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Centripetal force $\vec{F}$ is responsible for circular motion and is given by $$ \left|\vec{F}\right| = \dfrac{m\left|\vec{v}\right|^2}{r}$$

If the linear velocity $\vec{v}$ varies, the centripetal force also varies. Does this varying of centripetal force still produce circular motion? If so, why?

Now, come to the case of projectile motion. Projectile motion is almost like a circular motion. But one thing that prevented the projectile to follow circular motion was its linear(horizontal) velocity $\vec{v}$ which is small enough to make

$$\dfrac{m\left|\vec{v}\right|^2}{r} < mg$$

true, wasn't it? Thus it follows the parabolic path. But what would happen if we had $\dfrac{m\left|\vec{v}\right|^2}{r}> mg$ ?

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    $\begingroup$ I don't understand the first question. In mechanics, you are not given $v(t)$ and asked "what kind of motion does this produce?". Instead, you are given $F(\vec x,t)$ and use $\vec F = m\vec a$ to determine the kind of motion that occurs. So what do you mean by "If the linear velocity varies?" $\endgroup$
    – ACuriousMind
    Nov 21 '14 at 13:14
  • $\begingroup$ @ACuriousMind: Sir, I was telling that in case, when the linear velocity varies, will not then centripetal force vary? If it varies, will it produce a spiral motion?? I was just confused... $\endgroup$
    – user36790
    Nov 21 '14 at 16:17
  • $\begingroup$ ...also wanted to know why a body in a circular motion follows a circular path while in projectile motion, the body follows a parabolic path. Conditions are same in both cases: both horizontal velocity and downward force are present, still the paths they follow are different. Why?? $\endgroup$
    – user36790
    Nov 21 '14 at 16:34
  • $\begingroup$ If the conditions were the same, they would lead to the same motions. The force in circular motion is not downward, it is inward. $\endgroup$
    – ACuriousMind
    Nov 21 '14 at 16:40
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    $\begingroup$ Sure, inward (for me) means it always points towards the same point. Downward (for me) means it always points in the same absolute direction. Note that you are, in fact, correct that a falling object and an orbiting one are indeed really both circular/elliptical motion, but that one usually approximates the radial force as a downward force for falling objects (and the error is really really small in doing that, since the radius of the earth is so big). $\endgroup$
    – ACuriousMind
    Nov 21 '14 at 18:25
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The proper question should be "Can a varying centripetal force produce circular motion?" The answer is yes.

The (much-abused) expression $$m\frac{v^2}{r}$$ tells us what the instantaneous transverse (perpendicular) component of force must be in order for an object of mass $m$ must experience if it is travelling at instantaneous speed $v$ and is turning around an instantaneous center of curvature a distance $r$ away. It is not a force itself. It has force units, but it is the product of a mass times an acceleration. There must be some real forces which combine to result in the same net value as the product $m \times \frac{v^2}{r}$.

This is true every time the velocity of a particle changes direction.

If we apply it to circular motion, then $r$ is the radius of a circle. If we apply it to elliptical or parabolic or hyperbolic motion, then $r$ is constantly changing. If we apply it to straight line motion, then $r\rightarrow \infty$.

Back to circular motion, we can now write $${\large\Sigma}F_r=-m\frac{v^2}{r}.$$

This must be true instantaneously during every moment because the velocity direction is always changing. (The minus sign shows that the net force must be toward the center. Positive radial direction in polar coordinates is traditionally away from the center. This is important when considering a mixture of forces, some of which might, by themselves, pull the object away from the center.)

For uniform circular motion, the speed is constant, so the magnitude of the radial component of the net force must be constant. Notice that we could have two or more changing forces which still produce uniform circular motion as long as the net tangential component is zero (no speed change) and the net radial component adds up to $mv^2/r$ toward the center.

For non-uniform circular motion, we still have a constant $r$, but the speed is changing. That means that the radial component of the net force must change. How can this happen? Consider a mass attached to a string, swinging in vertical circle with non-constant speed. The mass moves fast at the bottom and slows down, but stays on the circular path at the top. The gravitational pull on the mass is constant from the Earth-bound reference frame. But it is not always a radial force. That means that the tension magnitude in the string must be changing in such a way that the net radial component always has a magnitude toward the center of $mv^2/r$.

