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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1633 A lOlb uniform wheel is at rest when it is placed in contact with
a moving belt as shown. The kinetic coefficient
of friction between the belt and the 16—in
diameter wheel is “k = 0.1 and the belt moves
with a constant speed of 30 ft/s. Determine the  number of revolutions that the wheel turns before semen? °°°°°° _mMmumw
it rolls without slipping on the moving belt, __.______________________________________________________________________________ Solution The wheel rotates about a fixed
axis through its mass center G. The equations of motion are +a EF = F — G = O
X +T SE = N — 10 = 0
Y C+ EM =—Ji— F = I a
G 12 G where a = a = 0 since the mass center of the wheel
Gx Gy does not move, 2
1 10 8 2
IG — 2 [ 322 ][ 12 J — 0.069013 slug ft and F = 1N while the wheel slips. Therefore 0.
N = 10 lb = constant
F = 0.1N = 1 lb = constant
a = 9.66 rad/s2 D = constant
Integrating the angular acceleration to get the angular
velocity and angular position gives w = 9.66t + 01 = 9.66t rad/s 6 = 4.83::2 + C2 = 4.83t2 rad where the constants of integration are both zero since the wheel starts from rest when 9 = 0. The wheel stops slipping when points on the surface of the wheel move at the same speed as the belt v = rt) = 72—{9.66t} = 30 ft/s
= 4.65839 5
9 = 104.814 rad = 16.68 rev . ...... .. ..... ........ Ans. 657 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges = 1636* A horizontal force F of 250 N is applied
to a cable that is wrapped around the inner drum
of the compound pulley being used to lift block
8. The 20kg pulley has a radius of gyration
with respect to the axis of rotation of 160 mm.
If block 8 has a mass of 10 kg, determine the
angular acceleration of the pulley and the tension in the cable connected to block 8. Solution Separate free—body diagrams must
be drawn of the pulley and the block
B since their motions are different. The equations of motion are
2
= 2SO{O.15) — 0.25T 20(0.160} a T — 10(9.31) 10aB where the accelerations are related by a = 0.25m
8 Therefore, combining the equations of motion T 98.10 + 2.58 = 150  2.048a 2
11.41 rad/s D 126.6 N ENGINEERING MECHANICS _ Dynamics W.F. Riley & L.D. Sturges 1639 The 50~lb solid circular disk A rotates
about a smooth pin at 0. Block B weighs 20 lb.
During motion of the systemIr determine The angular acceleration “A of disk A. The tension T in the cable. The horizontal and vertical components of
the force exerted on disk A by the pin at 0. Solution Separate free—body diagrams must
be drawn of the disk A and the block
8 since their motions are different. The equations of motion are +6 2F A
X 10 12 2O _ 2T 32.2 3 where the mass center of the disk does not move a = a = O
AGX AGy and the accelerations 33 and a are related by 10
2&3 ‘ 12 a Therefore, combining the equations of motion T 0.64700ﬂ = 10  0.12940a 2
12.88 rad/s D
= 8.33 lb ...... = 0 lb
= 58.3 lb T ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1674* The 220—mm diameter bowling ball has a mass of 7.25 kg. The
instant that the ball comes in contact with the alley, it has a forward velocity V of 7 m/s and a back spin w of 6 rad/s. If the kinetic coefficient of
friction between the ball and the alley is
0.15, determine the elapsed time and the
distance traveled before the ball begins to roll without slipping. Solution The equations of motion of the ball are +—)EF = B=7.5
Ax f 2 a0 +
_>
M ”3
ll 0 B — 7.25(9.81)
Ay n 0.1103 I u
f G 2 I = —5— {7.25)(0.110)2 2 "* 0.03509 kg°m Therefore 3n = 71.1225 N = constant and since the ball is slipping on the alley B = 0.153n = 10.6684 N = constant 2 2
1.47150 mfs = constant {= 1.47150 mfs 9—} 9.]
ll 2
u = 33.4432 radfs D = constant
Integrating the accelerations to get the velocity, angular velocity, and position of the ball as functions of time VG = 7 — 1.47150t m/s w = 33.4432: — 6 rad/s 2
XG = 7t  0.73575t m where the constants of integration have been chosen so that VG = 7 m/s, ll m = —6 rad/s, and xG = 0 when t 0. The ball will stop sliding and start rolling without slipping when VG 0.110w 7  1.47150t 0.110(33.4432t — 6] t = 1.487 s ................................... Ans. at which time its position will be = .78 ...................................
KG 8 m 723 ...
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 Spring '08
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