19.1 Linear and Angular Momentum

60 Pages·2015·7.9 MB·English

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–1.
The rigid body (slab) has a mass m and rotates with an mvG
angular velocity V about an axis passing through the fixed V
point O. Show that the momenta of all the particles
hcoamvinpgo sain mg atghnei tbuoddey m cavn abned r eapctriensge nthterdo ubgyh ap osiinngt lPe ,vceacllteodr vG rP/G P
G
the center of percussion, which lies at a distance
rP>G = k2G>rG>O from the mass center G. Here kG is the rG/O G
radius of gyration of the body, computed about an axis
perpendicular to the plane of motion and passing through G.
O
SOLUTION
HO = (rG>O + rP>G)myG = rG>O(myG) + IGv, where IG = mk2G
rG>O(myG) + rP>G(myG) = rG>O(myG) + (mk2G)v
k2
G
rP>G = yG>v
y
However, yG = vrG>O or rG>O = vG
k2
G
rP>G = r Q.E.D.
G>O
985
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–2.
At a given instant, the body has a linear momentum
L = mvGand an angular momentum HG = IGVcomputed mvG
about its mass center.Show that the angular momentum of
thebody computed about the instantaneous center of zero G
I V
velocity IC can be expressed as HIC = IICV, where IIC G
represents the body’s moment of inertia computed about
the instantaneous axis of zero velocity.As shown,the ICis
located at a distance r away from the mass center G.
G>IC r
G/IC
SOLUTION IC
HIC = rG>IC (myG) + IGv, where yG = vrG>IC
= rG>IC (mvrG>IC) + IGv
= (IG + mr2G>IC)v
= IIC v Q.E.D.
986
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–3.
Show that if a slab is rotating about a fixed axis perpendicular V
tothe slab and passing through its mass center G,the angular
P
momentum is the same when computed about any other
point P.
G
SOLUTION
Since yG = 0,the linear momentum L = myG = 0.Hence the angular momentum
about any point Pis
HP = IGv
Since vis a free vector,so is H . Q.E.D.
P
987
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–4.
The 40-kg disk is rotating at V = 100 rad s. When the force P (N)
P is applied to the brake as indicated by the graph. If the
coefficient of kinetic friction at B is mk>= 0.3, determine 500
the time t needed to stay the disk from rotating. Neglect the
thickness of the brake.
t (s)
2
P
300 mm 300 mm
B
Solution V
150 mm
200 mm
Equilibrium. Since slipping occurs at brake pad, Ff = mkN = 0.3 N. O
Referring to the FBD the brake’s lever, Fig. a, A
a+ ΣMA = 0; N(0.6) - 0.3 N(0.2) - P(0.3) = 0
N = 0.5556 P
Thus,
Ff = 0.3(0.5556 P) = 0.1667 P
Principle of Impulse and Momentum. The mass moment of inertia of the disk about
its center O is IO = 12 mr2 = 12 (40)(0.152) = 0.45 kg#m2.
t2
IOv1 + Σ MOdt = IOv2
Lt1
It is required that v2 = 0. Assuming that t 7 2 s,
t
0.45(100) + [-0.1667 P(0.15)]dt = 0.45(0)
L0
t
0.025 P dt = 45
L0
t
P dt = 1800
L0
1
(500)(2) + 500(t - 2) = 1800
2
t = 4.60 s Ans.
Since t 7 2 s, the assumption was correct.
Ans:
t = 4.60 s
988
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–5.
The impact wrench consists of a slender 1-kg rod ABwhich
is580mm long,and cylindrical end weights at Aand Bthat
each have a diameter of 20 mm and a mass of 1 kg.This
assembly is free to turn about the handle and socket,
C
which are attached to the lug nut on the wheel of a car.If
B
the rod AB is given an angular velocity of 4 rad>s and it
strikes the bracket C on the handle without rebounding,
determine the angular impulse imparted to the lug nut.
300 mm
SOLUTION A
300 mm
Iaxle = 112(1)(0.6 - 0.02)2 + 2c21(1)(0.01)2 + 1(0.3)2d = 0.2081 kg#m2
Mdt = Iaxle v = 0.2081(4) = 0.833kg#m2>s Ans.
L
Ans:
M dt = 0.833 kg#m2 s
L
>
989
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–6.
The airplane is traveling in a straight line with a speed of
300 km>h,when the engines A and B produce a thrust of
TA = 40 kN and TB = 20 kN, respectively. Determine the
angular velocity of the airplane in t = 5 s.The plane has a T (cid:31) 40 kN 8 m
A
mass of 200 Mg,its center of mass is located at G,and its
A
radius of gyration about Gis kG = 15 m.
