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1

To get to a final rotational velocity $\Omega$ is takes the same amount of energy regadless of where you push. Why? Well the final kinetic energy is $K=\frac{1}{2} I \Omega^2$ (where $I$ is the mass moment of inertia about the hinge) and this value does not depend on where you push. This is kind of a boring result. What does differ is how much you need to ...


4

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$ F_{\text{slide}} > \mu m g $$ where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...


2

If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...


0

Newton's Law, when applied to rotating systems such as yours, are different than the translational law you're probably used to seeing. The 2nd Law is $$ \vec{N} = \frac{d\vec{L}}{dt} = \sum_i \vec{r_i}\times \vec{F_i} $$ $\vec{N}$ here denotes torque, which is equal to the time derivative of angular momentum. $\vec{r}_i$ are the position vectors from the ...


1

As other posters have noted, the linear or translational momentum will not change when you apply the force in a way you described. This is made most apparent by observing the (non)movement of the centre of mass, but you can also 1) split the object into smaller parts, 2) draw vectors of acceleration and velocity of each, and 3) notice that the latter sum ...


11

This apparent paradox is actually not a paradox at all. Infact, it is because of Newton's second law, that we can say that the object will rotate exactly about it's centre of mass. This is true for any body on which net external force is 0, but net torque is present. In Newtonian mechanics, centre of mass serves to simplify calculations, for exactly the ...


6

No ,the centre of mass would not accelerate,the object will rotate about the centre of mass. If the object is on a table(or anywhere here on the earth) and if you apply the forces as shown in the fig it may not be rotated about the centre of mass because of the frictional forces acting on the object,but in space where there is no friction the object will ...


0

Let's do this using angular momentum as a vector. This should clear up the question on using both axes separately or one new one. The spinning around the y-axis will give an angular momentum in the y-direction: $\vec{L_{y}} = \hat{y} L_{y}$, while the spinning around z-axis gives angular momentum in the z-direction: $\vec{L_{z}} = \hat{z} L_{z}$. We can get ...


0

That looks lie a homework style question. So here the necessary hints. The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem


0

An object has a rocket (which, for simplicity, is fixed to the object so it always fires along a line perpendicular to the line between the rocket and the center of mass) which is offset from the center of mass by a distance d. The force from that rocket, assuming it is constant in magnitude, will produce the following amount of force: $$\vec{F}_{rocket} = ...


3

My fallacy was in thinking that resultant translational motion varies by the distance ($d$) between the point of force application and the center of mass. It was not a fallacy or misconception. It is simply impossible that the linear velocity of a rod is the same if the point of application A (A-CM = d) varies. If A coincides with the ...


2

If we treat the Earth as an isolated system then both its linear and angular momenta will remain constant. To answer your question you need only consider the angular momentum. The angular momentum is given by: $$ L = I\omega $$ Since $L$ is a constant, if the moment of inertia changes from $I_1$ to $I_2$ then we have: $$ I_1\omega_1 = I_2\omega_2 $$ and ...


0

Neither the linear momentum or the angular momentum will change (they are conserved quantities), unless you apply a force or a torque respectively. In this case, the change in radius will likely change the moment of inertia. And as the (constant) angular momentum depends on the product of moment of inertia and angular velocity, then the angular velocity ...


2

By adding the rubber band you did two things: increased the air drag unbalanced the fan The bearings of a fan don't like to be unbalanced - the friction goes up significantly because as the fan picks up speed there will be a large lateral force (centripetal force keeping the rubber band plus object in their circular orbit). When you balance the fan it ...


1

The load increased but the input power driving the fan remained same. Moreover in very accurate measurements, the air drag can also not be neglected, all this will hold for a very another reason that the blades rotated by the motor are of very much comparable mass to the taped rubber & stuff.


1

Consider two different regimes. First - when the ball is still slipping, the relative motion of the ground and ball causes a force on the ball which (a) slows down the center of mass and (b) increases the angular speed. Once the ball rolls without slipping this force disappears. Second - when the ball does not slip, the contact point must be stationary. ...


