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Unformatted text preview: Problem Set 2 Solutions 1. (a) Area of plate without hole = (0.2)2 = 0.04 m2
Area of hole = “(0.05)2 = 0.00735 m2
Area of plate with hole = 0.03215 m2
M1 = mass of plate without hole = 1.5(0.4)i’0.03215 = 1.8? kg
M2 = mass of material of hole = 1.5(0.00785)f0.03215 = 0.3? kg s— s’ Moment of inertia about axis aa‘ through the CM
ICM= M1a2f12  M21234
= 1.3?(02ﬂ12 — 0.37(0.05)2;4
= 0.0060 kgm2
Moment of inertia about axis bb' (from parallel axis theorem)
1 I ICM + (MI'MZXWQ)
= 0.0060 +1.5g0.1)2
= 0.0210 kgm (’0) Set 10: = F, with torque produced by the weight of the plate.
1 gram:2 = — Mg(af2)sin8 a Mg(af2)9 i9 diam2 = —(0.1Mg/I)0 = {.3020
Q9 ' m02=0.1(1.5)(9.81)f0.0210
(00 = 8.37 radfs M3 Period T = 211me = 0.75 s 2. Moment of inertia (by parallel axis theorem] I = mL2f12 + mxz, and 1a = F.
I 'n
Ix 03 :“
.MS (111sz 12 + m2)d28fdt2 = mgx sine as mngl
0302 = gX/(LZI‘IZ + x2)
To minimise the period T=2nfwm maximise {002 or minimise U0002
dfdx(L2f12x + x) = L2f(12x2) + 1 = 0
x2 = sz12
x = 0.29L = 29 cm 3. (a) If the springs are put under a tension T, their total extension should equal that of
a single spring of force constant k.
T = klxl = k2X1 5 1((X1 + X2)
(k1  10X; 2 lCXz and (k2 ' k)X2 = log
mix; = k!(k1 — k) = (k;  k)!k
 k2 = (k:  k)(k2  k) = klkz + k2  (k. + kak
k = klkﬂkl + k2) = 10(15)f25 = 6 Nr’m (b) Moment of inertia of a rectangular plate of dimensions a x b about a perpendicular
axis through its centre is Mtia2 + b )!12, and of a rod about its centre is 11112;" 12. .‘.Moment of inertia of rodplate system about its centre of mass is
P 1CM = 0.5[(0.3)2 + (0.3)21r12 + 0.25{0.3)2r12
= 0.0203 kgm2 Moment of inertia about P is
IP =1CM + 0350141,)2 = 0.1408 kgml
When the rod is displaced through angle 8, the spring is
extended by LG, where L is the distance from the pivot to
the springs; the springs therefore exert a force kLB and a
torque —kL29. I; When the system is pivoted at its CM the weight produces
no torque, but when pivoted at P the weight gives a torque
Mg(0.4sin9) z 0.4Mg9 (i) When pivoted at the CM:
[CM dzefdtz = (0.4)2k6
(139th = (0.16kﬂcM)e = @029
(002 = 0.16(6)r'0.0208 = 46.15
Period = 27:10:10 = 0.92 5 (ii) When pivoted at P:
1P (Prams2 = (0.4Mg + 0.64109
(of = (0.4Mg + 0.6%)pr
= [O.4(0.?5)(9.81) + 0.64(6)]!0.1403 = 48.2
030 = 6.94 rad/s
Period T = 21mg = 0.91 s 4. (a) At equilibrium, net torque on the rod is zero. Let the upward force exerted by the and be P.
senes mg{Lf2) — FL = 0
F = tug/2
If the spring is compressed by x0, F = kxo x{J = 05ka (b) If spring is compressed by x, torque acting is {ka  mgLQ] = k(x  xo)L
But x  x0 = LB
F =10: then gives Idinr'dt2 = kL26
diam) = (kL2tI)B = «9.326
frequency = (00/21: = (l!2n)\}(kL2H) 5. Tensions in the strings to the two masses are TI and T2. The string between the two
pulleys remains almost horizontal, so the tensions act on the rod in a horizontal
direction. Torque is produced by these tensions, and by the welght of the rod. IP = MLZB _
Pp: Mg(U2) sine T1L c059 + T21. c059
: Mgu2  TIL + TzL
Ipa = n (MLZB) cFa/dt2 = —Mg(uz)a + ('1‘2T1)L .................. (1) Consider the two masses m. If the rod is displaced to the right, the righthand mass
will have an acceleration upward, while the lefthand mass will have an equal
acceleration downward. mg  T1 = ma T2 — mg = ma T2  T1 = Zma
Acceleration of the masses is related to the angular acceleration of the rod by
a = Lot (note the Sign)
T2  T. = 2mL dzﬁfdtl ...................... (2)
From (1) and (2) (MLZIS + 2mm diam? = Mg(Lr'2)9 dzefdtz = mole
(1302 = MgKZMLB + 4mL) = 0.6(9.81)l[2(0.6)(1)f3 + 4(0f1)(l)]
(on = 2.?1 radfs
Period T = 2mm, = 2.3 s mum—1m
The CM of the rod moves up by h = (Ll2)~(L;’2)c056. _ The potential energy terms of the two masses cancel, because one moves up
the same distance as the other moves down. E = Ipmlrz + Mgh + 2(mv2f2)
dEfdt = o = 1pm dmldt + mg(L:’2)sin9 :19de + 2m dwdt
v=Lm, dBI'dt = a), and dw’dt = Ldmfdt
1pm dmfdt + Mg(Ll2) sinﬂm + 2mL2m dwfdt = 0
(MLEB + 2mL2) dwfdt = ~Mg(Lf2) sine z Mg(L:’2)B as above ...
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 Fall '10
 ProfessorSomi

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