes to above right hand side equivalent to the left hand side.


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This is all true because the ratio of a circle's diameter to circumference is constant (pi). The unit of radian is actually chosen so that l = r.theta. So if you go theta radians around a circle you travel r.theta. For the angular to linear velocities, think of a disc rotating at an angular speed omaga. then the further out from the centre of the disc you ...


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@Prayas Agrawal No I'm not missing that - I'm just explaining the original error. Of course the total work = the KE. In fact since the floor is frictionless, the disc remains stationary and just spins since the force has no line of action through the CoM. In the case of friction the force of the weight of the disc provides a torque on the ground via the ...


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"As we can see from the picture, both disks have the same force being applied to them and they also go the same distance d→" This is the erroneous assumption - the 2 discs do not go the same distance. Some of the distance that the rope is pulled will rotate disc 2 as it unravels. As a result the liner distance is less and the balance of work goes into ...


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While you are correct saying disk 2 has rotational kinetic energy, you are missing that no matter the situation, since ground is frictionless, the work done be external force(F, in this case), is same in both cases. Thus by work energy theorem, $$Work=change in KE$$.Thus since work in case 1 equals work in case 2 thus both disk have same kinetic energies.


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When connected to the centre of mass of disc 1 the force causes an acceleration of the centre of mass and the work done by the force is $Fd$ where $d$ is the displacement of the centre of mass and the force $F$. The translational kinetic energy of the disc increases by an amount $Fd$. When the force acts on the rim of disc 2 the centre of mass of the disc ...


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When youare talking about angular momentum about any axis(inluding one that passes through center of mass), you carry out the "formula":$\vec{L}=\vec{L_{com}}+I_{com}\vec{\omega}$ where direction of $\vec{L_{com}}$ and $I_{com}\vec{\omega}$ is to be kept in mind, during vector summation.While we calculate $\vec{L}$, about COM, $\vec{L_{com}}$ becomes zero ...


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When you talk about a rigid body the distance between any two points is fixed. That implies angular velocity w is same for all points. Now ony moment of inertia changes: I=Icm+md^2 [d is distance between cm and the point] This is the parallel axis theorem. Get I and angular momentum=Iw


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The angular momentum will be different: however, you will be able to calculate it with the parallel axis theorem. Measure the distance from your new axis to the center of mass and call it $d$. Your new rotational inertia $I$ can be calculated from the rotational inertia around the center of mass $I_{\mathrm{cm}}$ using the formula: $$I = I_{\mathrm{cm}} + ...


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The proof is very simple and comes right form the definition. let $R(t)$ be a rotation matrix as a function of time. $R$ is an orthogonal matrix so its inverse is equal to its transpose: $I = R(t)R^T(t).$ ($I$ is the identity matrix) The time derivative of the above equation is $0 = \frac{d[R(t)R^T(t)]}{dt} = ...


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The 1.5 m radius you'll use for the moment of inertia. The 0.1 m radius of the axle will relate the linear speed to the angular speed.


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But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ...


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I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since $\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$ and $\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$ the components of ${\rm d}u$ are perpendicular to $u$. Our first step is to choose cartesian axes, ...


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Assume that your external force $F$ was applied at some distance $r$ from the centre of mass of the spool. To satisfy the rolling condition the linear acceleration of the centre of mass $a$ must equal the radius $R$ of the spool times its angular acceleration about the centre of mass $\alpha$. $a = R \alpha$ Just suppose that there is no friction and ...



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