Elementary? - the cannonball and the carpet!

let me take a crack at this, but keep in mind i just got back from karting, water polo game, and 1 pitcher of molson dry...lol...Engineering!

Okay, so you have statically a cannon ball sitting on the carpet. The only forces acting at this point are gravity on the ball, and no forces on the carpet. If the carpet is accelerated horizontally to the right, the ball will eiter come with it, or roll counter-clockwise in place. I'm thinking that you have a frictional force acting between the ball and the carpet, being

First for the ball to move laterally with it, no rolling. The frictional force:

F= (coeff of static fric)*(Mass of cannon ball)*(acceleration due to gravity)

Acts in the direction of motion. I'm guessing if the force acting on the carpet is greater than this, then the ball will rotate.

Now for rotation. This force exerts a torque T on the ball. To have an acceleration "a"(rotational), is governed by newtons second for rotation:

T= I*a(rotational) ---or----- T=I*a(tangential)/R R=radius

I think that's a start. Up to that point i'm comfortable, but now i'm going to have a shower. Let me know if i'm right or wrong guys, i'm not afraid to take a crack at it.

-jp

I'll give this a shot...

Place a cannon ball on a plush carpet.

Push the top of the ball horizontally, and measure the force at which it begins to roll. That's the force that allows it to compress some new carpet, while releasing some squished carpet.

Now you have a number, redo the experiment. This time accelerate the carpet under the ball gently. If the acceration is insufficient to generate a static friction force on the ball sufficient to make it roll (i.e. less than the value of force you measured earlier), it just goes along for the ride. If the acceration is more brisk, the ball begins to roll.

The relation given by Newton was this: the force is proportional to the balls mass and to the size of the acceleration.

Clear as mud?

Brian Whatcott

Brian

Maybe this isn't so easy after all... :-) (like many "easy" things).

So it's the friction of the ball on the carpet that "decides" when it starts rolling? If the carpet were instead a sheet of glass then the ball should start rolling with a lesser force of accel needed? I was also wondering about this as it applies to that famous trick where someone pulls the tablecloth straight out from under a tableful of china cups! I suppose in that case the accel is so large that the cups are slipping (skidding?) on the cloth.

LOL, it is a little involved. Since last night i've been thinking about it. I think my attempt is halfway there, but is only the beginning. I did some trial runs with a bowling ball and a towel(yes, i am a loser) just to get a qualitative understanding of the system, and then i hit the physics textbooks.

From what i saw, when the ball rolled, it stayed in the same place, so it had no tranverse kinetic energy, just rotational energy because the center of mass stays pretty much in the same place. When the ball dosent roll it has only kinetic energy because it's center of mass is moving with velocity "v" in the same direction of the carpet. Now we know from newtons second that

F = ma ; so for the couple system of mass "m(sys)", we have: a = F/m(sys). If this acceleration is fast enough, the ball will roll while slipping on the carpet. So far that's what i understand.

acceleration is

sufficiently

(If this is another Professor-using-students-for-outside-fees-because-he-can't-get-it, have your professor send his consultant's check to Toys for Tots.)

1) you have the mass of the ball, which tends to hold the ball in place and thus, because the original frame of reference for rotation moves to reduce net acceleration, REDUCES the amount of available tangential force for rotation.

the mass moment of the ball, which, with the tangential force from the weight of the ball acting on the carpet-ball interface to resist rotation, determines the AMOUNT of rotation

the acceleration of the carpet relative to the ball, which always produces a rotation (this parameter is not a force which can be present and not displace- this is an acceleration which by definition displaces)

and the discontinuous rolling resistance of the ball on the inelastic carpet, which in its lower area does not deform and holds the ball in place and prevents acceleration, and in its upper area deforms and allows the ball to displace relative to the carpet, reduced by the inelastic value. Think of this as the ball having a flat spot, the larger the inelastic coefficient the bigger the flat spot. The input force needs to overcome the flat, i.e., to lift the ball over the edge of the flat on the ball so that the ball can roll, and it does that by grabbing the flat and flipping the ball around by lifting and squeezing the edge of the ball as it turns. (The end value here is when the ball is a block, and the force from trying to accelerate the block's mass has to be larger than the friction force for the block to separate from the carpet.) The value is proportional to the inelastic constant. But above a certain constant value, the force required to rotate is greater than static friction, and the ball will slide rather than rotate.

So IMHO, the ball always rotates in a perfectly elastic system. When the ball rotates in an inelastic system depends on the inelastic value, i.e., once the force required to overcome the deformation is achieved, the ball rotates. Little deformation, small carpet acceleration - lot of deformation, more carpet acceleration. When the force required to rotate the ball is greater than the force required to slide the ball, the ball will not rotate, it will just slide on the carpet

best guess.....

Rest assured, it is/was a perfectly genuine post - just for interest sake. I am not any kind of professor.

not a problem -

(perched firmly on the rectangular container for fat-alkalis compounds - the professor syndrome of which I speak is where they will consult on a matter for pay and then have students do all the work, work required of the students for advancement and gratis.

I may be oversensitive to the abuse of positi>

acceleration