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If the dipole is small enough, then the force on dipole would be: $$\vec{F}=\nabla(\vec{p}.\vec{E})$$ and consequently the torque would be: $$\vec{F} \times \vec{r}=\nabla(\vec{p}.\vec{E}) \times \vec{r}$$ where r is the length of the dipole


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The torque $ \tau $ on an electric dipole with dipole moment p in an electric field E is given by $$ \tau = p \times E $$ where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment. When p and E are parallel and anti-parallel, the torque is zero, so yes zero is possible. But the case in which p and E are anti-...


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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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Potential energy change of mass $m$ is equal to kinetic energy change of system ($m$ and $M$) plus wasted energy due to friction between spool and axle. $$mgy=\frac 12mv^2+\frac 12I\omega^2+W_f\tag 1$$ $\omega=\large{\frac{v^2}R}$ As the tension force of the string is constant, then the net force acting on mass $m$ will be constant ($F_{\textrm{net}}=mg-T$)....


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If the blade was spinning in a vacuum (and there was no friction and no motor), there would be no torque. The blade could spin by itself forever. Torque comes from spinning in air and pushing air around. Air is viscous. It resists the motion of the blades. It exerts a torque on the blades. The torque on all 4 blades is the torque on the quadcopter. ...


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First of all, I shall assume that the system is kept on a horizontal plane so as to simplify the calculations. Secondly, I shall assume that the line joining the mass $m$ and $M$ is perpendicular to the direction of motion of mass $3m$. Thirdly, I shall assume that the spheres to be point objects (in other words, you can consider it to be a perfect head-on ...


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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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You are way off in your thinking, and your 3 'tries' show that you are blindly guessing what is going on here. I think you need to go back to the basics of torque and rotational motion. The question asks you to find the angular acceleration $\alpha$ about the centre of mass (CM). This is related to net torque $\tau$ and moment of inertia $I$ by $\tau = I\...


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They ask you for the angular acceleration $\alpha$, not the linear acceleration. I could not understand what you did, but you know that torque $\tau=I \alpha$, to find $\alpha$ only divide the torque you have by the moment of inertia $I=mL^2/12$ which corresponds to the moment of inertia of a rod around its center of mass.


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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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For the first figure: You are misinterpreting the second law, it only says what happens to an object that is subject to a net force. In this case, if the impact force is larger than the friction, the object will accelerate. The second law doesn't say what happens to the object that produces the force. Now, the third law does, which means that the actuator ...


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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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The force is not magnified, is the torque that is transmitted; and is not an "amplification effect" of the interaction between atoms. The only role the interactions between atoms play is to hold the lever together. And as long as the forces involved using the lever, are smaller than the internally bounding forces of the lever, it will resist the strain and ...



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