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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

There are a few ways to justify it. First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly. Another way ...


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

Your idea of how rolling works seems (and probably is) correct. But that is not enough to give a mathematical or physical description. There is another way to see that. The motion of the individual particles of a rolling ball is pretty complicated, but that of its centre is very simple, motion with a speed $v$ in linear forward direction. So we think, how ...


3

Constant acceleration - whichever type of acceleration - means that the second derivative is constant. In the linear motion, you have that the second derivative of position is constant. In rotational motion, you have that the second derivative of the angle is constant. In both cases you have $$ f''(t)=c \tag{1} $$ where $f$ may be $r$ or $\theta$, and $c$ ...


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 ...


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

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

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 ...


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.


1

If the incline is frictionless then there's only gravity and normal force. That's it. None of these alter the rotation (since they both act through the rotational centre). So the rotation will never stop! the force that causes the ball to come to rest When you say this you mean that the translational motion comes to a rest, only. Because the ...


1

The only thing I've figured out, is that the object will rotate but not moving, if the center of mass is on the line defined by the $A$ point, and the normal vector of $\vec{F⃗}$. This is not necessarily true. Look at the free body diagram. Decompose $F$ into $x$ and $y$ components. Newton now tells us that: $$ma_x=\Sigma F_x$$ $$ma_y=\Sigma F_y$$ ...


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

The two sets of equations are mathematically equivalent. The angular velocity is defined as the rate of change of the angle: $$ \omega=\frac{d\theta}{dt}$$ Which is analogous to the equation for rectilinear velocity $$v = \frac{dx}{dt}$$ In the same way, the angular acceleration is defined as the rate of change of the angular velocity (so the second ...


1

That the moment of inertia about an axis passing through the CM is minimized, with respect to any other parallel axes, is a consequence of the quadratic (squared) dependence of the moment of inertia on distance. In other words, the ${r^2}$ term in ${I=mr^2}$ makes it so that masses at farther distances are preferentially weighted in their contribution to the ...


1

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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