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The main condition for Conservation of Angular momentum is that When the net external torque on a system is zero, then there is no change in angular momentum of the system. You mentioned the equation that led you to the desired result. If you focus on the equation carefully, you will notice that the "$mvr$" term is the angular momentum of the ...

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The external torque for the system consisting of the boy, the ball and the platform is $0$. The torque you mentioned is an internal torque. Notice that the angular momentum conservation equation includes the angular momentum of all the 3 components mentioned earlier. Internal torque cannot change the angular momentum of a system as internal torque is ...

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In general, yes the vector equation you wrote is correct, that is $\sum_i\vec{\tau}_i= I \vec{\alpha}$ (where $\vec{\tau}_i$ are the potentially multiple torques acting on an object, and $\vec{\alpha}$ is the angular acceleration of that object). However, as d_b noted in the comment, $\alpha_z$ is the component of a vector and not a vector itself. If you ...

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Same thing! Just replace your force $F$ by the desired torque $M$ to get a point moment.

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It should be noted that $F \delta(\boldsymbol{r})$ is not the force, but is called the Force Density (dimensionaly as well as physically), simply because integrating this over the volume gives you the force. Now, Torque is defined as $r \times F$ where $r$ is the distance from the axis of rotation to the point of application of force. Now to have $r$ you ...