1

Your assumption that the ball will be faster if it rotates faster has a very logical explanation and is completely valid. If your ball has a circumference C it will cover that distance with each rotation. Let us say your C was 1m and the period was 2 rps (revolutions per second) then your ball would cover 2 metres in 1 second. If your period is doubled to be ...


1

The ball rolls because: the friction present between the surface pushes the atoms/particles in contact with surface or ground backwards, hence with centre of mass moving in a direction, the ball's bottom surface(ie. The surface in contact) moves in opposite direction or backwards, causing the ball to roll. The ball slips because there is no friction that ...


7

Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is? When you push a body it will always rotate around the center of mass (white arrows) if you apply a force at the handle you are ...


0

As you mentioned that the friction is absent, ball(it's a solid spherical ball, moment of inertia = (2/5)m*r^2) will slip all the way and will not be able to roll, hence will not rotate, only translatory motion will account for. If you had friction on the track, Energy would surely have been distributed between kinectic translational and rotational.


2

First, two remarks: The melting of the ice in the Arctic ocean would have zero effect. What matters is the ice over Greenland and Antarctica. There is a very long-term secular slowing of the Earth's rotation rate due to the recession of the Moon from the Earth. This answer ignores that effect. (Alternatively, this answer has this secular effect subtracted ...


1

Torque $\tau = \vec{r} \times \vec{f} = I \vec{\alpha} $ $r$ here is moment of arm, ie. the distance(& perpendicular distance) from the axis of rotation of the body. You can only have and define torque if you have an axis about which the subject will rotate. It will be an absolute force driven motion, with NO torques compelled to induce.


0

If you open a door by gripping it near the hinge, you apply GREATER energy for a SHORTER time than when you grip it near the outside edge, which requires LESSER energy for a LONGER time. As the friction of the hinge and the weight of the door are equal in both cases, the total energy applied is the same. It's only the time applied and the intensity of ...


5

To a first-order effect, there would be no change. But one consequence of melting is that the water moves to other places. Water that moves from the poles to other areas on the surface of the earth would serve to (slightly) increase the moment of inertia of the planet. This is because the mass of the water would be farther from the rotational axis. The ...


2

When you calculate $\frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$ of a particle of mass, m, having a linear momentum of $\vec{\mathrm{p}}$ in an inertial frame via a rotating frame or rotating body where the acceleration is directed towards the origin, you get ...


8

The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ...


0

Torque=(R) x (F) Energy req. to rotate=(T).(Theta), Ta=Tb, ONLY Fa is less than Fb. & T at Hinge=0, it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point). [Ta means torque applied at A]


1

As @Gautham said, W = Torque * angular displacement which is also equal to the energy needed. here the angular displacement is same as we know and torque will be also same(Large distance implies less force needed,less distance implies large force needed but torque will be same) So energy needed will be same in both cases.


3

The answer is NO. Change of energy is work i.e $W = \Delta E$ and here the work done is $$W = \text{Torque} \cdot \text{angular displacement}$$ which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius. $$\text{Force at "b"} > \text{Force at "a"}$$ but not the work done or ...


14

You are right. To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$ where $r$ is the distance from the point of contact to the axis. However the ...


2

There are two senses to $v_{\mathrm{min}}$. Firstly it means that the speed of the particle must be greater or equal to $v_{\mathrm{min}}$ at $\theta = \pi$ or else it will not complete the cycle. Instead it will follow a ballistic trajectory until such time that tension comes back into the string. Secondly, the speed of the particle is not constant: it is ...


0

Yes, you are correct that in the case of the oscillating disk, the suspending string is not acting exactly at $r=0$. The attachment to disk is spread over a finite distance and is capable of supplying off-axis forces. If you hypothesize some sort of infinitesimally thin wire, you then have the problem of what it means for such an object to twist or rotate. ...


2

The example given is certainly misleading. Rotation in 3D is hard to visualize. In the case of the green ring, you can say it has an instantaneous axis of rotation which is the vector sum of the rotation about the two axes at a specific moment in time. But the support structure of the green ring itself rotates about the vertical axis, and so the ...


3

I guess he was talking about rotation through two stationary axes, which makes the box example incorrect. If the inside rotation axis rotates, there's no reason for it to be a constant rotation. That being said, if you first rotate through one axis and then another, the combination will give you a rotation through a third axis. If you want the math behind ...


