Physics 140
HOMEWORK Chapter 10A
Q4. Figure 10-21b is a graph of the angular position of the rotating disk of Fig. 10-21a. Is the angular
velocity of the disk positive, negative, or zero at (a) t = 1 s, (b) t = 2 s, and (c) t = 3 s? (d) Is the
angular acceleration positive or negative?
———
The graph is of θ vs t, so a positive slope corresponds to a positive ω, and correspondingly.
(a) positive
(b) zero
(c) negative
(d) negative. The angular velocity ω goes from positive to negative.
Q5. In Fig. 10-22, two forces and act on a disk that turns about its center like a merry-go-round. The
forces maintain the indicated angles during the rotation, which is counterclockwise and at a constant
~ 1 without changing the magnitude of F
~ 1 . (a) To keep
rate. However, we are to decrease the angle θ of F
~ 2 ? Do forces
the angular speed constant, should we increase, decrease, or maintain the magnitude of F
~ 1 and (c) F
~ 2 tend to rotate the disk clockwise or counterclockwise?
(b) F
———
~ 1 , so we must also reduce
(a) decrease. Increasing θ reduces the magnitude of the torque caused by F
~
the magnitude of the torque caused by F2 .
(b) Clockwise; a negative torque.
(c) Counterlockwise; a positive torque.
Q6. In the overhead view of Fig. 10-23, five forces of the same magnitude act on a strange merry-goround; it is a square that can rotate about point P , at midlength along one of the edges. Rank the
forces according to the magnitude of the torque they create about point P , greatest first.
———
τ5 > τ2 > τ4 > τ1 > τ3 = 0.
The torque from a force is the magnitude of the force multiplied by the moment arm (see text re
moment arm).
P3. When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it
falls. If the distance to the floor is 76 cm and for rotation less than 1 rev, what are the (a) smallest
and (b) largest angular speeds that cause the toast to hit and then topple to be butter-side down?
———
The toast
p rotates at constant ω as it falls. The time it takes to fall is
t = (2 · 0.76 m)/(9.8 m/s2 ) = 0.394 s. (This should be easy).
It starts out butter-side up, and is pushed off the counter with the leading edge first. If it hits the floor
on the leading edge with the slice vertical, it will flop over and be butter-side down. This corresponds
to 1/4 of a revolution, and represents ωmin . If it rotates 3/4 of a revolution, it will hit on the trailing
edge and flop over leaving butter side up (never happens in real life), representing ωmax . So:
(a) ωmin = (0.25 rev)/(0.394 s) = (1.57 rad)/(0.394 s) = 3.99 rad/s.
ωmax = (0.75 rev)/(0.394 s) = (4.71 rad)/(0.394 s) = 12.0 rad/s.
P6. The angular position of a point on the rim of a rotating wheel is given by θ = 4.0 t−3.0 t2 +t3 , where
θ is in radians and t is in seconds. What are the angular velocities at (a) t = 2.0 s and (b) t = 4.0 s?
(c) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at
t = 4.0 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of
this time interval?
———
We have θ(t), and will need ω(t) and α(t).
θ(t) = 4.0 t − 3.0 t2 + t3
ω(t) = dθ/dt = 4.0 − 6.0 t + 3 t2 .
α(t) = dω/dt = −6.0 + 6 t.
(a) ω(2 s)) = 4.0 − 12.0 + 12.0 = 4.0 rad/s.
(b) ω(2 s)) = 4.0 − 24.0 + 48.0 = 28 rad/s.
(c) αavg = (ωf − ωi )/∆t = (28 rad/s − 4 rad/s)/(2 s) = +12 rad/s2 .
(d) α(2 s) = −6.0 + 12 = +6 rad/s2 .
(e) α(4 s) = −6.0 + 24 = +18 rad/s2 .
P13. A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
(a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its
angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?
———
If rotational kinematic equations look foreign, write out the corresponding equations it terms of v, v0 ,
∆x, a, and t.
We are given ω0 = 1.5 rad/s, ωf = 0, and ∆θ = 40 rev = 251.3 rad (like cm to m, g to kg, convert this
immediately!)
(a) Even though we are asked for the time, I will start with
ω 2 − ω02 = 2α∆θ and solve for α:
α = (ω 2 − ω02 )/(2∆θ = (0 − (1.5 rad/s)2 )/(2 · 251.3 rad = −4.477 × 10−3 rad/s2 .
∆t = ∆ω/α = (−1.5 rad/s)/(−4.477 × 10−3 rad/s2 ) = = 335 s.
(b) We already got α = −4.477 × 10−3 rad/s2 .
(c) From ∆θ = ω0 t + (1/2)αt2 , substitute numbers (20 rev=125.7 rad).
125.7 rad = 1.5t − 2.238 × 10−3t2 ⇒ t = 98.12 s. Pick the solution of the quadratic which is less
than 335 s.
P18. A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits
a light beam. We receive a radio pulse for each rotation of the star.The period T of rotation is
found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation
of T = 0.033 s that is increasing at the rate of 1.26 × 10−5 s/y. (a) What is the pulsar’s angular
acceleration α? (b) If α is constant, how many years from now will the pulsar stop rotating? (c) The
pulsar originated in a supernova explosion seen in the year 1054. Assuming constant α, find the initial
T.
