Rotation-1.
(Taken from HRW Problem 10.7).
The wheel in the associated figure has eight equally spaced spokes
and a radius of 30 cm. It is mounted on a fixed axle and is
spinning at 2.5 rev/s. You want to shoot a 20 cm long arrow
parallel to this axle and through the wheel without hitting any
of the spokes. Assume that the arrow and the spokes are very
thin.
(a) What minimum speed must the arrow have?
(b) Does it matter where between the axle and rim of the wheel
you aim? If so, what is the best location?
Solution.
(a) To avoid touching the spokes, the arrow must go through the wheel within the time it takes the
spokes to turn 1/8th of a turn when it has an angular velocity ω. This is just θf − θi = ωt or
t=
0.125rev
= 0.05 s
2.5rev/s
The minimum speed of the arrow is such that it will move 20 cm in this time, or
v=
0.2m
= 4.0m/s
0.05s
(b) No, there is no dependence on radial position in the above computation. Although the spokes
are moving faster at the outside, they are also further apart, and the time to sweep out some angle
θ is constant, and given by the angular velocity which is the same throughout the body.
Rotation-2.
A bicycle slows down uniformly from vo = 6 m/s to rest over a distance of 100 m. Each wheel and
tire has an overall diameter of 65 cm. Determine:
(a) the initial angular velocity of the wheels,
(b) the total number of revolutions each wheel makes before coming to rest,
(c) the angular acceleration of the wheels,
(d) the time taken by the bicycle to come to a stop.
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Solution.
Let the angular velocity of the wheel be ω. Then a point on the hub will move with an instantaneous
velocity v = ωr about the hub. For pure rotation, in one rotation, a point on the hub moves one
circumference. For one rotation the wheel also ”unrolls” one circumference, so the velocity of the
bicycle with wheels turning at an angular velocity ω is the same as the velocity of a point on the
rim in pure rotation.
6
= 18.5rad/s
(a) Hence, vbike = ωr = 6 m/s, and r = 0.325 m. Hence ω = vr = 0.325
(b) The total distance travelled is 100 m. One revolution is 2πr = 2π(0.325) = 2.04 m. Hence the
100 m
= 49.0rev.
number of revolutions is 2.04m/rev
2
2
(c) ωf = ωi + 2α(∆θ). Hence
0 = (18.4rad/s)2 + 2α [(49rev)(2πrad/rev)]
Then 0 = 338.56 + 615.4α and solving gives:
α=−
338.56
= −0.56rad/s2
615.4
(d) ωf = ωi + αt. Solving for t gives:
t=
0 − 18.5rad/s
ωf − ωi
=
= 33s
α
−0.56rad/s2
Rotation-3.
A CD has inner and outer radii of 2.50 and 5.80 cm respectively. It is scanned at a constant linear
speed of 130 cm/s, starting from the inner edge and moving outward. The spiraled scan lines are
1.60 µm apart.
(a) What is the total length of the scan?
(b) What is the playing time?
Hint: find an expression for the radius r as as a function of the angle turned, the total angle turned,
and then integrate to get the total scan length.
Solution.
(a) The inner radius ri = 0.025m. For every turn, the radius will increase by 1.6 × 10−6 m, so for
any given angle turned θ we will have
r = ri + (1.6 × 10−6 )
θ
2π
The total angle turned will be found as follows:
rf − ri = 0.058 − 0.025 = 0.033m = (1.6 × 10−6 )
Hence
θf = rf − ri = 0.058 − 0.025 =
2
θf
2π
(2π)0.033m
= 129591rad
1.6 × 10−6
The total path length will then be just
Z
θf
Z
rdθ =
0
0
θf
R θf
θ=0
rdθ This can be written as:
θ
ri + 1.6 × 10
2π
−6
θ2
dθ = ri θ + 1.6 × 10
(2)2π
−6
θf
0
θf2
1295912
ri θf + 1.6 × 10
= 0.025(129591) + 1.6 × 10−6
= 5378
(2)2π
4π
−6
Total path taken is 5378 m, or 5.38 km!
(b) The total playing time can be found from the linear velocity and the total path taken. The
linear velocity is given as: v = 1.3m/s, and the distance is 5378 m. Hence the time to scan an entire
5378m
= 4137s = 69min
cd is: t = 1.3m/s
Rotation-4.
The figure above shows a hoist. All the pulleys are on fixed axles. Pulley A (Radius rA ) is motor
driven at angular velocity ωA and acceleration αA . Pulleys C (radius rc ) and D (radius rd ) are
welded together. Pulleys A and C are coupled by a belt B which does not slip. The cord supporting
mass M is wound around pulley D as shown.
(i) If A is rotating with angular velocity ωA and acceleration αA , find the angular velocity and
acceleration of pulley C, and the linear velocity and acceleration of M.
(ii) Pulley A accelerates with constant angular acceleration αA = 0.075rad/s2 and, at time t = 0,
ω0A = 9.0 rpm. Find the time t taken for C to reach an angular velocity ωC = 35 rpm. What is the
velocity of M at this point? Take rA = rD = 0.12 m and rc = 0.24 m. [ 85.2 s, 0.44 m/s]
Solution.
(a) Since the speed of the belt vB is the same at all locations, it follows that the speed of points on
the circumference of disks A and C have the same speed. Hence vB = ωA rA = ωC rC . By applying
the same logic for the acceleration of the belt, or by noting the radii are constant and differentiating
each side leads to: aB = αA rA = αC rC .
Then
αA rA
ωA rA
ωC =
and αC =
rC
rC
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For the mass M note that the string attached to the mass must move with the same linear velocity
as a point on the rim of disk D. vM = vD . But vM = vD = ωD rD . Since disks C and D are joined,
they must have the same angular velocity (and angular acceleration). Hence vD = ωC rD ,
vD = ωC rD =
ωA rA
rD
rC
The acceleration is found by same argument, or differentiation:
aD = αC rD =
αA rA
rD
rC
(9rev/min)(2πrad/rev)
= 0.94rad/s.
60s/min
(35rev/min)(2πrad/rev)
=
is equal to
60s/min
(b) The initial angular velocity of A is ω = 9.0 rpm. This is equal to
Likewise, the final angular velocity of disk C is ω = 35.0 rpm. This
3.665rad/s.
Since we know the angular acceleration of A, we will calculate the final angular velocity of A using
the information about C and ωC = ωArCrA .
ωA =
(3.665rad/s)(0.24m)
ωC rC
=
= 7.33rad/s
rA
(0.12m)
Then we use the equation of motion: ωAf − ωAi = αA t to give:
t=
ωAf − ωAi
7.33rad/s − 0.94rad/s
=
= 85.2s
αA
0.075rad/s2
The velocity of the mass M at this point is given by (from above)
vM = vD = ωC rD = (3.665rad/s)(0.12m) = 0.44m/s
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