6

An object is dropped from the top of a circular curve with radius r and rolls down the curve until it reaches the bottom. What would be the equation that would give the velocity of the object at any given time? Assume that the gravity of earth is acting upon it (approx 9.8m/s²). Ignore the effects of friction.

enter image description here

Community
  • 1
Jacob
  • 2,387
  • 12
  • 20
  • By top do you mean the part furthest left or right (because if you drop it from the highest point it would fall straight down)? – Gabriel Jan 27 '15 at 00:57
  • This may give you an idea: https://www.physicsforums.com/threads/ball-rolling-down-smooth-curve.395901/ ; he derives a formula for the velocity based on the angle – Gabriel Jan 27 '15 at 01:33
  • http://en.wikipedia.org/wiki/Circular_motion#Non-uniform and http://en.wikipedia.org/wiki/Centripetal_force also have a lot to say about this. – Gabriel Jan 27 '15 at 01:38

3 Answers3

2

Let $g=9.8\,\mathrm{m}/\mathrm{s}^2$ be the acceleration due to gravity, $r$ the radius of the circular arc, $s$ the distance the object has traveled from the start, and $\theta$ the angle of declination of the object from the center of the circular arc.

Then at angle $\theta$ the acceleration in the direction of the path would be $g\cos\theta$. Therefore,

$$\frac{d^2s}{dt}=g\cos\theta$$

Substituting $\theta=\dfrac sr$,

$$\frac{d^2s}{dt^2}=g\cos\left(\frac sr\right)$$

We have the initial conditions $\theta(0)=0$ and $\frac{d\theta}{dt}=0$. So we just have an initial value problem to solve to find $s$. Differentiate that to get your desired speed.

This is similar to the formula for a pendulum. In fact, If you replace $s$ with $\dfrac{{\pi}r}2-s$ (which is the distance to the bottom of the arc), then divide that by $r$ to change the unit, you get a the "Arbitrary-amplitude period" pendulum formula with special initial values. See the linked Wikipedia article for the solution to that--the solution involves elliptic integrals, which are beyond me.

Rory Daulton
  • 32,288
  • This doesn't make sense. Why would the acceleration depend on the radius of the arc? This would mean at the starting point, the acceleration would be a fraction of gravity, as it's divided by the radius of the arc, whereas the acceleration should simply be gravity, since it's falling straight down. – Cerin Aug 01 '17 at 06:28
  • @Cerin: You failed to notice that the "acceleration" in my final equation is that of the angle subtended from the center of the circle and not of position or distance. The does indeed depend on the radius--if the circle is large, a large distance covered on the circular arc with make only a small change in the angle. – Rory Daulton Aug 01 '17 at 09:49
  • @Cerin: I changed my answer to explicitly show the usual acceleration rather than that of the angle. Is that more clear? – Rory Daulton Aug 01 '17 at 12:30
  • Yes, thank you. – Cerin Aug 01 '17 at 14:26
0

The acceleration, $\frac{\mathrm{d}v}{\mathrm{d}t}$, is proportional to the sine of the angle of descent, $\frac{\mathrm{d}y}{\mathrm{d}s}$. That is, $$ \frac{\mathrm{d}v}{\mathrm{d}t}=-a\frac{\mathrm{d}y}{\mathrm{d}s} $$ Since $v=\frac{\mathrm{d}s}{\mathrm{d}t}$, we get $$ \frac12v^2=\int v\,\mathrm{d}v=\int\frac{\mathrm{d}v}{\mathrm{d}t}\,\mathrm{d}s =-\int a\,\mathrm{d}y=a(y_0-y) $$ where $y_0$ is the altitude at which the speed is $0$.

robjohn
  • 345,667
0

Answers above are for an object sliding down the arc. The heading clearly states that the object rolls down the arc. The question itself just says "dropped". So does it roll or slide?

If it rolls we obviously need to know the shape of the object--suppose it is a ball or a cylinder and is symmetrical about a horizontal axis.

If I were allowed to comment on other answers, which I am not, I would have said that, in the case of rolling, it is also necessary to take account of the rotational kinetic energy. This complicates the problem. The answer would then depend on the moment of inertia. This depends on whether the ball or cylinder is solid and uniform or a hollow shell etc.

To solve this, equate the potential energy to the sum of the translational kinetic energy and the rotational kinetic energy. It turns out that the equation is the same as for sliding except that the constants are different--depending on both the mass and the moment of inertia as well as the radius of the object.

Terry Moore
  • 11