Chapter 13 Solutions - Mosinee School District

Chapter 13
v
r
outer
outer, max vinner, max
rinner
15.0 mm 0.25 cm s 1.3 cm s
3.00 mm
13.75 (a)
k
m
500 N m
2.00 kg
15.8 rad s
(b)
Apply Newton’s second law to the block while the elevator is accelerating:
Fy Fs mg ma y
With F kx and a g 3 , this gives kx m g g 3 , or
s
y
x
4 mg 4 2.00 kg 9.80 m s
3k
3 500 N m
2
5.23 10
2
m 5.23 cm
13.76 (a) Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius
L y f . Also, observe that the y-coordinate of the object at this point must be negative y f 0 so the spring is
stretched and exerting an upward tension force of magnitude greater than the object’s weight. This is necessary
so the object experiences a net force toward the pivot to supply the needed centripetal acceleration in this
position. This is summarized by Newton’s second law applied to the object at this point, stating
mv2
Fy
ky f mg
L y
f
(b) Conservation of energy requires that E KEi PEg, i PEs, i KEf PEg, f PE s,
f
, or
1
2
1
E 0 mgL 0 mv mgy
2
f ky f , reducing to 2mg L y
2 2
f mv ky f
2 2
(c) From the result of part (a), observe that mv2 ( L y )( ky mg ) . Substituting this into the result from
part (b) gives 2 mg( L y ) ( L y )( ky mg)
ky
2
. After expanding and regrouping terms, this
f f f f
becomes (2 k) y2
(3 mg kL) y ( 3 mgL ) 0 , which is a quadratic equation ay2 by c 0 , with
f
f
f
f
f
f
a 2k
2 1 250 N m 2.50 103
N m
b 3mg kL 3 5.00 kg 9.80 m s2 1 250 N m 1.50 m 1.73 103
N
and
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