Problem#1
A hawk flies in a horizontal arc of radius 12.0 m at a constant speed of 4.00 m/s. (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc but increases its speed at the rate of 1.20 m/s2. Find the acceleration (magnitude and direction) under these conditions.
Answer:
(a) centripetal acceleration is given by
ac = v2/r
ac = (4.00 m/s)2/(12.0 m) = 1.33 m/s2
(b) the acceleration (magnitude and direction) under these conditions is given by
a = [ac2 + at2]1/2
a = [(1.33 m/s)2 + (1.20 m/s)2]1/2 = 1.79 m/s2
at angle tan θ = (ac/at)
then, θ = tan-1(1.33/1.20) = 48.00 inward
Problem#2
A pail of water is rotated in a vertical circle of radius 1.00 m. What is the minimum speed of the pail at the top of the circle if no water is to spill out?
Answer:
From Newton’s second law
∑Fy = may
n + mg = mv2/r
But n = 0 at this minimum speed condition, so
mv2/r = mg
v2 = gr = (9.80 m/s2)(1.00 m)
v = 3.13 m/s
Problem#3
A 0.400-kg object is swung in a vertical circular path on a string 0.500 m long. If its speed is 4.00 m/s at the top of the circle, what is the tension in the string there?
Answer:
At the top of the vertical circle,
∑Fy = may
T + mg = mv2/r
T = m(v2/r – g)
= (0.400 kg)[(4.00 m/s)2/(0.500 m) – (9.80 m/s2)]
T = 8.88 N
Problem#4
A roller coaster car (Fig. 4) has a mass of 500 kg when fully loaded with passengers. (a) If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the car at this point? (b) What is the maximum speed the vehicle can have at B and still remain on the track?
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| Fig.4 |
Answer:
(a) at A: v = 20.0 m/s, R = 10.0 m and n = force of track on roller coaster, then
∑F = mac
nA – mg = mvA2/rA
nA = m(g + vA2/rA)
nA = (500 kg)[(9.80 m/s2) + (20.0 m/s)/(10.0 m)] = 24.9 kN
(b) at B, ∑F = mac
nB + mg = mvB2/rB
The max speed at B corresponds to nB = 0, then
mg = mvB2/rB
vB,max = √(grB) = √[(9.80 m/s2)(15.0 m)]
vB,max = 12.1 m/s
Problem#5
A roller coaster at the Six Flags Great America amusement park in Gurnee, IL, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop (Fig. 6). The cars ride on the inside of the loop at the top, and the speeds
are high enough to ensure that the cars remain on the track. The biggest loop is 40.0 m high, with a maximum speed of 31.0 m/s (nearly 70 mi/h) at the bottom. Suppose the speed at the top is 13.0 m/s and the corresponding centripetal acceleration is 2g. (a) What is the radius of the arc of the teardrop at the top? (b) If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top? (c) Suppose the roller coaster had a circular loop of radius 20.0 m. If the cars have the same speed, 13.0 m/s at the top, what is the centripetal acceleration at the top? Comment on the normal force at the top in this situation.
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| Fig.6 |
Answer:
(a) the radius of the arc of the teardrop at the top is
ac = v2/r
r = v2/ac = (13.0 m/s)2/(2 x 9.80 m/s2) = 8.62 m
(b) Let n be the force exerted by the rail. Newton’s law gives
n + mg = mv2/r
n = m(v2/r – g) = m(2g – g) = mg (downward)
(c) ac = v2/r = (13.0 m/s)2/(20.0 m) = 8.45 m/s2
If the force exerted by the rail is n1
then
n1 + mg = mac
n1 = m(ac – g)
which is < 0 since ac = 8.45 m/s2
Thus, the normal force would have to point away from the center of the curve. Unless they
have belts, the riders will fall from the cars. To be safe we must require n1 to be positive. Then ac > g. We need
v2/r > g
or
v > √(gr) > √[(9.80 m/s2)(20.0 m)]
v > 14.0 m/s
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