# Circular motion in a cone

Imagine a vertical cone that is at an angle $$θ$$ with the ground.

In one of the cone's sides we place an object of mass $$m$$ giving it a initial velocity of $$v_0$$. Is it possible that our object, under only the effect of the normal force exerted to it by the cone and its weight, performs circular motion? If it is can this motion be uniform?

I feel that this is impossible because the normal force would always be less than the weight of the body and therefore there would always be a force dragging the object towards the center of the cone. However I have seen in textbooks that given an initial velocity large enough the object can perform circular motion. How is this possible?

• This is an ideal banking problem in disguise batesville.k12.in.us/physics/phynet/mechanics/circular%20motion/… – Farcher Mar 9 '17 at 23:31
• This is exactly what i am asking. Thanks alot. Could you explain how in this case the normal force is bigger in magnitude than the weight. I thoight that the normal force was the reactiom of the weight – TheNotoriousSc Mar 9 '17 at 23:38
• Look at the FBD in the link where the vertical component of the normal reaction must be equal to the weight. So the normal reaction must be greater than the weight. – Farcher Mar 9 '17 at 23:43
• Yes I see that. My question is since what is pushing the car in that case towards the ground to generate a force that is bigger than the weight – TheNotoriousSc Mar 9 '17 at 23:45
• The car's inertia as it wants the car to move in a straight line. – Farcher Mar 9 '17 at 23:46

Decompose the gravity force in two directions: one perpendicular to the cone surface and another tangential to it. The component perpendicular to the cone surface will be counterbalanced by the normal force. The component tangential to the surface will be just enough to keep the object moving in a circle at the velocity your textbook talks about.

If the velocity of the object were higher than this one, the object would slide up the cone. If it were smaller, it would fall to the center of the cone.

• Wouldn't the tangential force your are talking about point towards the center? Wouldn't that mean that the accelaration vector would also be towards the center? – TheNotoriousSc Mar 9 '17 at 23:05
• @TheNotoriousSc, indeed, the tangential force I am talking about points to the center and the acceleration vector is also towards the centre. Remember that the velocity vector of the object is changing, even though its magnitude is not changing. If the velocity vector is changing, there is an acceleration. Read about uniform circular motion. – toliveira Mar 10 '17 at 9:51

When referring to spinning objects that maintain a force component perpendicular to the direction of the angular velocity, look into the concept of angular momentum. The conservation of angular moment with an object traversing a circular path is what keeps 2 wheeled bikes upright and allows tops to spin. It is the cross product of the position and momentum (thus perpendicular to the position vector normal component and the velocity vector normal component).

• So it is angular momentum that in this case keeps the ball from going towards the center of the cone? – TheNotoriousSc Mar 9 '17 at 23:46
• Granted I it has been a year since I have taken Analytical Mechanics (junior physics course), though from a quick glance while walking I would say yes. If you take a bike wheel and hold it with the tire facing you while standing on a rotatable surface, then begin to spin the bike wheel, you will begin to spin on the rotating surface itself due to that same phenomena. The System is attempting to conserve angular momentum, when you limit it (in a circular system, you are 'limiting' the components to this circular path), it must be conserved elsewhere. – alexandros300 Mar 9 '17 at 23:51
• This doesn't answer the question. – Ben Crowell Oct 17 '18 at 22:49

For a frictionless object on a flat surface, the normal force vector is the negative of the projection of the force of gravity vector to the normal. Newton's third law is not broken in the case of an object sliding in circles either. According to the Wikipedia article Centrifugal force, the reactive centrifugal force is real. The object exerts a reactive centrifugal force on the cone and that's balanced by the larger than usual normal force.

The short answer to your question is that the tangential velocity $$v_0$$ causes the cone to exert a centripetal force on the ball, which can be large enough to perfectly counteract the gravitational force acting on the ball.

The long answer can be understood by looking at the different forces acting on the ball. Let's first set up a coordinate system to consider the system, which I'll define using the three unit vectors $$\hat{r},$$ $$\hat{z},$$ and $$\hat{\theta}$$. $$\hat{r}$$ points from the point of the cone outwards, along the direction of $$L$$ in your diagram. $$\hat{z}$$ points perpendicular to the surface of the cone, in the direction the normal force is acting. $$\hat{\theta}$$ will point in a radial direction of the cone, so along the dotted line in your diagram, and if I understood your diagram correctly it is in this direction that the ball begins moving with speed $$v_0$$.

The force of gravity acting downwards on the ball is $$F_g = mg$$ and the centripetal force acting horizontally on the ball is $$F_{cent.} = m\frac{v_0^2}{R} = m\frac{v_0^2}{r(t)\cos\theta},$$ where the radius of the circle the ball is going around is dependent on its initial position $$L$$ and its subsequent movements along the $$\hat{r}$$-axis.

We can make the following deductions about the total forces in the directions of the basis vectors:

• The ball won't spontaneously fly off the surface of the cone, so the force along the $$\hat{z}$$ direction must be zero.
• Assuming there are no frictional forces, the ball's tangential velocity $$v_0$$ (in the $$\hat{\theta}$$ direction) won't change either. The normal, gravitational, and centripetal forces acting on the ball are all perpendicular to the $$\hat{\theta}$$ basis vector, a.k.a. acting in the $$\hat{r}-\hat{z}$$ plane, so there are no other forces that could change the tangential velocity. Therefore, the force in the $$\hat{\theta}$$ basis must also be zero.
• For the ball to move in a circle, its velocity along the $$\hat{r}$$ axis must be zero. Some people mentioned that this is like a banking problem, however not exactly. In a banking problem we assume the car is so heavy that frictional force will ensure there is no sliding along the $$\hat{r}$$ axis, but for the ball we cannot make such an assumption. In fact here I assume there is no friction, otherwise the ball's tangential velocity would always decrease and then the ball would always fall to the bottom of the cone.

The crux of the problem comes from this last point.

A FBD (sorry I didn't have time to upload one) shows that the net force acting on the ball in the $$\hat{r}$$ direction, which is the only one we care about, is $$F_r^{NET} = m\frac{v_0^2}{r(t)} - mg\sin\theta.$$ However, for the ball to be going in a circle, the total forces along the $$\hat{r}$$ direction must be zero, as I argued earlier. Therefore, this would simplify to $$\frac{v_0^2}{L} - g\sin\theta = 0.$$ If the initial velocity has the value $$v_0 = \sqrt{Lg\sin\theta}$$, then the ball should travel in a circle around the cone. This can be intuitively understood as a result of the centripetal force required to keep the ball moving around the cone at speed $$v_0$$ perfectly counteracting the gravitational force trying to pull the ball down.

• The acceleration has a component perpendicular to the cone, so there actually is a net force in the direction of your defined $\hat z$ axis. – Aaron Stevens Nov 20 '18 at 17:43
• Also, the net force in the $\hat r$ direction isn't $0$. The acceleration also has a component parallel to the surface. You are falling into the common misconception that a radial force means a change in $r$, and this is not the case. – Aaron Stevens Nov 20 '18 at 17:55
• In an ideal banking problem you assume the car is frictionless, so this is the same problem. – Chris Nov 20 '18 at 17:58
• @AaronStevens In a rotating reference frame, including the fictitious centrifugal force, the net forces in both directions are indeed zero. – Chris Nov 20 '18 at 18:01
• @Chris Right. This answer says nothing about working in a rotating reference frame. – Aaron Stevens Nov 20 '18 at 18:03