5

As pointed out by lemon, two angles are enough to specify a direction in a three dimensional coordinate system, but another is needed to specify a complete coordinate transformation. You can think of a rotation transformation in three dimensions as a mapping between two different coordinate systems. Two angles are needed to specify the relative pointing ...


0

You are correct. Start with the height difference between AB and the center. I have $h=R \sin\left(\frac{\phi}{2}\right)$. Now you calculate the torque applied to the ball center as $\tau_C = (2 F) h$ Sum of forces along groove $$ M g \sin \theta - 2 F = M a $$ Sum of forces perpendicual to groove $$2 N - M g \cos \theta = 0 $$ Sum of torques about the ...


6

Use the spatial inertia to relate linear/angular momentum to changes in linear/angular speed $$\begin{pmatrix} \vec{L}\\ \vec{H}_A \end{pmatrix} = \begin{bmatrix} m {\bf 1}_{3×3} & -m [\vec{c}\times] \\ m [\vec{c}\times] & I_{cm}-m [\vec{c}\times][\vec{c}\times] \end{bmatrix} \begin{pmatrix} \Delta\vec{v}_A\\ \Delta \vec{\omega} \end{pmatrix}$$ ...


0

You're presumably talking about objects dropped away from the equator. At the equator it causes a lower effective gravity. Away from the equator it causes a deflection towards the equator.


1

I think what you are trying to get to is $${\rm d}{\bf I}_G = -[{\bf r}\times][{\bf r}\times]{\rm d}m$$ where ${\rm d}{\bf I}_G$ is the contribution to the mass moment of inertia of a mass ${\rm d}m$ located a distance ${\bf r}={\bf R}-{\bf R}_G$ from the center of mass. The 3×3 skew symmetric cross product operator $[{\bf r}\times]$ is defined as ...


2

When you fix a reference point (take it to be the origin of your reference frame) you can write the position as $$\vec{r} = r \hat{r}$$ where $\hat{r}$ is the unit vector pointing toward the particle. Deriving you obtain $$\vec{v} = \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}$$ The first term is the radial component of the velocity, the second one is ...


2

If you attached two flywheels through a motor to the disk at the positions you show, and the motors start spinning in the same direction, then conservation of angular momentum tells us that as the flywheels spin clockwise, the disk must (and will) rotate counterclockwise. However - if you attach the motors to an external structure, you are preventing ...


0

By saying point(c) (or any of the other points) applies a torque on the disc, it sounds like point(c) is a small physical body (If point(c) exerts a torque on the disc, then disc must exert a torque on point(c)) I'm going to assume that the mechanism by which point(c) causes a torque on the disc is a motor connecting point(c) and the disc. So, if points (a) ...


0

On a free floating body, if a pure torque is applied (with net zero force) then the body is going to rotate about it's center of mass (see http://physics.stackexchange.com/a/81078/392). This is regardless of where the torque is applied, or how many torques are applied. If the net force is zero then the center of mass will not move. Now if C or A or B are ...


4

There is a difference - but not exactly why you think. There are prevailing winds around the earth - these used to be called the "Trade Winds" because traders, knowing the direction of the wind, knew how best to navigate the globe. Basically, on the equator (in the tropics) they flow from east to west, and at higher latitudes they flow from west to east: ...


1

Start digging at the equator and move all the dirt to the polar regions. This will decrease the moment of inertia of the planet about its spinning axis. Due to the conservation of angular momentum this will result in an increase in angular velocity, akin to a figure skater who retracts her arms while spinning.


2

Cover it in mirrors that are highly reflective on one side and painted black on the other. Position the mirrors so that the "faces" are perpendicular to the surface. A sketch is below (I have only shown three mirrors, the idea is that you would cover the planet with them, but they will be most effectively placed close to the equator). The plan is that each ...


1

I believe your second interpretation is spot on. Whenever anything rolls without slipping, it means that the point of contact of the ball with the frictional surface is instantaneously stationary. Therefore, the frictional force applied to this point does not do any work, and so there is no frictional dissipation. Also, because the ball is still rolling ...



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