———
First, note that T is the reciprocal of the number of revolutions per second, so that ω = 2π/T . Take
d/dt of this equation:
dω/dt = α = −2π(dT /dt)/T 2 .
dT /dt = 1.26 × 10−5 s/yr = 1.26 × 10−5/(3.156 × 107 ) = 3.993 × 10−13 .
Note that there are 3.156 × 107 s in a year, and that dT /dt is dimensionless.
(a) α = −(6.28)(3.993 × 10−13 )/(3.3 × 10−2 s)2 = −2.30 × 10−9 rad/s2 .
(b) With ω = 2π/T = 190.4 rad/s and ω(tstop = 0 = ω0 + αtstop ,
tstop = −ω0 /α = −(190.4 rad/s)/(−2.30 × 10−9 rad/s2 ) =
= 8.26 × 1010 s = 2620 yr.
(c) The time from 1054 to 2011 (time of measurement) is 957 yr or 3.02 × 1010 s.
ω0 = ω − αt = 190.4 rad/s − (−2.30 × 10−9 rad/s2 )(3.02 × 1010 s) = 260 rad/s.
T0 = 2π/ω0 = 0.0242 s.
Using the current year instead of 2011 will make only a tiny difference.
P22. An astronaut is being tested in a centrifuge. The centrifuge has a radius of 10 m and, in starting,
rotates according to θ = 0.30 t2 , where t is in seconds and θ is in radians. When t = 5.0 s, what are the
magnitudes of the astronaut’s (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and
(d) radial acceleration?
———
(a) ω = dθ/dt = 0.6t ⇒ ω(5 s) = 3.0 rad/s.
(b) v = rω = (10 m)(3.0 rad/s) = 30 m/s.
(c) α = dω/dt = 0.60 rad/s2 ⇒ atang = rα = 6 m/s2 .
(d) The radial acceleration is the same as centripetal acceleration.
acent = rω 2 = (10 m)(3 rad/s)2 = 90 m/s2 .
Note how radians disappeared at will.
P28. In Fig. 10-28, wheel A of radius rA = 10 cm is coupled by belt B to wheel C of radius rC = 25 cm.
The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2 . Find the time
needed for wheel C to reach an angular speed of 100 rev/min, assuming the belt does not slip. (Hint:
If the belt does not slip, the linear speeds at the two rims must be equal.)
———
From the hint, we see that the angular velocities of the wheels are inversely related to their diameters,
so that
ωA = 2.5ωC .
We want ωC = 100 rev/min ⇒ ωA = 250 rev/min = 26.18 rad/s.
ω = ω0 + αt = 0 + (1.6 rad/s2 ) t = 26.18 rad/s ⇒
t = (26.18 rad/s)/(1.6 rad/s2 ) = 16.4 s.
P32. was not assigned, but you might find helpful A car starts from rest and moves around a
circular track of radius 30.0 m. Its speed increases at the constant rate of 0.500 m/s2 . (a) What is the
magnitude of its net linear acceleration 15.0 s later? (b) What angle does this net acceleration vector
make with the car’s velocity at this time?
———(a) “Linear acceleration” means ~a; the word linear is to distinguish it from the angular acceleration α.
This acceleration has a centripetal component from v 2 /R and a tangential component, perpendicular
to the centripetal component, of 0.5 m/s2 . The magnitude of the acceleration is by Pythagoras on these
two components.
vtang = v0 + atang t = 0 + (0.5 m/s2 )(15 s) = 7.5 m/s
acent√
= v 2 /R = (7.5 m/s2 )2 /(15 m) = 3.75 m/s2 atang = 0.5 m/s2 .
a = 3.752 + 0.52 = 3.78 ms2 .
(b) We can work in a coordinate system where the axes are in the tangential and centripetal directions.
To be definite, take ı̂ as tangential and ̂ as centripetal. Then ~v is purely tangential, and ~a has the
components derived above:
~v = (7.5ı̂ + 0 ̂) m/s
~a = (0.5ı̂ + 3.75 ̂) m/s2 .
Find the cosine of the angle between these via dot product:
cos φ = (~v · ~a)/(va) = ((0.5)(7.50) + 0)/[(7.5)(3.78)] = 0.1321.
φ = Cos−1 (0.1321) = 82.4◦ .
P38. Figure 10-32 shows three 0.0100 kg particles that have been glued to a rod of length L = 6.00 cm
and negligible mass. The assembly can rotate around a perpendicular axis through point O at the left
end. If we remove one particle (that is, 33% of the mass), by what percentage does the rotational inertia
of the assembly around the rotation axis decrease when that removed particle is (a) the innermost one
and (b) the outermost one?
———
Note that d on the figure is 0.02 m. Let IT be the rotational inertia of the complete assembly, and Ia
and Ib be the rotational inertias for parts (a) and (b) respectively.
I0 = (0.01 kg)[(0.02 m)2 + (0.04 m)2 + (0.06 m)2 ] = (0.01 kg)[4 × 10−4 m2 + 16 × 10−4 m2 + 36 ×
−4
10 m2 ] = (0.01 kg)(56 × 10−4 m2 ) = 5.6 × 10−5 kg m2 . The ratio of kinetic energies is the same as
the ration of the moments of inertia.
(a) Ia = 5.2 × 10−5 kg m2 , so the decrease is 4/56 = 7.14 %
(b) Ib = 2.0 × 10−5 kg m2 , so the decrease is 36/56 = 64.2 %