A uniform solid disk of radius R is set into rotation with an angular speed ?i a

Question

A uniform solid disk of radius R is set into rotation with an angular speed ?i about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and immediately released as shown in the figure below.

(a) What is the angular speed of the disk once pure rolling takes place? (Use the following as necessary: ?i. ) ?f = ?i? 3? Correct: Your answer is correct.

(b) Find the fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs.

?E E = ?

(c) Assume the coefficient of friction between disk and surface is ?. What is the time interval after setting the disk down before pure rolling motion begins? (Use the following as necessary: R, ?i, ? and g. )

?t = ?

(d) How far does the disk travel before pure rolling begins? (Use the following as necessary: R, ?i, ? and g. )

?x =?

Explanation / Answer

let,

radius of the disk is R

mass of the disk is m

initial angular speed is Wi

a)

final angular momentum is Wf

by using law of conservation of angular momentum,

L_initial = L_final

I*Wi=(I*Wf+(m*R^2)*Wf)

(1/2*m*R^2)*Wi=(1/2*m*R^2*Wf+m*R^2*Wf)

(1/2)*Wi=(1/2*Wf+Wf)

(1/2)*Wi=(3/2*Wf)

=========> Wf=(Wi/3)

b)

fractional chaange in K.E is, (K2-K1)/K1

here,

K1=1/2*I*Wi^2

K2=(1/2*I*wf^2+1/2*m*V^2)

K2=(1/2*I*wf^2+1/2*m*(R*Wf)^2)

now,

(K2-K1)/K1=(1/2*I*Wf^2+1/2*m*(Vf)^2-1/2*I*Wi^2)/(1/2*I*Wi^2)

(K2-K1)/K1=(1/2*I*Wf^2+1/2*m*(R*Wf)^2-1/2*I*Wi^2)/(1/2*I*Wi^2)

(K2-K1)/K1=(1/2*I*Wf^2+I*(Wf)^2-1/2*I*Wi^2)/(1/2*I*Wi^2)

(K2-K1)/K1=(Wf^2+2(Wf)^2-Wi^2)/(Wi^2)

(K2-K1)/K1=(3*Wf^2-Wi^2)/(Wi^2)

(K2-K1)/K1=(3*(Wi/3)^2-Wi^2)/(Wi^2)

(K2-K1)/K1=(1/3-1)

(K2-K1)/K1=-2/3

fractional chaange in K.E is (K2-K1)/K1=-2/3 <----------------

c)

frictional coefficient between disk and surface is u

change in momentum,

dP=f*dT

m*v=(u*mg)*dT

v=(u*g)*dT

(R*Wf)=(u*g)*dT

(R*Wi/3)=(u*g)*dT

===> dT=(R*Wi/(3u*g))

time interval, dT=(R*Wi/(3u*g))   ( before pure rolling)

d)

by using enegry relation,

W=1/2*I*Wf^2 (before pure rolling)

(f*x)=1/2*(m*R^2)*(Wi/3)^2

(u*(m*g)*x)=1/2*(m*R^2)*(Wi/3)^2

(u*g*x)=(R^2*Wi^2)/18

====> x=(R^2*Wi^2)/(18*u*g)

distance travelled by the disk, x=(R^2*Wi^2)/(18*u*g)

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