Identical cylinders of weight and radiusare pushed by a series of moving arms. The coefficient of friction between all surfaces is . (Updated to add: you may differentiate between and if necessary). Denoting by the magnitude of the acceleration of the arms, derive an expression for (a) the maximum allowable value of if each cylinder is to roll without sliding, (b) the minimum allowable value of if each cylinder is to move to the right without rotating. Part (c) Let, and between all surfaces (updated to add: you may assume if necessary). Determine the horizontal component of the force exerted on each cylinder when the acceleration is (1) to the right and (2) 10 to the right. Benchmark solutions for part (c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right. For part (c), your solution should be coded into MATLAB so it can easily be run for alternate cases. Part (d) Plot the horizontal component of the force exerted on each cylinder for different ranges of acceleration, which illustrate the various case of rolling and sliding obtained in parts (a) and (b).

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Only need A, B, and C please.
Identical cylinders of weight and radiusare pushed by a series of
moving arms. The coefficient of friction between all surfaces is
. (Updated to add: you may differentiate between and if
necessary). Denoting by the magnitude of the acceleration of
the arms, derive an expression for (a) the maximum allowable
value of if each cylinder is to roll without sliding, (b) the
minimum allowable value of if each cylinder is to move to the
right without rotating.
Part (c) Let, and between all surfaces (updated to add: you may
assume if necessary). Determine the horizontal component of
the force exerted on each cylinder when the acceleration is (1) to
the right and (2) 10 to the right. Benchmark solutions for part
(c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right.
For part (c), your solution should be coded into MATLAB so it
can easily be run for alternate cases.
Part (d) Plot the horizontal component of the force exerted on
each cylinder for different ranges of acceleration, which
illustrate the various case of rolling and sliding obtained in parts
(a) and (b).
Transcribed Image Text:Identical cylinders of weight and radiusare pushed by a series of moving arms. The coefficient of friction between all surfaces is . (Updated to add: you may differentiate between and if necessary). Denoting by the magnitude of the acceleration of the arms, derive an expression for (a) the maximum allowable value of if each cylinder is to roll without sliding, (b) the minimum allowable value of if each cylinder is to move to the right without rotating. Part (c) Let, and between all surfaces (updated to add: you may assume if necessary). Determine the horizontal component of the force exerted on each cylinder when the acceleration is (1) to the right and (2) 10 to the right. Benchmark solutions for part (c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right. For part (c), your solution should be coded into MATLAB so it can easily be run for alternate cases. Part (d) Plot the horizontal component of the force exerted on each cylinder for different ranges of acceleration, which illustrate the various case of rolling and sliding obtained in parts (a) and (b).
SAM
A cord is wrapped around the inner drum of a wheel and pulled hori-
zontally with a force of 40 lb. The wheel weighs 100 lb and has a radius
of gyration of 3.5 in. Knowing that μ, = 0.20 and = 0.15, determine
the acceleration of G and the angular acceleration of the wheel.
a. Assume Rolling without Sliding. In this case, we have
à = rα = ₂a
By comparing the friction force obtained with the maximum friction
force, we shall determine whether this assumption is justified. The mass
and moment of inertia of the wheel are
111 =
- ΣF, = md:
Equations of Motion
+) EMc = Σ(Me)ett:
100
32.2
I = mk²
649
= 3.11 lb-sec²/ft
+) EM, = la:
=
100 3.5
32.2 12
(40)(2) ma() + Ia
6.67= (3.11)(a)()+ 0.264a
a +8.30 radians/sec²
à ra = () (8.30)= 3.46 ft/sec²
mā
F + 40
F+40
F= 29.3 lb
N
100 0
+1EF, = 0:
Maximum Available Friction Force
= 0.264 lb-ft-sec²
(3.11)(3.46)
Fmax = μ,N= (0.20) (100) = 20 lb
Since F> Fmax, the assumed motion is impossible.
b. Rolling and Sliding. Since the wheel must roll and slide at
the same time, we draw a new diagram, where a and a are inde-
pendent and where
F = 29.3 lb-
N = 100 lb ↑
F = F₁ = ₂N = (0.15) (100) = 15 lb
From the computation of part a, it appears that F should be di-
rected to the left.
42F, = må:
40 15 3.11a
à +8.04 ft/sec²
à 8.04 ft/sec²
(40) () = 0.264a
(15))
a = 14.2 radians/sec² a = 14.2 radians/sec²
100 lb
F= 15 lb
G
IN
100 lb
OG
с
5 in.
IN
=
40 lb
2 in.
3 in.
r= 5 in.
C
40 lb
5.0₂
40 lb
12 in.
ā
mā
5 in.
ma
5 in.
Transcribed Image Text:SAM A cord is wrapped around the inner drum of a wheel and pulled hori- zontally with a force of 40 lb. The wheel weighs 100 lb and has a radius of gyration of 3.5 in. Knowing that μ, = 0.20 and = 0.15, determine the acceleration of G and the angular acceleration of the wheel. a. Assume Rolling without Sliding. In this case, we have à = rα = ₂a By comparing the friction force obtained with the maximum friction force, we shall determine whether this assumption is justified. The mass and moment of inertia of the wheel are 111 = - ΣF, = md: Equations of Motion +) EMc = Σ(Me)ett: 100 32.2 I = mk² 649 = 3.11 lb-sec²/ft +) EM, = la: = 100 3.5 32.2 12 (40)(2) ma() + Ia 6.67= (3.11)(a)()+ 0.264a a +8.30 radians/sec² à ra = () (8.30)= 3.46 ft/sec² mā F + 40 F+40 F= 29.3 lb N 100 0 +1EF, = 0: Maximum Available Friction Force = 0.264 lb-ft-sec² (3.11)(3.46) Fmax = μ,N= (0.20) (100) = 20 lb Since F> Fmax, the assumed motion is impossible. b. Rolling and Sliding. Since the wheel must roll and slide at the same time, we draw a new diagram, where a and a are inde- pendent and where F = 29.3 lb- N = 100 lb ↑ F = F₁ = ₂N = (0.15) (100) = 15 lb From the computation of part a, it appears that F should be di- rected to the left. 42F, = må: 40 15 3.11a à +8.04 ft/sec² à 8.04 ft/sec² (40) () = 0.264a (15)) a = 14.2 radians/sec² a = 14.2 radians/sec² 100 lb F= 15 lb G IN 100 lb OG с 5 in. IN = 40 lb 2 in. 3 in. r= 5 in. C 40 lb 5.0₂ 40 lb 12 in. ā mā 5 in. ma 5 in.
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