Reference solution for Quiz #2

Phys 270: Planar rigid body dynamics

Problem 1 (5 points): The first picture seems the most accurate. From it we can infer the separations between two tracks, and the approximate amplitude A of the oscillation,

s ≈ 94µm, A ≈ 20µm.

Assuming a oscillation with a single frequency ω = 2π 440 Hz (not the closest approximation to mu-sic), then the oscillation amplitude is x(t) = A sinωt. Hence, ax(t) = −Aω2 sinωt, and the typicalacceleration is

atypical ≈Aω2

√2

= 102 m/s2 = 10g.

The numerical value is 110 m/s2 but due to the very coarse approximations and also measurementinaccuracies, the result is not accurate at the percent level. Any result within a factor of 3 should becounted as correct, if the reasoning is.

The reason that the needle doesn’t jump out of the track is that only the tip is accel-erating that hard, not the whole needle with the heavy read head at the other end.

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Consistency checks:

1. We can compute the distance between the tracks in a different way. Namely, we know that thevinyl record runs for 25 min and revolves 33 times per minute. In the course of this, the needlemoves 10 cm inwards (=(outer diameter - inner diameter)/2). Thus, the distance s between thetracks is s = 10 cm/(25 ∗ 33) = 121µm. Within our measurement accuracy, this is consistent withthe above stated value.

2. For the second picture (or the third), we can estimate the frequency of the sound played. Forthe second, depending on which track we look at, the period of oscillation is 3 to 6 times theseparation between tracks, i.e., 0.3 mm to 0.6 mm. Depending on whether the tracks we look atare from the inner or outer part of the record (we don’t know), this translates into an angle of0.002 rad - 0.012 rad. Given that the record is played at 33 revolutions per minute, these anglescorrespond to oscillation times 5.8 × 10−4s - 3.5 × 10−3s, which, in turn, implies a frequency ofsound in the range 290 Hz− 1730 Hz. Within the uncertainties given, this is compatible with theA4 of 440 Hz. The human ear can hear frequencies between 50 Hz and 16 kHz.

Problem 2 (5 points):

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The position of point C is fixed. Point A moves along the horizontal line and is confined to the line.Point B undegoes general plane motion but cannot slide along the rod DC itself (it’s a fixed hinge).

a) Determine vB and aB.

b) Determine ωAB and vA.

c) Determine αAB and aA.

Solution: (a) By geometry, BC = AB sin 50◦

sin 30◦ = 4.596m, and

~rB/C = (−3.98i+ 2.298j)m. (1)

Then, the velocity of point B is ~vB = (~vA = ~0) + ω × ~rB/C . Hence

~vB = (11.49i+ 19.90j)m/s . (2)

Similarly, the acceleration of point B is ~aB = (~aA = ~0) + α× ~rB/C − ω2~rB/C . Hence

~aB = (105.1i− 47.92j)m/s2 . (3)

(b) By geometry, ~rA/B = −AB(i cos 50◦ + j sin 50◦

), and ~vA = ivA. Also, the velocity of point A is

related to the velocity of point B (which we already know) by ~vA = ~vB + ~ωAB × ~rA/B. Equating thetwo expressions for ~vA and sorting by components, we obtain

i : vA = vB,x +AB ωAB sin 50◦,

j : 0 = vB,y −AB ωAB cos 50◦,(4)

This is a linear system of equations for the two unknowns ωAB and va. We obtain

ωAB = 10.32/s, ~vA = 35.2im/s . (5)

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(c) The acceleration of point A is horizontal, ~aA = aAi. Also, ~aA is related to the acceleration ~aB ofpoint B (which we already have computed) via ~aA = ~aB + αAB × ~rB/A − ω2

AB~rB/A. Equating the twoexpressions for ~aA and sorting by components, we obtain

i : aA = aB,x +AB αAB sin 50◦ + ω2ABAB cos 50◦,

j : 0 = aB,y −AB αAB cos 50◦ + ω2ABAB sin 50◦,

(6)

This is a linear system of equations for the two unknowns αAB and aA. We find

αAB = 102.07/s2, ~aA = 545.1im/s2 . (7)

(The acceleration of point A is about 55g. The rods better be sturdy.)

Problem 3 (3 points): Denote the instantaneous bottom point of the outer cylinder by C, theinstantaneous bottom point of the inner cylinder by D (this is the last point of contact between theinner cylinder and the string), and the centre of the cylinder by E.

vstring

r.w.s.

rRA

B

CD

E

C is the ICR. Therefore, vD = v = ω(R− r), from which we obtain ω, then ~vE = iv RR−r . Hence,

~vA = vR

R− r(i+ j), ~vB = 2v

R

R− ri . (8)

Furthermore, the point E has no acceleration, and, using the relative motion formulae,

~aA = v2R

(R− r)2i, ~aB = −v2 R

(R− r)2j . (9)

Limits: When r −→ R, the velocities approach infinity which is unphysical. As D approaches C, itbecomes harder and harder to satisfy the r.w.s. assumption. When C = D, it is impossible to satisfyit: the string requires C = D to move with velocity v, whereas the r.w.s assumption requires C = D tobe at rest. Contradiction.

Problem 4 (5 points):

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aC vC

Block C slides along the bar and is traveling away from A. The bar is connected to the wheel whichrotates without slipping.

a) Calculate vC , and sketch it in the diagram.

b) Calculate aC , and sketch it in the same diagram.

Solution. (a) By geometry, for the reference frame rotating with the wheel & bar

i′ =

√3

2i+

1

2j,

and~rC/A = (0.7m+ 0.5m)(i

√3/2 + j/2) = (1.04i+ 0.6j)m. (10)

Furthermore, the angular velocity and angular acceleration of the wheel are

ωAC = −vA/R = 3/s(ccw),αAC = −aA/R = 0.5/s2(ccw).

(11)

Then, the velocity of point C is ~vC = ~vA + ~ωAC × ~rC/A + (~vC)x′y′ . Using the given numbers,

~vC = −1.5im/s+

∣∣∣∣∣∣i j k0 0 3

1.039 0.6 0

∣∣∣∣∣∣m/s+ 10.39im/s+ 6j m/s. (12)

Hence,

(7.09i+ 9.12j)m/s .

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(b) Similarly, the acceleration at point C is given by

~aC = ~aA + ~αAC × ~rC/A − ω2AC~rC/A + 2~ωAC × ~rC/A + (~aC)x′y′ .

Using the given numbers and ωAC , αAC from Eq. (11), we obtain

−37.5im/s2 + 62.5j m/s2 .

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