HW 6
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Homework 6
1. 1 revolution = 2π radians and 1 minute = 60 seconds. Therefore, 56 revolutions per minute = 56 × 2π/60
rad/s= 5.86 rad/s.
2. a)x = rθ, or θ = x/r radians. Therefore, θ = 5000/4000 = 1.25 radians.
1 radian=180/π, therefore, 1.25 radians = 1.25 × 180/2π = 71.62◦ .
b) Angular velocity = angular distance/time. The plane takes 8 hours, thus, ω = 1.25/(8 × 60 × 60) =
4.34 × 10−5 rad/s.
3. Your roommate is working on his bicycle and has the bike upside down. He spins the 56 cm (or 0.56 m)
diameter wheel, and a pebble in the tread goes by 6 times every second. Thus, the wheel’s angular speed
is 6 rev/s or 6 × 2π = 37.7 rad/s. Tangential speed, v = rω, thus, v = (0.56/2) × 37.7 = 10.56 m/s.
radial acceleration,
ac =
10.562
v2
2
=
= 397.942 m/s
r
0.28
4. Mass of the ball = 28 g = 0.028 kg
The track is 42 cm in diameter or 42/2 = 21 cm in radius.
Speed of the ball = 60 rpm = 60 × 2π/60 rad/s = 6.283 rad/s.
Centripetal acceleration of the ball= ac = ω 2 r = 6.2832 × 0.21 = 8.29 m/s2 .
The force for this acceleration comes from the track. F = mac = 0.028 × 8.29N = 0.232N .
5. Diameter of the crankshaft = 3.0 cm. Therefore, radius = 3.0/2 = 1.5 cm = 0.015 m.
Initial rotation speed = 2400 rpm = 2400 × 2π/60 rad/s = 251.327 rad/s.
The wheel comes to a stop in 1.5 s.
a) Therefore, its angular acceleration can be calculated using,
ωf = ωi + αt
0 = 251.327 + α × 1.5
α = −167.552 rad/s
2
Now, tangential acceleration, at = αr = −167.552 × 0.015 = −2.513 m/s2 .
b) The angular distance it covers before coming to a stop is given by,
θ=
ωi + ωf
t
2
251.327 + 0
1.5 = 188.496 rad
2
or θ = 188.496/(2π) = 30 rev
θ=
6. Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 34 s to
speed up from rest to its top speed of 1 rotation every 1.3 s. The radius of the circle in which the astronaut
moves is 6.8 m.
a) The astronaut’s top angular speed is 1 revolution per 1.3 s. Thus, its speed is 1 × 2π/1.3 rad/s =
HW 6
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4.833 rad/s. We can calculate his angular acceleration using, ωf = ωi + αt, or 4.833 = 0 + α × 34, or
α = 0.142 rad/s2 . The tangential acceleration, at = αr = 0.142 × 6.8 = 0.9667 m/s2 .
b) When the device is rotating at its top speed, the centripetal acceleration of the astronaut is given by
ar = ω 2 r = 4.8332 × 6.8 = 158.848 m/s2 . In terms of g, it is, 158.848/9.8 = 16.21g.
7. A 485 g (0.485 kg) ball swings in a vertical circle at the end of a 2.1 m-long string. At the bottom of the
circle, the tension in the string is 21 N. The weight of the ball is mg = 0.485 × 9.8 = 4.753N . Therefore,
the net force on the ball is T − mg = 21 − 4.753 = 16.247N .
The acceleration produced by this force is the centripetal acceleration of the ball.
ac = F/m = 16.247/0.485 = 33.5 m/s
2
Now, centripetal acceleration,
ac =
v2
r
33.5 = v 2 2.1
v = 8.387 m/s.
8. A 5.0 g is placed 15 cm (0.15 m) from the center of a turntable. The coin has static and kinetic coefficients
of friction with the turntable surface of µs = 0.8 and µk = 0.5. The turntable very slowly speeds up to 60
rpm. Does the coin slide off?
That the turntable speeds up very slowly means that we can ignore any tangential acceleration of the coin.
All we have to consider is the centripetal acceleration. The force for centripetal acceleration is provided by
the force of static friction.
mω 2 r = µs N
⇒ mω 2 r = µs mg
⇒ ω 2 r = µs g
⇒ ω 2 = 0.8 × 9.8/0.15 = 52.267
or ω = 7.23 rad/s.
Converting to rpm, 7.23 rad/s=7.23/2π × 60 rpm = 69 rpm. Thus, at 60 rpm the coin will not slide off the
turntable.
9. A car drives over the top of a hill that has a radius of 38 m. At the speed at which the car loses contact, the
normal force from the ground vanishes. Thus, the acceleration due to gravity is the centripetal acceleration
of the car, or
v2
=g
r
v 2 = 38 × 9.8 = 372.4
v = 19.298 m/s.
10. A 450 g (0.45 kg) steel block rotates on a steel table while attached to a 1.2 m long hollow tube. A thrust
force of 4.8 N perpendicular to the tube is exerted by compressed air. The maximum tension the tube can
withstand without breaking is 50 N. The tension in the tube provides the centripetal acceleration of the
block. Thus, the angular velocity of the block at the instant the tube breaks off is given by,
ac =
Tmax
= ω2 r
m
50
= ω 2 × 1.2
0.45
ω 2 = 92.593
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ω = 9.623 rad/s
The tangential acceleration of the block can be calculated by the thrust of the compressed air. at = F/m =
4.8/0.45 = 10.67 m/s2 . Thus, the angular acceleration is α = at /r = 10.67/1.2 = 8.89 rad/s2 . Knowing the
acceleration and final angular speed we can calculate the angular distance covered,
ωf2 = ωi2 + 2αθ
9.6232 = 02 + 2 × 8.89θ
θ = 5.208 rad
Converting to revolutions, 5.208 rad= 5.28/2π = 0.829 rev.