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If a stone is rotated with the help of a rope, then both centripetal force and centrifugal force will be liberated. The magnitude of that forces are same, but direction of them are opposite to each other.

Again, We know that, centripetal force works toward centre of circle. So why stone does not come back/reach to the centre of circle?

Suppose, for any time $t$ during rotating, stone is situated at point $A$ on the circumference of circle. By drawing a tangent on point $A$ we can show the direction of velocity of stone for that point. Again suppose that after very short interval of time stone reaches at point $B$, and by sketching another tangent on point $B$ we can show the direction of velocity of stone for that point. Will anybody tell me from which two velocity vector this resultant velocity vector for point $B$ will be obtained? If it is possible, please show that with image, or picture.

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You have two questions.

Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?"

The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your second question.

"Will anybody tell me from which two velocity vector this resultant velocity vector for point B will be obtained?"

Assume point B is an extremely small distance along the circle away from point A. You will note, after you draw this, that the Vector B is ever so slightly angled toward the center of the circular path. The tip of Vector A touches the tail of the extremely tiny vector that joins Vector A to Vector B. In other words, Vector A plus the tiny Vector result in Vector B. Keep in mind we're talking about an infinitesimally small increment.

If you draw this on paper, you'll see that, theoretically, the very tiny vector points directly to the center of the circular path. That tiny vector represents the change in velocity or acceleration. The stone is accelerating toward the center of the circle. It's constantly moving toward the center of the circle, but it never reaches it. The acceleration $\frac{v^2}{r}$, can be derived geometrically from the radius vectors and the velocity vectors.

The stone never reaches the center of the circle because as it makes a move toward the center of the circle due to the slight vector rotation, it is also tending to move in a straight line tangentially, taking it away from the center of the circle. This constant move inward is opposed by a constant move outward resulting in no progress to the center of the circle. The stone constantly wants to move in a straight line away from the center, but the force in the string is pulling it inward an equal amount for each increment of time.

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When you swing a rope over your head, your arm, and the stone, and the rope are all traveling in circles. That is, all the various pieces have tangential velocities ( tangent to circles ). The rope transfers the forces between your hand and rock such that the rope is always in tension.

If the applied force increases, the the Centripetal and Centrifugal force both increase, and the tangential velocity increases, and the tension force in the rope increases.

There are two things happening here. The forces acting on the rock are equal and opposite acting along the axis to the rope. 90 degrees to this are two more equal and opposite " Reactions " one is the Tangential velocity which is opposed by the Inertia of the rock ( and air drag ), which does not want to be accelerated.

So if you draw Vectors ( four arrows pointed out from the rock ), they are all 90 degrees apart. Two are forces, two are inertial. They are in equal and opposite pairs.

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The parametric equation of a circle (for simplicity take $r=1$) is given by $x=\cos(\theta)$ and $y=\sin(\theta)$ thus the radial vector $\vec{r}(t)$ is given by $$\vec{r}=\cos(\omega t)\hat{x}+\sin(\omega t).$$

Since the acceleration is the second derivative of the position vector, it is given by $\ddot{\vec{r}}=\vec{a}=-\omega^2\cdot\vec{r}$, from which you arrive at the centripetal force.

What I am about to say may look like circular reasoning but if you start with this differential equation and solve it you will see that indeed the equation of motion is given by a circle. The reason why I went backwards is because I didn't want to solve a vector ODE.

Furthermore you should not confuse centripetal and centrifugal forces, the latter being a fictitious force, which can only be observed in a non-inertial frame. To quote Wikipedia

In classical mechanics, the centrifugal force is an outward force which arises when describing the motion of objects in a rotating reference frame.

In conclusion there is no centrifugal force in an inertial frame and the forces do not cancel hence there is a net acceleration. I guess this has answered your first question.

The velocity vector is given by $$ \dot{\vec{r}} = \vec{v} = -\omega\sin(\omega t) \hat{x} +\omega \cos(\omega t) \hat{y}$$

Let the ball be at $A$ at time $t=t_0$ and at $B$ at time $t=t_1=t+\tau$ ($\tau$ being small) then the vector $\vec{c}$ which shows the change in the velocity vector $\vec{v_A}$ is the following $$\vec{c}=\vec{v_B}-\vec{v_A}=\vec{v}(t_1)-\vec{v}(t_0)$$ $$=-\omega(\sin(\omega t_1)-\sin(\omega t_0)) \hat{x} +\omega (\cos(\omega t_1)-\cos(\omega t_0)) \hat{y}$$ $$=2\omega\cos\frac{\omega(t_1+t_0)}{2}\sin\frac{\omega(t_1-t_0)}{2} \hat{x} - 2\omega \sin\frac{\omega(t_1+t_0)}{2}\sin\frac{\omega(t_1-t_0)}{2} \hat{y} $$ $$\approx -\omega^2 \tau\cos(\omega t_0) \hat{x} - \omega^2 \tau\sin(\omega t_0)\hat{y} $$ $$= \tau \vec{a}(t_0),$$ which is roughly acceleration times time as expected.

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In the stone and rope example you have a typical example of Newtow's third law. The stone moves on a circular trajectory because it is being pulled towards the centre by the rope. By Newton's third principle, the stone is now pulling the rope with an opposite force which equals the former in magnitude. That the work is zero comes from the fact that the centripetal force that keeps the stone on a circular trajectory is perpendicular to the velocity vector.

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  • $\begingroup$ But I have asked other two questions also. Please try to answer of them, especially last question. $\endgroup$ – Nousher Ahmed Jan 1 '15 at 16:32

protected by Qmechanic Mar 10 '16 at 9:08

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