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1

No its SPEED will remain same, velocity has a direction, it will change, both while entering and leaving. On the curve, frictional force will balance the centrifugal force. Assuming constant coefficient of friction throughout the track and constant force provided to engine.


3

First of all, you need to be careful when saying the train does not accelerate. In Physics, acceleration is the rate of change of velocity. As velocity is vector, if the velocity of the train changes direction, as it does on the curved portion of the track, then the velocity is changing over time (changing in direction). Therefore, there must be an ...


8

The train does accelerate as it goes around a curve. Velocity is a vector, with magnitude and direction. Speed is the magnitude. The train changes direction. Acceleration is caused by a force. If the force causes a change in direction with no change in speed, it must be perpendicular to the direction of velocity. For example, A planet in a circular orbit ...


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 ...


0

They say that for a rolling body, the velocity of the contact point is zero. Indeed. In other words, as indicated in several other anwers already: Any particular point (or smallish piece of surface) of the wheel which, at some particular instant, makes contact with the pavement has vanishing instantaneous speed with respect to the pavement, at that ...


0

A planar body moving along a single axis has two degrees of freedom. Translation of the center and rotation about the center. The combination makes each part of the body to have different velocities according to the rule $$\vec{v} = \vec{v}_{cm} + \vec{\omega} \times \vec{r} $$ Rolling by definition is motion where the velocity at the contact point is zero ...


0

An intuitive way to think about this is to imagine the wheel falling with the point of contact as the pivot. To exaggerate this intuition imagine a "wheel", which is an equilateral triangle rolling by falling about a vertex before rising about the next vertex and falling again. As it is rising and falling about a vertex, the vertex is stationary with respect ...


1

Have less than 50 rep, so can't write these in comments: The question is phrased "How can it be zero when it's in continuous motion?" @terry's answer is essentially: From the point of view of the surface (that the body rolls on), the motion of given points on the body's circumference is not continuous. Their velocity lowers until the time of contact, at ...


43

What luck! Just yesterday I was thinking about this exact same phenomenon whilst watching the film 'The Imitation Game'; the title sequence contained a moving tank, more on that later. When I was little, I used to observe this all the time; not in wheels however, but in caterpillar tracks: Notice how, when a segment of the track touches the ground, it ...


2

To me, this is most easily seen with the teeth of a gear that is running a conveyor belt. This is basically the optimal non-slipping wheel; it rotates, and moves along the conveyor belt (really the belt moves along the wheel, but you can imagine the opposite). Each time a tooth enters the conveyor belt's track, the wheel effectively pivots on that tooth. ...


1

This answer is very simple and should completely clear your doubt:When a body rolls on a horizontal surface without slipping the horizontal velocity of the contact point is zero. This is true not only for horizontal surfaces but any surface where the body rolls without slipping.When does a block placed on a surface has a velocity with respect to the ...


0

You're looking for velocity of points on the wheel's circumference with relation to the ground. Imagine that at each instant, each point on the circumference is connected to the point of contact by a "lever" which is a chord of the circle that forms the wheel. At point of contact the "lever" has zero length - in other words it is the only point on the ...


10

Consider a point $P$ on the surface of the wheel. If you look at the horizontal velocity of that point in the frame of reference of the wheel (axis stationary), then for a wheel of radius $r$ with angular velocity $\omega$ that point will have horizontal component of velocity $$v_h = r\omega\cos(\omega t)$$ The linear velocity of the wheel $v = \omega r$. ...


8

you wrote: "How can [the velocity of the contact point] be zero when it's in continuous motion?". However, you should keep in mind that motion is relative and therefore your question should be actually read as: "How can the velocity of the contact point be zero relative to the contact surface, when it's in continuous motion relative to its axis of ...


2

If not, it would skid. [these characters are written to reach the minimum length of an answer]


0

There are several effects. The most obvious one is that, on the equator, your ball will experience about 0.2% less downwards acceleration than at a pole (because of the centrifugal pseudoforce). There are less obvious pseudoforces, too, the Coriolis forces. When you throw the ball straight up at the equator, it will seem to lag behind the earth's rotation: ...


3

The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ...


1

So we are to assume that the planet is not gravitationally locked for some period of time. This could possibly happen for a neutron star because it's gravitational bulging would be minimal owing to its structural integrity. We also have to assume that $t_{rot}\ll t_{orbit}$, because of $g^{tt}$ varies appreciably from one side of the planet to the other, ...



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