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Yes I agree, the solution in the manual misses out some terms in the equation of motion, as well as the factor $\frac12$ in front of $k$. Probably some approximations have been made without explanation. Your equation seems correct to me. When the ball arm is vertical, and there is no rotational motion in the vertical plane, then balancing moments about O ...


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No - any simultaneous rotation on multiple axis is identical to a single rotation along another axis. It is similar to moving straight along two different vectors at the same time - that is just a movement along a new (combined) vector. The same is true for rotation, it is just not as easily imaginable.


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You have a few questions here. They mostly concern the energy loss due to friction. In this problem there is no energy loss because of friction. Friction is present between the rope and the pulley. This is static friction. It makes the pulley rotate. Energy is lost due to friction (dissipated as heat) only when there is relative motion between the ...


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For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...


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Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...


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There is another issue that must be clarified before making any definite conclusion. The question is:"Is rigid body in free state or is it constrained"? This question must be asked when dealing with rotors constrained by gravity in the bearings, constituting a rotor/ bearing system. Then next question arises:"Does this system has a resonance (natural motion) ...


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First analysis is correct. Because current problem is a rigid body dynamics problem and you cannot model a rigid body by a single particle. If it was possible, there was no need to rigid body dynamics. If you want to solve the problem by particles dynamics, you must model the rigid body by infinite number of particles. So, reduction of potential energy of ...


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If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \...


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The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ...


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The torque supplied by $F_m$ results in the torque due to $F_v$, so these torques are equal : $\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either $\vec A \times \vec B = (AB \sin\theta) \hat k$ where $A$, $B$ are magnitudes and $\theta$ is the angle between (b) or $(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ...


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You can easily find direction of friction force by drawing free body diagram for the cylinder. FBD of the cylinder is shown below (Note that I have drawn diagrams for counter clockwise rotation) As you see in figure above, friction force $f$ opposes to rotation of the cylinder. Equations of cylinder motion are as below $$f=ma_G\;\tag 1$$ $$N=mg+F\;\tag 2$...


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If the cylinder is not accelerating and the rolling no slip condition $v_{cm} = R \omega$ is satisfied then the frictional force between the cylinder and the ground is zero. The horizontal force on the cylinder is zero as it is not accelerating in that direction. The weight of the cylinder is equal an opposite to the normal reaction pn the cylinder due to ...


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galaxy come in many different sizes: some of the small-er ones do rotate ["orbit"] around the edge of a large galaxy ... one can also visualize galaxy-clusters, in which the entire cluster rotates .....


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Although you do not need it you have miscalculated the position of the centre of mass of the arrangement of squares. About corner $O$ the two left-hand squares produce a anticlockwise torque whilst the right-hand square, the normal reaction of the squares due to the ground $N$ and the force $F$ all produce a clockwise torque about corner $O$. Without the ...


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The object will topple about the lower RH corner. Take moments about this corner. The net clockwise moment must be > 0 for the object to topple. There is no need to find the CM. You can calculate moments for each cubical component of the object.


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You claim the horizontal distance of the CoG ($X$) from the bottom left corner is $\frac{5a}{12}$ (I've not verified this). That is the distance $|OP'|$ in my version of the diagram. The distance $|PP'|$ is therefore: $$|PP'|=\frac{a}{2}-\frac{5a}{12}=\frac{a}{12}$$ When you start exerting the force $F$, a torque about the point P, which is $F\times \frac{...


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You need to consider the moments of the forces involved. Moment is force times perpendicular distance from the axis. In order to turn the dumper bucket, the clockwise moment of the force from the hydraulic cylinder needs to be at least as much as the anticlockwise moment of the bucket's weight (when loaded). The weight of the bucket acts through its centre ...


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The body will start to rotate when the torque is barely above zero. As you apply the force, the normal force made by the floor will move to the right to prevent rotation, until it reaches the right corner of the bottom square. After that it cannot keep moving. Thus to solve the problem make force diagram to get $N$ in terms of $F$, and impose a torque equal ...


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if the body is rolling on a plane, then its degree of freedom is 2: one for rotation about the body's axis and one for translation of its center of gravity in forward and backward direction. if there is no slipping, the translation can be calculated from the rotation and the radius of the block. Thus the degree of freedom is degenerated to 1.


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Is vertical uniform circular motion even possible? No, it isn't. Because magnitude of velocity isn't constant and we know that in a uniform circular motion the object moves with constant speed. $\large{\frac {\mathrm d}{\mathrm dt}}v=g\sin\alpha\neq 0$ ($v$ is the speed (magnitude of the velocity vector $\vec v$) of the object) Is this analysis ...


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The total torque about some axis is defined as: $\vec{\tau}_\text{tot}=\Sigma \left(\vec{r}_i \times{}\vec{f}_i\right) $ If you change to a parallel axis located $\vec{r}$ away from the first one on the plane of the forces, the new torque will be: $$\vec{\tau}_\text{tot}'=\Sigma \left[(\vec{r}_i+\vec{r}) \times{}\vec{f}_i\right] =\vec{\tau}_\text{tot}+\vec{...


