Question

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Can someone please help me with this question?

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Answer 1

Explanation:

In the problem, the rigid body is in translation equilibrium but there is an angular retardation. We first sketch the free body diagram of the cylinder. Obviously, the friction forces, acting on the cylinder, are kinetic. From the condition of translational equilibrium for the cylinder.

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Answer 2

Question:-

a uniform cylinder of radius r is spinned about its axis to the angular velocity w0 and then placed into a corner (fig. 1.58 ) the coefficient of friction between the corner walls and the cylinder is equal to how many turns will the cylinder accomplish before it's stops .

${\red{\underline{\underline{\bold{Answer:-}}}}}$

From the condition of translational equilibrium for the cylinder.

mg=N 1 +KN 2 ;NN =KN 1

N 1 = mg/1+kk ;N 2 =K mg/ 1+k2

For pure rotation the cylinder about its rotation axis,

N 2 =Iβ 2

or,−KN 1 R−KN 2R= mR 2/2 β 2

or KmgR(1+K)/− 1+K 2 =mR2/2 β 2

or,β 2 = -2K(1+K)g/ (1+K 2 )R

Now from the kinematical equation

w 2 =w 02 +2β 2 Δϕ

Δϕ= w 02 (1+K 2 )R/4K(1+K)g

because w=0

Hence the sought number of turns,

n= Δϕ/2π =w 02 (1+K 2 )R/ 8πK(1+K) g