Problem#1
A space station, in the form of a wheel 120 m in diameter, rotates to provide an “artificial gravity” of 3.00 m/s2 for persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions per minute) that will produce this effect.
Answer:
Standing on the inner surface of the rim, and moving with it, each person will feel a normal force
exerted by the rim. This inward force causes the 3.00 m/s2 centripetal acceleration:
ac = v2/r
3.00 m/s2 = v2/(120 m)
v = 13.4 m/s
The period of rotation comes from
v = 2πr/T
13.4 m/s = 2π(120 m)/T
T = 28.1 s
so the frequency of rotation is
f = 1/T = (1/28.1 s)(60 s/1 min) = 2.14 rev/min
Problem#2
An amusement park ride consists of a rotating circular platform 8.00 m in diameter from which 10.0-kg seats are suspended at the end of 2.50-m massless chains (Fig. 1). When the system rotates, the chains make an angle θ = 28.0° with the vertical. (a) What is the speed of each seat? (b) Draw a free-body diagram of a 40.0-kg child riding in a seat and find the tension in the chain.
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| Fig.1 |
Answer:
(a) The mass at the end of the chain is in vertical equilibrium. Thus
∑Fy = 0
T cosθ = mg (*)
and
∑Fx = mv2/r
T sinθ = mv2/r (**)
with
r = 4.00 m + 2.5 m sin 28.00 = 5.17 m
then, from (*) and (**)
tanθ = v2/rg
v2 = rg tanθ
v2 = (5.17 m)(9.80 m/s2) tan28.00
v = 5.19 m/s
(b) from (*), we get
T = mg/cosθ
T = (50.0 kg)(9.80 m/s2)/cos28.00 = 555 N
Problem#3
A piece of putty is initially located at point A on the rim of a grinding wheel rotating about a horizontal axis. The putty is dislodged from point A when the diameter through A is horizontal. It then rises vertically and returns to A at the instant the wheel completes one revolution. (a) Find the speed of a point on the rim of the wheel in terms of the acceleration due to gravity and the radius R of the wheel. (b) If the mass of the putty is m, what is the magnitude of the force that held it to the wheel?
Answer:
(a) The putty, when dislodged, rises and returns to the original level in time t. To find t, we use
vf = vi + at
–v = v – gt
t = 2v/g
where v is the speed of a point on the rim of the wheel.
If R is the radius of the wheel,
v = 2πR/t
v = 2πR/(2v/g)
v = √(πRg)
(b) The putty is dislodged when F, the force holding it to the wheel is
F = mv2/R
F = πmg
Problem#4
An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough such that any person inside is held up against the wall when the floor drops away (Fig. 3). The coefficient of static friction between person and wall is µs, and the radius of the cylinder is R.
(a) Show that the maximum period of revolution necessary to keep the person from falling is T = (4π2Rµs/g)1/2. (b) Obtain a numerical value for T if R = 4.00 m and µs = 0.400. How many revolutions per minute does the cylinder make?
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| Fig.3 |
Answer:
Normal force behaves as centripetal force, therefore
n = mv2/R
and in vertical motion we get a relationship
f – mg = 0
f = mg
with
f = µsn, then
mg = µsmv2/R
v2 = gR/µs
because, v = 2πR/T, then
(gr/µs)1/2 = 2πR/T
T = 2π(Rµs/g)1/2
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