Question

Please Help : Q.8.57


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Answer 1

Heyy!!
Here is your answer :

Horizontal acceleration of bottom row of cylinder is (√2-1)a.

Same Type of Question was asked in

=> International Physics Olympiad held in last 20 years before as long as I can recall.

Such a nice Question :

Concepts involved :
1) Newtons Laws
2) Rotational Motion around fixed axis of rotation.

Please dont fear about that number of Planks are infinity.

It is just a trap in which the Questioner wants us fall.

Let the forces that the planks exert on the cylinders be F .i.e., The force on the given cylinder .

and G is the force from the plank above it with positive directions as defined in figure.

For Calculation see pic uploaded.

Steps :

1) Since , planks are massless,
So.apply
Newtons law on plank.

Name it as eq.(1).

2) Our aim will be to solve for linear amd angular acceleration of each cylinder in terms of the cylinders above it.

Since, we want to solve for two quantities , we will need to produce two equations relating to the acceleration of two cylinders .

=> One equation will come from a combination of F = ma , torque and Non-dlipping equation.

Name them as eq.(2) and (3), respectively

as shown in clip attached.

3) Most important step:::

Consider the ratio of the acceleration of given Plank to the force that this plank exerts on the cylinder above it.

Since, any Plank can be considered to be tje bottom plank of an infinite system. and since accelerations depend linearly on forces , this gforce is same for all levels.

We will use this fact for the first two levels to determine Rα in terms of A1.(acceleration of cylinder bottom most )

Finally , after obtaining F1 , A1 , F 2 , A 2 .

Equate to their coresponding ratios ..

At the end we will be getting , Rα1 = √2a1

Use this in Non slipping condition .

And then, acceleration of cylinder 1 will be (√2-1)a.

Answer 2

plz refer to the attachment for answer here

thanks