7 Kinematics and kinetics of planar rigid bodies II
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7 Kinematics and kinetics of planar rigid bodies II
7.1 In-class A rigid circular cylinder of radius a and length h has a hole of radius 0.5a cut out. The density of the cylinder is ρ. Assume that the cylinder rolls without slipping on the floor. Compute the ki- netic energy and the potential energy of the cylinder using the generalized coordinate θ shown.
Solution
v = vi
I. Choose the reference system Set the reference system as shown below.
II. Moment of inertia The mass of the cylinder is
M = ρh [ πa2 − π(0.5a)2
4 πa2ρh . (7.1)
The moment of inertia of a cylinder of radius R without the hole is
IO = π
7 Kinematics and kinetics of planar rigid bodies II 7-2
Considering a new reference system as shown below,
the center of mass of the cylinder is
XC = 0 because of symmetry , (7.3)
YC = πa2(a)− π(a
2 )2(a+ a
6 a . (7.4)
The center of mass of the cylinder without the hole and the center of mass of the hole are denoted by O, O′ respectively. Applying the parallel axis theorem and the additive property of the moment of inertia, we get
IC = ρh
[ π
2
96 πa4ρh . (7.5)
III. Potential and kinetic energy The potential energy of the cylinder is
V = MgyC = 3
3
T = 1
2 Mv2C +
ω(t) = −θ(t)k . (7.8)
Since point A is the instantaneous center of rotation (vA = 0)
|vC(t)| = θ(t)|rAC | . (7.9)
Note that |rAC | depends on time. Considering the geometry (7.9) becomes
|vC(t)| = θ(t) √ |rOC |2 + |rOA|2 − 2|rOC ||rOA| cos θ(t) , (7.10)
7 Kinematics and kinetics of planar rigid bodies II 7-3
|vC(t)| = θ(t)
√(a 6
√ 37
T (t) = 1
7 Kinematics and kinetics of planar rigid bodies II 7-4
7.2 In-class A rod of mass m, length 2a and centroidal moment of inertia IC = 1 3 ma2 is dropped onto the edge of a table
as shown. The rod is horizontal, has zero angular velocity and has downward veloci- ty v0 at the moment just before touching the table.
(a) Determine, in terms of v0, the angular velocity of the rod just after impact, assuming that energy is conserved in the collision.
(b) Under the same assumptions, determine the velocity of the end of the rod that touched the table just after the impact. Does your result seem reasonable? Explain.
Solution
v = vk
Note that A denotes the contact point on the table and A′ denotes the contact point on the rod. I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Draw the configurations
7 Kinematics and kinetics of planar rigid bodies II 7-5
IV. Conservation of energy Assume that the collision occurs at t = t1. A vertical impulse (Py) acts on the end of the rod (point A′) at t = t1. As a result, velocity of the center of mass would be v1 and angular velocity ω1 just after the impact. Energy is conserved in the collision1
T (t−1 ) + V (t−1 ) = T (t+1 ) + V (t+1 ) . (7.14)
Since gravity does not have enough time to act
V (t−1 ) = V (t+1 ) . (7.15)
Substituting (7.15) into (7.14) yields
T (t−1 ) = T (t+1 ) , (7.16)
1
2 + 1
3 a2(ω1)
2 . (7.18)
V. Angular momentum principle In order to get a second relation, we apply angular momentum about point A on the table 2
HA + vA × P = M ext A . (7.19)
Since vA = 0 and gravity has no time to act, angular momentum is conserved about point A.
HA(t−1 ) = HA(t+1 ) . (7.20)
VI. Angular momentum transfer formula Applying angular momentum transfer formula to point A and C
HA = HC + P × rCA = HC +mvC × rCA . (7.21)
Using (7.20) and (7.21), we get
HC(t−1 ) +mvC(t−1 )× rCA(t−1 ) = HC(t+1 ) +mvC(t+1 )× rCA(t+1 ) , (7.22)
0−mav0k = −1
v0 = v1 + 1
3 aω1 . (7.24)
1Note that this statement is not valid in general, only if the collision is totally elastic. Elastic collision is defined as a collision in which kinetic energy is conserved. In several problems, this is a fair approximation. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.
2Please remember: we can choose any arbitrary point as reference to apply angular momentum prin- ciple. The point is not necessarily on the body (lecture notes page 52).
7 Kinematics and kinetics of planar rigid bodies II 7-6
Using (7.24) and (7.18), we get
v1 = v0 2 . (7.25)
The angular velocity of the rod just after the impact
ω1 = 3v0 2a
. (7.26)
(b) VII. Velocity transfer formula In order to determine the velocity of point A just after the impact, we use velocity transfer formula with respect to point A and C.
vA′(t + 1 ) = v1 + ω1 × rCA . (7.27)
vA′(t + 1 ) = −v1j + (−ω1k)× (−ai) . (7.28)
vA′(t + 1 ) = −v0
) j = v0j . (7.29)
We know that just before the impact the velocity of point A was −v0j. Therefore our result seems reasonable. Since the magnitude of the velocity of A on the rod is conserved and changes just the direction in the collision. This shows an elastic collision which is expected when energy is conserved.