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Quoting David Morin : Let the position of the origin be $\mathbf{r}_{0}$ (see Fig. 8.16), and let the positions of the particles be $\mathbf{r}_{i}$. The vectors $\mathbf{r}_{0}$ and $\mathbf{r}_{i}$ are measured with respect to a given fixed coordinate system. The total angular momentum of the system, relative to the(possibly accelerating) origin $\mathbf{... 2 In the section "torques relative to the right edge of the bar:" The torque from the$N$should be included...best of luck getting the two ways to match. 2 The first solution is wrong. When you calculate the torques relative of the right edge of the bar you forgot about the torque of force$N$. The correct equation should be $$f\cdot L \sin\theta + N \cdot L\cos\theta = P \frac{L}{2}$$ and then you'd get the result that agrees with the second solution. 1 In my opinion, the best definition of torque is the work per unit angle which can be done by a force which is acting in a manner that tends to cause a rotation. If motion occurs then τ = +dU/dθ where U is the total energy of the rotating system (not just the potential energy). Your expression refers to the torque coming from an external source, such as a ... 3 Short answer is yes, as long as your system is conservative. Longer answer is that the Euler-Lagrange equations tell you what the equations of motion really are for these systems, so you setup the Lagrangian$L(r, \theta,\dot{r},\dot{\theta}) = T(\dot{r},\dot{\theta}) - U(r,\theta)$, where$T$is the kinetic energy and$U$is potential energy. Then calculate:... 0 A hinged body is constrained to rotate about the hinge and therefore needs to generate reaction forces in order to accelerate the center of mass. The only time a body purely rotates about the center of mass is when a pure torque is applied, or the net forces acting add up to zero. Consider a hinged body whose center of mass is located at$\boldsymbol{c}$... 0 If the object has a finite mass, it also has a finite size. The horizontal string cannot bring the center of mass to a radius of zero. -1 If you will see it from the centre of the mass frame then it will seem as if it is rotating around you and the frequency will be the same as frequency doesn't depend upon the frame of reference.In fact, this would be true for any point on the disc. I am taking into account that alpha=0 -1 Newton's Second Law for Rotation reads$\vec{\tau} = \frac{d\vec{L}}{dt},$where$\vec{L}=\int \vec{r}\times\vec{p}$. Newton's Second Law holds in any inertial frame for all objects. The reason why we think of two angular momenta separately in Gyroscope precession is that the expression for$\vec{L}$can be simplified as$\vec{L}=\vec{L}_{p}+\vec{L}_{cm}$... 2 If there is no friction, then in reality there is a limit to how far the mass$m$can be pulled towards the hole.$L = mvr$, so $$v=\frac{L}{mr}$$ The centripetal force needed to move the mass in a circle is $$F=\frac{mv^2}{r} = \frac{L^2}{mr^3}$$ So as the mass approaches the hole, and$r$decreases, there will be a position where the force$F$just ... 1 I have added the "new" angular momentum although I have not moved the image of the top itself as this would make the diagram more difficult to interpret. So you have$\vec L_{\rm new}= \vec L_{\rm old} + d\vec L$with the change in angular momentum,$d\vec L$, caused by the torque,$\tau$, applied on the top about point$O$due to the force of ... 0 It is the torque$\vec\tau$that gives rise to the precession. That is, it gives rise to the rotation of the axis$\vec L$(as in your diagram) about the z-axis. Torque is directly proportional to the timed rate of change of angular angular velocity (and identical to the timed rate of change of angular momentum) as is defined as $$\vec \tau=I\frac{d\vec \... -1 The angular momentum of precession about the origin O, is constant, it's a vector of constant magnitude along the z axis, see also \omega_p. However the angular momentum of the Gyro itself is changing, it's the one labelled L, the change is the orange arrow near the top labelled dL. Since it's acting at right angles to L the direction of L ... 1 You just need to realize, that speed of the mass increases to infinity as it approaches to the center:$$L=mvr\Rightarrow v=\frac{L}{mr},$$L being the conserved angular momentum and v velocity of the mass. So r approaching zero does not mean L is approaching zero also. 1 Your are right in thinking that work is done by the indicated applied force on the pulley. The torque from the force continuously increases its rotational kinetic energy, an evidence of work being done. You argue that since the portion of the rope that applied torque during the time [t,t+dt), thus inducing an angular change d\theta in the orientation of ... 0 Torque and angular momentum is summed about the COM (point) but using common basis-vectors (orientation) as a non-rotating frame. These problems assume there is an inertial coordinate system coincident with the COM at each time frame. Sometimes this is called a co-moving reference frame and it is valid so long at it is not accelerating and not rotating. 1 This equation is true in an inertial frame, but the centre of mass is rotating and hence non inertial, then how is the equation correct? The centre of mass in itself is just a point, there is no unique frame attached to the centre of mass. Usually when we say COM frame, there is no rotation and we take a frame which has centre of mass at origin in all times.... 0 f_1 and f_2 are the centrifugal forces in the non-inertial frame. Once you also include the non-inertial forces then that equation is correct. 1 General 3D rotation is not simple. Physics mechanics texts address this approach in detail, but the mathematics requires some understanding of tensors; for example, see Mechanics by Symon or Classical Mechanics by Goldstein. As @David Hammen stated, using angles of rotation will not work (see one of the physics mechanics texts). These texts explain the ... 