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The answer is no. The total torque on the system given in terms of the center of mass coordinates is $$\vec\tau=\vec R_{cm}\times\vec F^{ext}+\sum_i\vec r_i'\times\vec F_i^{ext}$$ where $M$, $\vec R_{cm}$ and $\vec F^{ext}$ are respectively the mass of the system, the center of mass position vector and the total external force on the system. The vector $\...


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Any rigid body is composed of particles bonded together. This bonding allows forces applied on one part to transferred to all other parts of the body. Thus when a net force is applied, all the particles will translate together at the same time. We describe this motion is a translation of the center of mass. In addition, the bonded particles try to maintain ...


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Consider the midpoint of the rod that divides the rod into a left half and a right half. We are applying a force to the right end in the upward direction. Now consider the left half. It has to (overall) go in the direction of the force applied - the rod doesn't break, right? So if we think of just the left half then there must be some net force on it in the ...


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It is not clear what your problem is. You say that you understand intuitively what will happen, and that Farcher's answer comes close to explaining it for you. What exactly do you still not understand? You ask about picturing the individual particles of the rod. Lucas and Bruce address that. But you also say If I cause part of the rod to move up, ...


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I think the confusion is that a single push at the corner doesn't "feel" like you're "doing a rotation". Here's an explanation of why it must be, by symmetry. Suppose that pushing it up on the right end, as shown in your diagram, gives the rod angular velocity $\omega$ and linear velocity $v$. (We have not assumed $\omega \neq 0$, we're going to show that.)...


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It is easier to explain in terms of torque and angular momentum, but it can also be seen with some difficulty using only internal forces. Imagine that instead of a continuous rod you have instead a bunch of particles attached by cords. If you push the mass at the edge this one will tend to move up first, but due to the cord will also try to rotate around ...


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Yes, it rotates but it's center of mass (i.e. The full rod) also translates. The force causes ma on the center of mass, and rotation around the center of mass from torque = I x Omega. You need both equations to get the motion. This is physics 101


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Assume that the rod is made by some particles (or molecules). The force $F$ is acting on one particle of the rod as you see in the figure above. If we isolate that particle, we will have figure below (free body diagram): Force $f_1$ is applied by the particle next to the first particle. As you see in the figure, there is a distance between $F$ and $f$ ...


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Consider to figure below (there is no friction): If we calculate the net torque about point $\textrm O$, we will have: $$\Sigma M_{\textrm O}=-FL$$ Can we say that the block will rotate about point $\textrm O$? Yes, we can. But, this rotating isn't equal to rolling. This rotation is a particle rotation about a point. Because particles of the sphere don't ...


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The question answered by @SatwikPasani is related but not quite a duplicate. The apparent paradox is resolved by realising that using a frame of reference relative to the sphere which is accelerating down the slope is a non-inertial frame of reference. If there is friction and the no slipping condition is satisfied then the frame of reference attached to ...


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In such a hypothetical situation in which there is no friction between the sphere and plane, there can be no tangential force acting on the sphere, and hence no torque. The only force acting on the sphere would therefore be its weight, and the component of that force acting perpendicularly to the plane would be responsible for its translation down the plane, ...


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In this case, the rod cannot experience only pure translation. Because resultant torque acting on the rod isn’t equal to zero. If acting points of the forces are fixed, the rod will rotate counter clockwise until it is parallel with the forces. So, until that final state of the rod is established (without damping forces, this state will never be ...


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It is not clear to me exactly what you are asking for : a derivation such as given by Lawrence B Crowell, or a recommended reading list either for the mechanics of macroscopic bodies or for the microscopic properties of materials. Your question "how does it know?" seems rather a naive one for an electrical engineer to be asking, because you must have asked ...


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Think of a single particle with mass $m$ at $\vec r$ from a coordinate origin. Now suppose this body is in circular motion with angular velocity $\omega$ around this origin. The momentum of the particle is then $$ \vec p = m{\vec v} = m\frac{d\vec r}{dt}. $$ The velocity is then $\vec v = \vec\omega\times\vec r$. Now let us consider the angular momentum of ...


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Your problem cannot be solved without further information. This is because the board and pins in your model are ideal, not real : they are perfectly flat, rigid and inflexible. As an ideal problem it is statically indeterminate : there are infinitely many solutions which are compatible with the given conditions. It is possible to increase some reactions ...


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Yes. An easy way to think about this, is that the mass moves as a whole. So when balancing, it doesn't "know" whether the triangular space above the left weight is filled or not. The attachment point doesn't matter. It could even be attached to the right of the support. What matters in mechanical terms is the total moment around the support (it'll rotate ...



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