Sliding motion along a frictionless circular incline
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This article describes the sliding motion of a small block placed on a frictionless circular incline. This is similar to the situation of sliding motion along an inclined plane, except that the inclined plane has constant slope whereas in the case of the circular incline, the slope is constantly changing.
The more general version with friction is computationally much harder in terms of its kinematic evolution, and is discussed under sliding motion along a circular incline.
If denotes the angle that the radial line to the circular incline makes with the vertical, then is also the angle made by the tangent line with the horizontal. Locally, the analysis when the block is at such a point on the circular incline looks very similar to the analysis of sliding motion along a frictionless inclined plane where the plane makes an angle of with the horizontal. The key difference is that, since the motion is circular, the normal component of acceleration is not zero; rather, it is given by where is the speed and is the radius.
- 1 Basic components of force diagram
- 2 For a block initially placed at rest with radial making angle with vertical
Basic components of force diagram
A good way of understanding the force diagram is using the coordinate axes as the axis tangent to the circular incline and normal to the circular incline (which in this case is radial). For simplicity, we will assume a two-dimensional situation, with no forces acting along the horizontal axis that is part of the circular incline (in our pictorial representation, this no action axis is the axis perpendicular to the plane).
The two candidate forces
Assuming no external forces are applied, there are two candidate forces on the block:
|Force (letter)||Nature of force||Condition for existence||Magnitude||Direction|
|gravitational force||unconditional||where is the mass and is the acceleration due to gravity||vertically downward, hence an angle with the normal to the incline and an angle with the incline|
|normal force||unconditional||adjusts so that there is no net acceleration perpendicular to the plane of the incline||outward normal to the incline|
Taking components of the gravitational force
The most important thing for the force diagram is understanding how the gravitational force, which acts vertically downward on the block, splits into components along and perpendicular to the incline. The component along the incline is and the component perpendicular to the incline is . The process of taking components is illustrated in the adjacent figure
Component perpendicular to the circular incline
In this case, assuming a stable surface of contact, and that the circular incline does not break under the weight of the block, and no other external forces, we get the following equation from Newton's first law of motion applied to the axis perpendicular to the circular incline:
where is the normal force between the block and the circular incline, is the mass of the block, and is the acceleration due to gravity. acts inward on the circular incline and outward on the block.
Component along (down) the circular incline
For the axis down the circular incline, the gravitational force component is . There are no other forces in this direction, so, if denotes the tangential acceleration (which is also the change in speed) measured positive in the downward direction, we get:
After cancellation of , we get:
Note that if the block is sliding upward (for instance, if given an initial upward velocity) this acceleration functions as retardation, whereas if the block is sliding downward (which may happen if the block is placed at rest, or given an initial downward velocity, or of it turns back after sliding upward) then the acceleration increases the speed.
The overall system of equation is thus:
where we have:
For a block initially placed at rest with radial making angle with vertical
If the block is placed right on top (), it is in equilibrium, but the equilibrium is an unstable equilibrium, so even a very minor perturbation will cause the block to slide down. We assume that it starts sliding down with a zero initial velocity and deduce the kinematics.
We have that:
Eliminating , we get:
Integrating, we get that:
This can also be deduced from energy considerations: the height difference between angles and is , so the loss in potential energy is . This becomes kinetic energy, since there is no loss due to friction, and we get:
Solve to get above.
In the case , this becomes:
When the speed is sufficiently high, the block flies off the circular incline. To determine this, we need to set . We get:
This simplifies to:
Cancel the and and we get:
So we get:
At this point, the speed is:
In particular, when , we get and .
Reintroducing time: kinematics
The time taken to reach from to can be computed using the following integral:
This integral need not be computable by elementary means, though it is so for some values of . The case is discussed below:
Fill this in later