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001
An athlete swings a 5:72 kg ball horizontally
on the end of a rope. The ball moves in a
circle of radius 0:906 m at an angular speed of
0:696 rev=s.
What is the tangential speed of the ball?
Answer in units of m=s.
002
What is its centripetal acceleration? Answer
in units of m=s2.
003
If the maximum tension the rope can with-
stand before breaking is 50:9 N, what is the
maximum tangential speed the ball can have?
Answer in units of m=s.
004
An airplane travels 94 m/s as it makes a
horizontal circular turn which has a(n) 2.3 km
radius.
What is the magnitude of the resultant
force on the 66 kg pilot of this airplane? An-
swer in units of kN.
005
In the spin cycle of a washing machine, the
tub of radius 0.237 m rotates at a constant
rate of 733 rev/min.
What is the maximum linear speed of the
water inside the machine? Answer in units of
m/s.

2007-04-02 05:37:37 · 1 answers · asked by elquida25 1 in Science & Mathematics Physics

answers and formulas would help

2007-04-02 12:19:58 · update #1

1 answers

001 v = rω
v = (0.906 m)(0.696 rev/s)(2π rad/rev)
v ≈ 3.962026 m/s

002 a = v^2/r
a = (3.962026 m/s)^2/(0.906 m)
a ≈ 17.32632 m/s^2

003 F = mv^2/r
v = √(rF/m)
Ignoring gravity,
v = √((0.906 m)(50.9 N)/(5.72 kg))
v ≈ 2.839390 m/s
(the speed and acceleration in 001 and 002 would break the rope. If gravity is figured in the speed is even lower.)

004 Fr = √((mv^2/r)^2 + (mg)^2)
Fr = m√((v)^2/(r))^2 + (g)^2)
Fr = (66 kg)√((94 m/s)^2/(2,300 m))^2 + (9.80665g)^2)
Fr ≈ 0.6951318 kN

005 v = rω
solves the same as 001 . . . remember to include 2π rad/rev

2007-04-02 19:43:23 · answer #1 · answered by Helmut 7 · 0 0

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