2. Gravitational potential energy-Kinetic energy:A chain with total length L is placed on africtionless table surface. One end of theL-achain is drooping with a length of a, seethe right Figure. Assuming the chain isinitially at rest, find the speed at themoment when the chain starts tocompletely leave the table.bood

Name: 2. Gravitational potential energy-Kinetic energy:A chain with total l
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Description: 2. Gravitational potential energy-Kinetic energy:A chain with total length L is placed on africtionless table surface. One end of theL-achain is drooping with a length of a, seethe right Figure. Assuming the chain isinitially at rest, find the speed

Answered: Gravitational potential energy-Kinetic…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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2. Gravitational potential energy-Kinetic energy:
A chain with total length L is placed on a
frictionless table surface. One end of the
L-a
chain is drooping with a length of a, see
the right Figure. Assuming the chain is
initially at rest, find the speed at the
moment when the chain starts to
completely leave the table.
bood

Expert Solution

Step 1

Let mass per unit length be $λ$ .

Initially the positing of center of mass = $\frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} = \frac{(λ (L - a)) \times 0 + (λ a) (\frac{- a}{2})}{λ L} = \frac{- a^{2}}{2 L}$

i.e., the center of mass is a distance $\frac{a^{2}}{2 L}$ below the table.

When chain leaves the table, at that instant, the hanging part of the chain is of length L and its center of mass is a distance L/2 below the table.

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