Beetle on Sphere
A Beetle starts from rest at the top of a sphere of radius R and slides* on the sphere under the force of gravity. How far below its starting point does it get before flying off the sphere?
* assume sphere is frictionless.
* assume sphere is frictionless.
Labels: funphysics
posted by Rajesh Lal at
3:09 PM
11 Comments:
If you're counting only vertical distance (like I percieve)then the distance is R
PQ
Well, with no friction, the beetle will start sliding off as soon as it's center of gravity moves from top dead center. This is not exactly "flying", but since most beetles can fly, it can fly off from top dead center, with no sliding at all.
if you want the distance traveled on the sphere it would be (2*pi*R)/4
Verticle distanse travelled is r/3
The vertical drop is R/3. The beetle will have moved through an angle of approx 48.2 degrees; it is not a "pretty" angle, so the surface distance travelled is not pretty either.
The beetle comes off the sphere when the component of gravity perpendicular to the surface is no longer sufficient to provide the centripetal force required to hold the beetle onto the sphere.
The calculation is very simple, but needs a geometric diagram to explain it.
This post has been removed by the author.
yes using the mgR*2=mgR1+mV^2/2 and mg*(R1-R)/R=mV^2/R because at the very moment of leavin the sphere N=0 u get the solution R1=5/3*R and the drop of R2=2R-R1=R/3
As milos' answer is overly concise and non-explanatory, I'll try to flesh it out.
Let h be the vertical drop that we are seeking (R2 in milos' answer).
Then when the beetle has dropped through height h, it will have gained KE = mgh = (1/2) m v^2.
=> v^2 = 2gh.
Let A be the angle that the beetle is at (measured from the centre of the sphere. Angle = 0 at top.
The component of g normal to the sphere's surface is g*cos(A).
But cos(A) = (R-h)/R {adjacent side = R-h, hypoteneuse = R}.
The centripetal force holding the beetle on = mv^2/R = 2ghm/R.
Equating to mg*cos(A) =>
2ghm/R = mg(R-h)/R => 2h = (R-h) => h = R/3 is the answer.
chris quite a great explanation but i didnt feel like putting all the explanations because of the cos etc...
chris quite a great explanation but i didnt feel like putting all the explanations because of the cos etc...
Hi milos, Thanks for the compliment. I feel a bit mean now. I've just been ripping your "weight problem" solution into shreds.
Post a Comment
Links to this post:
Create a Link
<< Home