If we define $\theta$ to be the angle measured from a vertical line passing through the center, with $\theta=0$ being at the bottom of the circle, then $$mg\cos\theta-F_\text{ten}=-m\frac{v^2}{r}$$

$$F_\text{ten}=m\left(g\cos\theta+\frac{v_{\theta}^2}{r}\right)$$ where $v_{\theta}$ is the speed at angle position $\theta$ (not the angular speed). We also know that the tension force will not affect the speed (it acts perpendicularly to the velocity), but the gravity force will change the speed like $$\frac{1}{2}mv_\text{bottom}^2=\frac{1}{2}mv_{\theta}^2 + mgr(1-\cos\theta).$$

From there it's an algebraic exercise to show that the tension force must change, and that there is a minimum required speed at the bottom for the mass to stay on the vertical circle.

Summary: for non-uniform circular motion, the radial magnitude of the net force must change, and it can.

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When an object undergoes uniform circular motion, it's velocity changes under the effect of a centripetal force which 'pulls an object toward the centre'. By uniform circular motion, we mean that the speed of the object does not change, only it's direction changes as it traverses a circular path. In this case, the magnitude of the centripetal force is constant, but it's direction changes as the position on the object moves around the circle so as to pull it towards the centre. The centripetal force always acts in the direction perpendicular to the velocity along the axis towards the centre.

In projectile motion, the force due to gravity is a constant force (constant direction and magnitude), pulling the object down toward the earth. An object with a horizontal speed will maintain this horizontal speed, but it's vertical position will change to follow a parabolic path as it falls back toward the Earth.

There is a speed at which the object will no longer fall back down to Earth, but will go into orbit! This is called the 'escape speed', given by:

$v_e=\sqrt{\frac{2GM}{r}}$

where $G$ is the gravitational constant ($=6.67\times10^{-11} Nm^2/kg^2$,

$M$ is the mass of the Earth (=$5.976x10^{24}kg$)

and $r$ is the distance to the centre of gravity (radius of the Earth at the equator is $6378km$).

so, the escape speed for the Earth is $v_e=3370.9 m/s=12,135 km/h$

If an object were to reach this speed, it would go into orbit around the Earth (ignoring air drag). That is, instead of falling back down to the surface of the Earth (as in projectile motion), it would continue around the Earth in circular motion, with the force of gravity now providing the (constant) centripetal force.

enter image description here

In this case, the orbit velocity of an object with mass, m, can be calculated by equating centripetal force with the gravitational force.

That is:

$\frac{mv^2}{r}=\frac{GmM}{r^2}$

which gives

$v=\sqrt{\frac{GM}{r}}$

So the object will travel in orbit with a speed of

$v=\frac{v_e}{\sqrt{2}}$

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  • $\begingroup$ Sir, I want to know, if circular motion is non-uniform, the centripetal force will vary,right?? In what case, the circular motion becomes a spiral motion?? $\endgroup$
    – user36790
    Nov 21 '14 at 16:04
  • $\begingroup$ If the magnitude of the centripetal force increases, it will pull the object in a spiral towards the centre at an ever increasing speed. $\endgroup$
    – theo
    Nov 21 '14 at 19:39
  • $\begingroup$ Sir, that means there is no non-uniform circular motion?? But, that sounds crazy as my book mostly deals with non-uniform circular motion where centripetal force varies. But they aren't spiral motion:-[ $\endgroup$
    – user36790
    Nov 22 '14 at 2:19
  • $\begingroup$ When the magnitude of the force varies, the motion will not be circular.Non-uniform circular motion happens when object has a tangential force component. That is, in uniform circular motion, the centripetal force is perpendicular to the velocity (ie: centripetal force is toward the centre of the circle). If the force has a constant magnitude with a tangential component (ie: the force is not directly toward the centre), then it will accelerate (or decelerate) in a circular motion. This is non-uniform circular motion. $\endgroup$
    – theo
    Nov 22 '14 at 3:46
  • $\begingroup$ Yes, you are correct. But centripetal force is related to linear velocity $v$ by $$\dfrac{mv^2}{r}$$ . In non-uniform circular motion, $v$ varies, so also centripetal force. So, how can it be constant?? $\endgroup$
    – user36790
    Nov 22 '14 at 3:58
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For an object to follow a circular path its radial acceleration has to be always equal to $\frac{v^2}{r}$ now in the case of non-Uniform Circular Motion (U.C.M) due to tangential acceleration, velocity of the object varies. Therefore, radial acceleration has to vary so that object can perform circular motion.

Therefore in non U.C.M the tangential as well as radial acceleration are both non-constant.

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