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at the highest point on the path. ote that the gravitational force has the same magnitude

and direction at each poi nt on the circular path.The tension force varies in magnitude at

different points and isalways directed toward the center of the path.

(c) Atthetop ofthecircle,= mv2/r = T + Fg,or

T=m---F=m---mg=m -- g)

r g r r

20

= (0.275 kg)[(5. mfs? - 9.80 m/s2]= 16.05 Nl0.850 m

(d) At the bottom of the circle,Fe = mv2/r = T - Fg = T - mg, and solving for the

speed gives

v2 = ;,(T-mg)= r(- g) andIf thestring is at thebreaking point at thebottom of the circle, then T=22.5N, and the

speed of the object at thispoint must be

v = (0.850 m)22.5 N

( 0.275 kg- 9.80 m/s2

) = 17.82 m/s l

48. In a home laundry dryer, a cylindrica l tub containing wet

clothes is rotated steadily about a horizontalaxis as shown

in Figure P6.48.So that the clothes wiH dry uniformly, they

are made to tumble. T he rate of rotation of the smooth-

FigureP6.48

walled tub is chosen o that a small piece of cloth will lose

contact with the mb when the cloth is at an angle of() =68.0 above the horizontal. If the radius of the tub is r =

0.330 m, what rate of revolution is needed?

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P6.48 When the cloth is at a lower angle 9,the radial component ofLF = ma reads. mv

2

n+mgsmO=-r

p

mg

pC:U in68

mgcos68

ANS FIG.P6.48

v2

At8 = 68.0, the normal force drops to zero and g sin 68 =-.r

The rate of revolution is

angu lar speed = (1.73 m/s)(1rev)( t1Cr ))=I 0.835 rev/s I = 50.1 rev/tnin

2JCr 2JC 0.33 m