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53

No, a car cannot steer on a frictionless surface. This has little to do with gyroscopic action and more to do with conservation of momentum: to turn, even when conserving its speed, the car needs to accelerate at right angles to its motion, which changes the total momentum of the motion. This change in momentum requires a force which, in normal roads, is ...


47

If the wheels had spun fast enough for a gyroscopic effect to become noticeable, the only result on a frictionless surface (which would be the same without a surface at all) is that when you turn the wheels, the rest of the car would rotate instead of just the front wheels :) You need some reaction force to alter the trajectory, like a sail or surface ...


27

Yes you can It is actually possible with a real car, but you would have to be very patient to steer a little bit. Suppose you have built a car with power on the big front wheels to induce a gyroscopic effect. If you rotate the wheels, the direction in which the center of mass is going will not change directly, but the angle in which the rest of the body ...


10

Since there is no friction, then it will not affect any other forces that may act on the car. The direction of wind blowing on the car may change its trajectory, as any driver will attest when driving in high winds. Turning the car wheels may have a slight affect on the resultant direction of the force. If the car has curved roof, then it may acts as ...


7

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...


6

There is a general and simple formula to calculate the moment of inertia with respect to some axis if the moment of inertia with respect to another axis is known. I am pretty sure you'll be able to find it in your textbook. The theorem is parallel axis theorem.


6

Actually, your book is correct. Even if the most usual uses of angular momentum involve circular or rotating motion, this is not the general case. An object moving in a straight line has angular momentum in a reference frame in which the origin does not fall on the the line. To see this simply remember the definition of angular momentum ...


5

What would happen if axis of rotation pass through centre of mass of an object? Will the object rotate when we will apply force to the object? For the sake of future readers, I'll reply to the original question: Every body has a Centre of Mass, whatever its form. If an object has a regular shape and uniform, homogeneous distribution of mass its ...


4

It is fully explained in this answer Angular momentum $ L = m*v*r$, since angular velocity is linear velocity divided by the radius, $ω = v/r \rightarrow v = \omega *r$ , then $L = \omega*[r*r*m]$ since $L = \omega*I \rightarrow I = m *r^2$ In the second sketch m=2, v = 3, r = 2 $\rightarrow L = 2*3*2 = 12 Kg*m^2/s$ Since angular velocity is v/r = ...


4

If you solve the problem for the two forces the vertical and the horizontal force (which is required for the circular motion) you obtain the relation $$\omega^2=\frac{g}{L\cos\theta}$$ Hence the minimum required $\omega$ is $\sqrt{g/l}$, the same as the angular frequency for motion of a bob in a plane.


4

That's what they wrote. But I am really confused why they wrote so. In fact I can't imagine torque & angular momentum without circular motion. Why did they tell so? What is the cause?? Please explain. A planet in an elliptical orbit has angular momentum alright, there is no problem. The magnitude of angular momentum does not change, is conserved, ...


4

Friction is the only force that would cause the car to move along a different path. On a frictionless surface, the gyroscopic effect could change the orientation of the car a bit, but not the trajectory of the car. In other words, the front car would no longer point along the direction of travel, but would "skid". (That is, if you could call frictionless ...


3

On a completely frictionless floor, with the absence of other external forces, the centre of mass of the car will continue in the same trajectory for ever. Hence no steering is possible. However, irrespective of whether the front wheels are rotating or not, turning of the front wheels will produce a counter torque changing the orientation of the car, albeit ...


3

I see now your problem and I believe that I can help. Let's begin from the velocity formula $$v_i = v_r + Ω \times r .\tag{1}$$ Let's take the derivative of $v_i$ IN THE INERTIAL frame, $$a_i = \left(\frac{dv_r}{dt}\right)_i + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_i .$$ Here we use as much as we can our formula $(dF/dt)_i = (dF/dt)_r + Ω ...