0 When a demonstration of gyroscopic precession is given the demonstrator will invariably release the gyroscope gingerly. This is not done on purpose, it just feels natural to do so. However, by releasing gingerly nutation is suppressed. What in fact happens is that the demonstrator is providing critical damping. (Critical in the sense that nutation cycles are ... 0 Thus, why does lower gear generate less acceleration/speed of the car compared to higher gear? The wheels RPM depends on the motor RPM and gear ratio. N_w = \frac{N_m}{i}. The acceleration is the derivative with respect to time: \frac{dN_w}{dt} = \frac{\frac{dN_m}{dt}}{i}. So, for smaller i (higher gear), the (potential) acceleration is greater. The ... 1 Yes. The blade of a wind turbine is like a wing. Air blowing over it generates life force, which pulls the wing "upward". The wing is oriented so the "upward" force pulls the blade around the propeller axis. On the other hand, drag does not contribute a useful torque. Drag pulls the wing "backward", which is the direction the ... 3 why does lower gear generate less acceleration/speed of the car compared to higher gear? You are mixing up acceleration and speed. Acceleration and speed are different and behave differently in the lower and higher gears. Lower gears give higher acceleration and lower speed . Higher gears give lower acceleration and higher speed . To get an intuitive ... 1 5 Pa is a unit of pressure equal to F/A and is a scalar quantity. N\cdotm, when specified as a unit of torque, \vec\tau, is a vector quantity where$$\vec\tau=\vec r \times \vec F$$Where \vec F is the applied force \vec r is the position vector of the force relative to the point where the torque is calculated, and \times indicates the cross ... 4 It is a 1 newton force applied at a radius of 1 meter, or its equivalent. For example two newtons force at a radius of 0.5 meters would also equal a 1 N\cdotm torque and so forth. Force times the radius it was applied at equals torque. 1 Well, 1 N\cdotm is when, 1 N of a force is applied at a point on the arm, which is at a distance of 1 m from its axis of rotation. So the torque generated under this condition is 1 N\cdotm. 3 What you need is the whole torque curve (torque vs. speed of the motor). These are quite complex, but there are distinct stages. First, there is a constant torque, then at about 45 mph the motor transitions to constant power, and over 75 mph to some other form of less power. To calculate acceleration you have to know the power P produced at speed v and ... 0 The equipollent torque between two points A and B is defined as$$ \boldsymbol{\tau}_B = \boldsymbol{\tau}_A + (\boldsymbol{r}_A-\boldsymbol{r}_B)\times \boldsymbol{F} \tag{1}$$That is torque summed at B as a function of the torque summed at A in the presence of net force. If the net force is zero \boldsymbol{F}=0, then$$\boldsymbol{\tau}_B = \boldsymbol{... 1 The statement simply means that the net torque of all forces acting on an object in translational equilibrium (i.e. resultant of all forces on the object is zero or$\sum\vec F_{ext}=0$) is the same regardless of the axis chosen. This statement can be visualised in a better way if we consider two equal and opposite forces (say$F_1$) acting on a uniform rod ... 0 Provided the door accelerates the same way in both cases, this is simply a matter of work done -- if you push closer to the hinge, you apply a stronger force over a shorter distance. If you push further away from the hinge, you apply a weaker force over a longer distance. It's the same principles that pulleys or inclined planes use. 1 It takes the same amount of energy to close the door no matter where you push on it. But when you push farther away from the hinges, you spread out that energy transfer over a longer distance (the same angular distance, but a longer actual distance). Since energy transfer through work is equivalent to force integrated over distance, a longer distance ... 0 There is another way to accomplish this which does not require big rotating masses, as follows: In the ship is mounted a small gyroscope with roll and pitch sensors attached to it, just like in the attitude gyro in an airplane's autopilot. Sticking out of the sides of the hull are large fins whose pitch is controlled with accordingly large servomotors. The ... 2 If the mounting of the spinning disk is fixed then it will be no good. However, in the market of luxury motor yachts there are systems available that apply gyroscopic effects with a responsive mounting. You can internet search with search terms such as 'stabilizer', 'gyro' and/or 'gyroscopic' Strong actuators are in place that tilt the fast spinning gyro ... 0 Initially, there is a relative motion between the two discs. So, the frictional force is given by$ F_{fr} = \mu_k N $, where$N$is the normal force, and$\mu_k$is the coefficient of kinetic friction. At one point, the relative velocity between the two discs becomes 0. At this stage, the frictional force also be 0. Hope this answers your question. 1 Take an arbitrary point A and note the relationship between torque$\boldsymbol{\tau}_A$, change in angular momentum$\tfrac{\rm d}{{\rm d}t} \boldsymbol{H}_A$, velocity of the body$\boldsymbol{v}_A$and momentum$\boldsymbol{p}$$\boldsymbol{\tau}_A = \tfrac{\rm d}{{\rm d}t} \boldsymbol{H}_A + \boldsymbol{v}_A \times \boldsymbol{p}$$ Reference: ... 2 You've sort of proven your own answer in the statement of the question. If it is moving, then the force to the right must be greater than friction. Since you assume it is rolling without slipping, it will roll to the right, done. The more interesting question (which I think is conceptually what you're getting at) is under what conditions it will roll to ... 1 The negative sign indicates the tendency of the dipole to be aligned with the magnetic field; a dipole transverse to its local magnetic field has more energy than when it's parallel to the field. Looking at your computation, it looks like the confusion lies in how you've defined\theta$. In the torque expression:$\tau = \mu B\sin\theta$,$\theta\$ is ...

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