Sticky ball falling down conveyer

[EFuzzy]

While searching for a collection of interesting mechanics problems, I stumbled across this set of problems. http://www.mechanics-book.net/mechanics_problems.pdf Problem 57 and the given solution are attached. However, I think there is an error in the solution. They assume that velocity of the conveyor is v*cos(a/2), but I think that v*cos(a/2) gives the x-component of the velocity of the sticky ball, which is greater than the velocity of the conveyor. Am I missing something?

Thanks

 

[Doc Al]

I agree with you--the solution seems a bit off. I'd say that the velocity of the body with respect to the ground will equal its velocity with respect to the trianguar frame (at angle alpha) plus the velocity of the frame with respect to the ground. And the speed of the frame with respect to the ground must equal the speed of the body with respect to the frame (since it's just the speed of the conveyer belt with respect to the frame).

 

[EFuzzy]

OK, thanks a lot!

I'm kind of disappointed actually, because the book seemed to contain some interesting problems, but it's not really worth it if some of the solutions are wrong.

 

[Shooting Star]

Doc Al,

Could you explain this sentence given in the problem.

All interactions are so perfect that they don’t let any part of the system slide.

The ball can roll down, but not slide or what? Also, is the conveyor belt fixed, or can it rotate around the three masses, due to reaction of the ball M?

I have got a pertinent question, but this statement is somewhat confusing, and has to be cleared first.

 

[Doc Al]

I assumed that it was just a badly phrased statement. I took it to mean that there's no slipping between surfaces (between the falling mass and the belt and between the belt and the surface it moves across). The dot is not even a ball, it's some "point body" that sticks to the conveyer belt and pulls it along as it moves down the incline. The conveyor belt is not fixed, it is free to slide over the cylindrical masses. (No energy loss due to friction, of course!)

 

[Shooting Star]

Oh, it's glued to the belt. That changes everything.

 

[entphy]

I do agree that the solution provided in the 2nd attched diagram could have made a mistake. The horizonal component of the velocity of mass M is not the same as the velocity of the conveyor frame. The distance that M travels over the same period of time is longer than the conveyor frame.

The magnitude of the velocity and acceleration of M with respect to the conveyor frame along the slope is simply the same as the velocity and acceleration of the conveyor frame with respect to the ground, given that there is no slipping across all surfaces and no frictions about the 3 cylindrical masses of m.

The max velocity of M reaching the bottom at angle $\alpha$/2 is
$\sqrt{\frac{2Mgh}{cos(\alpha/2)[2M+3mcos(\alpha/2)]}}$