3

Definition of angular momentum A body B with velocity (and linear momentum) has a potential rotational momentum L with reference to/around any point/body O which does not lie on its trajectory. The magnitude of L can be found multiplying its linear momentum (p = m*v) by the distance of point O from the trajectory: $r$. In the full formula: $L = m * [v * ...


2

I think the Hamiltonian is not necessarily the energy for the following reason: you can demonstrate that the Lagrangian may be deduced from the D'alembert principle which is linked to the concept of force, etc. but it may be also deduced from the Hamilton's principle which is a pure mathematical concept applied to physics (a certain quantity has to be an ...


2

Expanding the correct answer of @SAKhan: Assume that the conical pendulum is rotating at an angle $\theta$ at an angular velocity $\omega$. Note also that the radius of the circle is given by $$r=L\sin(\theta)$$ For the point mass to move in a horizontal circle, the total vertical force is zero:$$T\cos(\theta)=mg$$ The net horizontal force must supply ...


2

It happens I think because rotation about axes which do not pass through the center of mass are usually inherently unstable and hence the rotation tends to "decay" to a more stable axis, i.e the one passing through its center of mass. Also, the moment of inertia of a body is usually the lowest through a certain principle axis through its center of ...


2

The discussion is mostly semantic. They are both calculated relative to a point, in the case of the torque the point has the additional meaning that if you put an axle trough the point, the object will start to rotatte around it if the net torque is not zero. It happens also that the torque will be the same if you chose any other point along the axis ...


2

All your thinking is very good and correct. BUT you are missing a point in the question :) Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True. BUT let's read the question: A door is hinged at one end and is free to rotate about ...


2

The moment of inertia is linked with the kinetic energy. $$E_c=\sum_i \tfrac12 m_i v_i^2 = \sum_i \tfrac12 m_i (\omega r_i)^2 = \tfrac12 \omega^2 \sum_i m_i r_i^2$$ (wikipedia) It seems quite logical to me. Other reason is dimensional.


2

with your definition of $dA$ you must integrate between r and 0, because you start at the center, and the rings grow in radius as you go for (r-x) from x=r to x=0


2

Nice question. First, look at the following Free Body Diagram. For the disk $$\text{T}\text{R}=\text{I}_{\text{cm}}\alpha $$ where $\text{T}$ is the tension in the string, $\alpha$ is the angular acceleration in the disk and $\text{I}_{\text{cm}}$ is the moment of inertia of the disk about it's center, which equals ...


2

In the cases of Bottle A and Bottle C, they are full and empty. So the amount of water in them (or not) can be considered to be a complete system along with the bottle, since there is no possible way in which the fluid in the full bottle would reduce its volume or overall distribution, certain properties like the system's center of mass,center of gravity and ...


1

First, there is no reason why a physical quantity must be linear in r. That only happens for some variables (perhaps the ones you are familar with). In the case of the moment of inertia, it is actually a multiplication of two linear variables in r, that is why it end result is quadratic in r. By definition: the moment of inertia is $I=L/\omega$, so ...


1

That the floor is not flat but circular, an the centrifugal force acts radially, and is the same at the same radius. If you cover the concavity of the floor with a planar surface or either make the stations of planar segments (a polygon instead of a circle), then you will measure different accelerations at different points of the floor, because they are not ...


1

what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured.


1

It is simply not true, as stated. A cube is as perfectly balanced around its center of mass as a sphere is. You have shown it mathematically, and you are perfectly correct. However, the principal axes may be chosen perpendicular to the faces of the cube. Of course, when the point is the center of mass, this is no better than any other choice, except for ...


1

The only outright requirement is that you compute all the angular momenta in your problem around the same center (modulo applying the parallel axis theorem to break the angular momentum of extended bodies into of-and-around the CoM parts). So you can freely chose any single point to use Now, as with most such "free" choices in physics there are generally ...


1

The idea is that if there are no forces on an object, then no matter how it rotates, its center of mass must move at a constant velocity. Then in the frame of the object, the center of mass appears stationary and everything else rotates around it. In general this cannot be said of any other point in the object. To see that the center of mass moves at a ...



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