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Assistant Professor: Zhou Yufeng

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Assistant Professor: Zhou Yufeng The 7 kg uniform slender rod BC connected to a disk centered at A and to a crank CD. Knowing that the disk is made to rotate at the ... – PowerPoint PPT presentation

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Title: Assistant Professor: Zhou Yufeng


1
Tutorial 11
Kinetics of Rigid Body (Newtons Second Law)
  • Assistant Professor Zhou Yufeng

2
  1. The 7 kg uniform slender rod BC connected to a
    disk centered at A and to a crank CD. Knowing
    that the disk is made to rotate at the constant
    angular velocity of 180 r/min (rotations per
    minute), determine for the position shown
  1. the resultant force of all the external forces
    acting on BC,
  2. the resultant moment about mass center of all
    the external forces acting on BC,
  3. constraint forces at B and C.

Rotation of the disk wD 180 r/min 18.85
rad/s, and aD 0 , Rod BC is in translation.
(explain why)
(a)
Resultant force acting on BC
(1)
(b) For rod BC a 0, so MG 0.
(2)
3
(c) Assume the constraint forces at B and C to
be
which include 4 unknowns.
Using (1) we have
Using (2) we have
4
  • A uniform slender rod, of length L 750 mm and
    mass m 2 kg and , hangs freely
    from a hinge at A. If a force P of magnitude 12 N
    is applied at B horizontally to the left (h L),
    determine
  • the angular acceleration of the rod,
  • the components of the constraint force at A.

Rod AB rotates about a fixed point A, so we have
(a) Use Principle of Ang. Momentum about fixed
point A
5
  • A disk of radius r 125 mm, mass m 7 kg, and
    , rests on a frictionless horizontal
    surface. A force of magnitude 4 N is applied to a
    tape wrapped around the disk as shown. Determine
  • (a) the angular acceleration of the disk,
  • (b) the acceleration of the center G of the disk.
  • (c) the length of tape unwound after 2 s.

Draw the kinematics Diagram and Free body diagram.
(a)
(b)
4 7 (aG) ? aG 0.5714 (m/s2)
(c)
s 0.5aA/G t2 0.5(1.1429)(4) 2.2858 (m)
6
  1. Each of the double pulleys shown has an moment of
    inertia IG 10 kg?m2, and is initially at rest.
    The outside radius is 400 mm and the inner radius
    is 200 mm. Determine (a) the angular acceleration
    of each pulley, (b) the angular velocity of each
    pulley at t 3 s, (c) the angular velocity of
    each pulley after point A on the cord has moved 3
    m. Neglect friction.

Consider system disk blocks
IO IG 10 (kg.m2) IO 10 100 (0.22)
IO 10 500 (0.22) IO 10 50 (0.42)
7
w a(3)58.86 (rad) w a (3) 42.043 w
a (3)19.62 w a (3) 32.7
Note This problem can be easily solved using
standard method by separating blocks and disk
8
  1. A pipe (thin ring) has a mass of 500 kg and
    radius of 0.5 m and rolls without slipping down a
    300 kg ramp as shown. If the ramp is free to move
    horizontally (frictionless), determine the
    acceleration of the ramp. Neglect the masses of
    the wheels at A and B.

9
Step 2 Free body diagrams of Pipe and Ramp
Pipe
(1)
(2)
(3)
Ramp
(4)
Using (3) in (4) gives
(5)
Substituting (3) and (5) into (1) and (2) yields
(7)
(6)
Solving (6) and (7) gives
10
Alternative solution
Step 1 The system is a conservative system
(why?). Assume the pipe and ramp are released
from rest. During the history when pipe travels s
relative to the ramp, the work done by the
applied force is
The kinetic energy of the system at state 1 is
where
11
i.e.,
Using (2) in (1) gives
(3)
(4)
Using (4) in (3) we have
From (2) we also have
12
  • The disk D of weight 50 N, radius r 0.15 m, IG
    1/2(mr2), and block B of weight 20 N are
    released from rest as shown. The friction at
    contact point between the disk and the surface is
    large enough that the disk rolls without
    slipping. Neglect the masses of pulleys and
    cable. Determine
  • the acceleration of block B at the instant,
  • the minimum value of the static friction
    coefficient ?s between the disk and the surface.

Step1 Kinematics analysis
(why? Because C is the instant center)
13
Step2 Free Body Diagram Kinetics analysis
(1)
Block B
Disk
Solve (1) (2) and (4) for a, T and F
a - 13.77 (rad/s2)
F 2.6315 (N)
T (Fill in by yourself)
14
Since F ms N, we have the minimum friction
coefficient
Good students are encouraged to try DAlembert
Principle in this problem. By applying inertia
force and inertia moment at G, we combine FBD and
KD as one dynamic equilibrium diagram as shown.
15
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