1. A bead can slide along a frictionless metal o-ring of radius 20.0 cm. The ring is in a vertical planeand rotates about a vertical axis through its centre at a constant angular speed , as shown

Given quantities: = 3 rev/s = 6pi rad/s and radius of ring = 0.2 m d

0.2m

~Fgr = 0 xmg y (1)~FN = ~FN sin() x+ ~FN cos() yN (2)

As it is in vertical equilibrium,

~Fx =

mv2

dx (3)

~Fy = ~0 (4)

Combining equation 3 with equations 1 and 2,

~Fx =

mv2

dx

~FN sin() x = mv2

dx

~FN = mv2

d sin()(5)

Combining equation 4 with equation 1 and 2, ~Fy = ~0

mg y + ~FN cos() y = ~0 ~FN cos() = mg (6)

Substituting equation 5 into equation 6,

mv2 cos()

d sin()= mg

v2 cot()

d= g

cot() =dg

v2

tan() =v2

dg(7)

We know that v = wd,Substituting that into equation 7,

tan() =d2

g(8)

Substituting d = 0.2 sin() into equation 8, (9)

1

tan() =0.2 sin()2

g

1

cos()=

0.22

g

cos() =g

0.22(10)

Substituting = 6pi into equation 10, we get = 82.073o. (b) for = 1rad/s,substituing it inequation 10 we see that the solution does not exist.

2.

Eki = 15J (1)

Ekf = 18J (2)

F (x) = 5 cx+ x2 (3)

Using the work-energy theorem,

W = Ek xfxi

Fdx = 18 15 32

(5 cx+ x2) dx = 3[5x c

2x2 +

x3

3

]3

2

= 3[15 c

29 +

33

3

][10 c

222 +

23

3

]= 3

c =10

3

2