# Tag Info

21

The fan motor provides a torque $\tau$ which has to accelerate $\alpha$ the fan blades whose moment of inertia is $I$: $$\tau=I\alpha$$ Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ...

10

A much simpler way of thinking about this is to consider energy. When the fan is spinning it has quite a lot of kinetic energy (try to stop it by putting your finger in the way to confirm this (don't actually do this!)). That kinetic energy goes as the square of the rotation rate, in fact. So as the fan starts, the motor needs to add energy to it. It ...

10

You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows: $$I = \iiint\rho(x,y,z)||r||^2\ dV$$ Or with uniform density, $$I = \rho\iiint||r||^2\ dV$$ From this formula, you can see that ...

8

It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.

5

simply the resistance of a body to rotate it over an axis? Gosh, I dislike the word resistance in this context since resistance is, in general, dissipative and, in particular, resistance to rotation would imply that an isolated object that is rotating would eventually stop. Think of moment of inertia (rotational inertia) about an axis as a measure of ...

4

Everything you have derived is correct. The reason for your perceived paradox is, I believe, a confusion between force and power. The same force can produce more power if it is being exerted at a greater velocity. When you exert a force at a radius r from the CM, the point of application of the force will accelerate more quickly than the CM, allowing the ...

4

I think that what your teacher has told you is that the angular momentum of a body can be split into two components: The spin angular momentum which is an intrinsic property of the body and is independent of the point about which you wish to find the angular momentum. $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v r$ where $I_{\text{cm}}$ is the ...

3

As an unbalanced force, $\mathbf{f}$ acts to accelerate the disk. Since it is located at the bottom of the disk, O must accelerate as well and is therefore in a non-inertial frame of reference. That non-inertial frame will have a fictitious forces appear that oppose acceleration. We can draw a force $\mathbf{f'}$ that acts through the center of mass in ...

3

The equation of motion $$\text{torque about stationary geometrical point O} = \text{moment of inertia w.r.t. O} \times \text{angular acceleration w.r.t. O}$$ is valid only if the motion of the body is planar rotation around an axis that passes through O. This is the case if the point O is taken to be point of contact of the body when rolling without ...

3

At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...

2

In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four. Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they ...

2

shouldn't we use the parallel axis theorem ... to compute the moment of inertia? You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis.

2

Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved.

2

The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ...

2

In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ...

2

Use conservation of angular momentum relative to the center of the ring. Since both the ring and the bug were initially at rest, the angular momentum of the system about the center of the ring was zero. Since no external torque is acting on the system, angular momentum of the system must remain zero even when the bug starts moving along the rim. How should ...

2

When doing this sort of problem you can add two forces acting at the centre of mass whose resultant is zero. This system of three forces can now be viewed in the following way. The frictional force $f$ is exactly equivalent to a force of the same magnitude whose line of action passes through the centre of mass of the disc (shown in blue) and a pair forces ...

2

Let the body rotate about the $z$-axis, then by the definition of angular momentum $$\vec{L}=\vec{\omega} I_z.$$ where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$: $$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$ Now ponder the terms in it. If I understand the ...

2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...

2

As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity. You can also derive this. formula like this: $$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } ... 2 The rotation will not necessarily be parallel to the ground. The general motion will be a combination of rotation in a horizontal plane (the conical pendulum) and oscillation in a vertical plane (the simple pendulum). If the support (pivot) is a fixed point, the motion you get depends on the starting conditions. If you launch the mass horizontally at the ... 2 Theese concepts usually arise in rigid body mechanics. So consider a rigid body which is a set of points in which the distance between any two points do not change. If this is too abstract you can just think of a piece of rock. One talks of translational motion when the body moves along a straight line, or more exactly when every point of the body travels ... 2 You mean go all the way around? It could if you had enough force to overcome gravity and like a tether ball swing all the way although most humans do not have the strength to apply the force needed to push another or them selves to a full revolution around the bar of a swing with out a jerk, but if the chain was replaced with a solid bar to prevent jerking ... 2 When the fan starts spinning, each blade starts from rest. Newton's first law of motion states that unless a force is applied to it, the [velocity][1] of a body does not change. That property is inertia. To speed up, the blades must accelerate. Newton's second law of motion states that the force necessary for an acceleration of a body is proportional to and ... 1 Since the hard drive platters are relatively heavy, and spin at a very high speed (7200 rpm for most hard drives, and even 10000 rpm or more for high-performance drives), the hard drive becomes one big gyroscope. If you've ever seen a gyroscope toy, you've observed gyroscopic precession firsthand. When you turn your hard drive along any axis other than the ... 1 All you have shown is that the instantaneous velocity of the particle at point D is can be equated to the translation of an arbitrary axis together with a rotation about that axis. The statement "while \vec \omega is independent from the chosen rotation axis, v depends on it" refers to all the particles in the body. So the all have the same ... 1 The Lagrangian is in fact an equation of \vec \Omega, however, in general it will be a quadratic function of \vec \Omega, as the rotational kinetic energy would be given by$$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the ...

1

What you are referring to is a special case of the Ehrenfest Paradox In its original formulation as presented by Paul Ehrenfest 1909 in relation to the concept of Born rigidity within special relativity,1 it discusses an ideally rigid cylinder that is made to rotate about its axis of symmetry. The radius R as seen in the laboratory frame is ...

1

I think that I understand what you are asking and I think that it is a good question. If the centre of mass of the disc and the point mass moves a distance $x$ then the work done by the force is $Fx$ in both cases and yet the disc has gained some extra kinetic energy because it is rotating. If the centre of mass of the disc moves a distance $x$ the point ...

1

You forgot to include angular momentum of center of mass. Thus add $MR^2\omega$ to your answer By the way angular momentum is not $I\omega$, it is $\vec{l_{com}}+I_{com}\vec{\omega}$ It is $I_{contact}\omega$ only for cases when point of contact is the point of instantaneous or fixed axis of rotation.

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