A block of mass M with semicircular track of radius R ,rests on a horizontal frictionless surface.A uniform cyllinder of radius r and mass m is released at the top (pont A)The cyllinder slips on the semicircular frictionless track.How far has the block moved when the cyllinder reaches the bottom (point B)of the track?[Ans :-m(R-r)/(M+m)] How fast is the block moving when the cyllinder reaches the bottom of the track?

Question

Archit Uttarwar · Answer

Here external force is 0 therefore, Let x be the displacement of the wedge, And r is the radius of cylinder and R is radius of semicircular path 0= m(R-r-x) - Mx x(M+m) =m(R-r) x=m(R-r)/(M+m)

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Mechanics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship