p11 032

background image

32.

(a) The angular speed in rad/s is

ω =



33

1

3

rev/min

 

2π rad/rev

60 s/min



= 3.49 rad/s .

Consequently, the radial (centripetal) acceleration is (using Eq. 11-23)

a = ω

2

r = (3.49 rad/s)

2



6.0

× 10

2

m



= 0.73 m/s

2

.

(b) Using Ch. 6 methods, we have ma = f

s

≤ f

s, max

= µ

s

mg, which is used to obtain the (minimum

allowable) coefficient of friction:

µ

s, min

=

a

g

=

0.73

9.8

= 0.075 .

(c) The radial acceleration of the object is a

r

= ω

2

r, while the tangential acceleration is a

t

= αr. Thus

| a| =



a

2

r

+ a

2

t

=



(ω

2

r)

2

+ (αr)

2

= r



ω

4

+ α

2

.

If the object is not to slip at anytime, we require

f

s,max

= µ

s

mg = ma

max

= mr



ω

4

max

+ α

2

.

Thus, since α = ω/t (from Eq. 11-12), we find

µ

s,min

=

r



ω

4

max

+ α

2

g

=

r



ω

4

max

+ (ω

max

/t)

2

g

=

(0.060)



3.49

4

+ (3.49/0.25)

2

9.8

=

0.11 .


Document Outline


Wyszukiwarka

Podobne podstrony:
p11 032
p11 015
p11 086
p11 083
P23 032
p11 046
032 Mostek Wheatstone'a ćwiczenieid 4668
p11 076
p11 058
p11 048
p11 093
p11 017
032
p11 049
p11 100
p11 022
p11 066
p11 044

więcej podobnych podstron