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Accelerating Cylinders !!Someone pls help !!

I am stuck in this question. Please help!!

Three identical cylinders are arranged in a triangle(one lying on top of the other two and all 3 are in contact), with the bottom two lying on the ground.The ground and the cylinders are frictionless. You apply a constant horizontal force (directed to the right) on the left cylinder. Let a be the acceleration you give to the system. For what range of a will all three cylinders remain in contact with each other ?

Note by Ch Nikhil
2 years ago

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were you doing david-morin at that time ? Shubham Dhull · 5 months, 4 weeks ago

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At the time when the upper cylinder just leaves the contact with the bottom cylinder.

N Sin60 = mg;
N cos60 = ma;
a = g cot60 = g \(\frac{1}{\sqrt3}\) Rohit Gupta · 2 years ago

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@satvik pandey @Kushal Patankar please help!! Ch Nikhil · 2 years ago

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@Ch Nikhil Hi Nikhil!

img

img

It's is obvious that the cylinder at the top will first loose contact with cylinder at right. The moment it looses contact with that the normal force acting between them would be zero. If we consider a frame of reference which is moving with the CoM of the sytem then pseudo force on top cylinder is \(ma\).

So

\(N_{1}cos(30)=mg\)....(1)

and \(N_{1}sin(30)=ma\)......(2)

Solving this we get a value of ,say \(a_{0}\). So the cylinders would loose contact for all accelerations greater than \(a_{0}\). Satvik Pandey · 2 years ago

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@Satvik Pandey Thanks a lot @satvik pandey !! Ch Nikhil · 2 years ago

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@Ch Nikhil You are welcome! Satvik Pandey · 2 years ago

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@Satvik Pandey Hi Satvik, Can you tell me how you draw the diagram and upload it..?? Rohit Gupta · 2 years ago

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@Rohit Gupta Hi Sir!! I draw these diagrams using paint.

First upload the diagram on imgur. Then type:

! [ anything] ( url of the image). Without spaces. :) Satvik Pandey · 2 years ago

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@Satvik Pandey Thanks.... ;) Rohit Gupta · 2 years ago

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@Rohit Gupta You are welcome, Sir! :) Satvik Pandey · 2 years ago

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