G
B
SOLUTION
T (cid:31) 20 kN
B 8 m
Principle of Angular Impulse and Momentum:The mass moment of inertia of the
airplane about its mass center is IG = mkG 2 = 200A103BA152B = 45A106B kg#m2.
Applying the angular impulse and momentum equation about point G,
t2
Izv1 + © MGdt = IGv2
Lt1
0 + 40A103B(5)(8) - 20A103B(5)(8) = 45A106Bv
v = 0.0178 rad>s Ans.
Ans:
v = 0.0178 rad s
>
990
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–7.
The double pulley consists of two wheels which are attached
toone another and turn at the same rate.The pulley has a
mass of 15 kg and a radius of gyration of kO = 110mm.If 2kN
the block at A has a mass of 40 kg, determine the speed
200 mm
of the block in 3 s after a constant force of 2 kN is applied
75mm
tothe rope wrapped around the inner hub of the pulley.The
O
block is originally at rest.
SOLUTION
Principle of Impulse and Momentum:The mass moment inertia of the pulley about
point O is IO = 15A0.112B = 0.1815kg#m2.The angular velocity of the pulley and
y
the velocity of the block can be related by v = 0.B2 = 5yB.Applying Eq. 19–15,
we have A
aasyst. angular momentumb + aa syst. angular impulseb
O1 O1-2
= aa syst. angular momentumb
O2
(a+) 0 + [40(9.81)(3)](0.2) - [2000(3)](0.075)
= -40yB(0.2) - 0.1815(5yB)
yB = 24.1m>s Ans.
Ans:
vB = 24.1 m s
>
991
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*19–8.
The assembly weighs 10 lb and has a radius of gyration
kG = 0.6 ft about its center of mass G.The kinetic energy
#
of the assembly is 31 ft lbwhen it is in the position shown.
0.8 ft
If it is rolling counterclockwise on the surface without 1 ft
G
slipping,determine its linear momentum at this instant.
1 ft
SOLUTION
IG = (0.6)2a3120.2b = 0.1118 slug#ft2
1 10 1
T = 2a32.2bvG 2 + 2(0.1118) v2 = 31 (1)
vG = 1.2 v
Substitute into Eq.(1),
v = 10.53 rad>s
vG = 10.53(1.2) = 12.64 ft>s
10 #
L = mvG = 32.2(12.64) = 3.92 slug ft>s Ans.
Ans:
#
L = 3.92 slug ft s
>
992
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–9.
The disk has a weight of 10 lb and is pinned at its center O.
Ifavertical force of P = 2lbis applied to the cord wrapped
around its outer rim,determine the angular velocity of the 0.5 ft
O
disk in four seconds starting from rest.Neglect the mass of
the cord.
SOLUTION
t2
(c+) IOv1 + © MOdt = IOv2 P
Lt1
1 10
0 + 2(0.5)(4) = c2 a32.2b(0.5)2dv2
v2 = 103rad>s Ans.
Ans:
v2 = 103 rad s
>
993
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19–10.
The 30-kg gear Ahas a radius of gyration about its center of
mass O of kO = 125 mm. If the 20-kg gear rack B is
subjected to a force of P = 200 N, determine the time 0.15 m
required for the gear to obtain an angular velocity of O
20 rad>s,starting from rest.The contact surface between the
gear rack and the horizontal plane is smooth.
A
B P (cid:31) 200 N
SOLUTION
Kinematics:Since the gear rotates about the fixed axis,the final velocity of the gear
rack is required to be
(vB)2 = v2rB = 20(0.15) = 3 m>s :
Principle of Impulse and Momentum:Applying the linear impulse and momentum
equation along the xaxis using the free-body diagram of the gear rack shown in Fig.a,
t2
A :+ B m(vB)1 + © Fxdt = m(vB)2
Lt1
0 + 200(t) - F(t) = 20(3)
F(t) = 200t - 60 (1)
The mass moment of inertia of the gear about its mass center is IO =
mkO 2 = 30(0.1252) = 0.46875 kg#m2.Writing the angular impulse and momentum
equation about point Ousing the free-body diagram of the gear shown in Fig.b,
t2
IOv1 + © MOdt = IOv2
Lt1
0 + F(t)(0.15) = 0.46875(20)
F(t) = 62.5 (2)
Substituting Eq.(2) into Eq.(1) yields
t = 0.6125 s Ans.
Ans:
t = 0.6125 s
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