7 Kinematics and kinetics of planar rigid bodies II 7-7
7.3 In-class A rigid, uniform flat disk of mass m and radius R is moving in the plane towards a wall with central velocity v0 while rotating with angular velocity ω1, as shown. Assuming that the collision in the normal direction is elastic and no slip occurs at the wall, find the velocity of the (center of the) disk after it collides with the wall.
Solution
ω = ωk
Note that B denotes the contact point on the wall and B′ denotes the contact point on the disk. I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Draw the configurations
7 Kinematics and kinetics of planar rigid bodies II 7-8
IV. Velocity transfer formula Assume the disk collides with the wall at point B, where an impulse (Px,Py) acts on it at t = t1. Collision in the normal direction (y) is elastic so the magnitude of the velocity in the normal direction is conserved
|(vC)y|(t+1 ) = |(vC)y|(t−1 ) , (7.30)
|(vC)y| = (|v1)y| = |v0| cos θ . (7.31)
Since no slip occurs at the wall
(vB′)x(t + 1 ) = 0 . (7.32)
Using the velocity transfer formula, we get
(vB′)x(t + 1 ) = (vC)x(t
+ 1 ) + (ω1 × rCB)x = 0 , (7.33)
(vC)x(t + 1 ) = (v1)x = Rω1 . (7.34)
V. Angular momentum principle Applying angular momentum principle about point B is
HB + vB × P = M ext B . (7.35)
Since there is no force which produces external torque on B and vB = 0,the angular momentum is conserved about B
HB = 0⇒ HB(t−1 ) = HB(t+1 ) . (7.36)
Using the angular momentum transfer formula (7.36) becomes
HB = 0⇒ HC(t−1 ) + P (t−1 )× rCB(t−1 ) = HC(t+1 ) + P (t+1 )× rCB(t−1 ) . (7.37)
lim t−1 →t1
|rCB| = R .
1
2 mR2ω1k +mR(v1)xk . (7.38)
Using (7.38) and (7.34) the x-component of the velocity of the disk after the impact is
(v1)x = 2
3 Rω . (7.39)
Using (7.31) and (7.39) the velocity of the disk after the collision is
|v1| =
√( 2
7 Kinematics and kinetics of planar rigid bodies II 7-9
7.4 Homework A cube with sides of length 2a and a mass M is moving with an initial speed v0 along a frictionless ta- ble. When the cube reaches the end of the table it is caught abruptly by a short lip and begins to rotate. What is the minimum speed v0 such that the cube falls off the ta- ble? (The collision is not elastic.)
Solution
0. Notation Note that B denotes the contact point on the lip.
I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram at t = t1
III. Angular momentum principle Applying angular momentum principle about point B
HB + vB × P = M ext B . (7.41)
Three external forces (N , F , mg) act on the cube while only N and mg produce torque about point B. Since forces N and mg are not impulsive, we get
M ext B = rBC × (N +mg)⇒
t+∫ t−
M ext B dt = 0 . (7.42)
Since vB = 0 angular momentum with respect to B is conserved so
HB(t−1 ) = HB(t+1 ) , (7.43)
Ma|v0| = IB|ω| . (7.44)
7 Kinematics and kinetics of planar rigid bodies II 7-10
From which the angular velocity of the cube just after the collision
|ω| = Ma|v0| IB
. (7.45)
For the block to tip over the lip, its center of mass must end up a distance a( √
2 − 1) above its original position.
The energy of the rotational motion 1 (just after the impact) has to be large enough to raise the center of mass with a(
√ 2− 1).
√ 2− 1) (7.46)
√ 2− 1)
Ma . (7.48)
The centroid moment of inertia of a cube which has mass m and edge k is
I = 1
IC = 2
v0 >
√ 28Ma2
2− 1) . (7.52)
1Note that the energy of the rotational motion transforms to potential energy, therefore ω continuously decreases till the center of mass reaches its maximum height.
7 Kinematics and kinetics of planar rigid bodies II 7-11
7.5 Homework A pendulum consists of a rod of length L with a frictionless pivot at one end. The pendu- lum is suspended from a flywheel of radius R which rotates with fixed angular velocity ω, as shown be- low.
(a) Determine the angular velocity of the rod in terms of ω and the generalized coordinate θ indicated in the sketch.
(b) Calculate the velocity of the mid point C of the rod
Solution
v = ve1
I. Choose the reference system Set the reference system as shown below.
II. Draw the reference and displaced configuration
III. Angular velocity of the rod To find the angular velocity of the rod, compare orientation of AB to A’B’
ωrod = [θ + ]k = [θ + ω]k . (7.53)
7 Kinematics and kinetics of planar rigid bodies II 7-12
IV. Draw the displaced configuration
V. Velocity transfer formula The velocity transfer formula
vC = vA + ωrodk × rAC , (7.54)
where vA is
Now the velocity of point C is
vC = ωReψ + (ω + θ) L
2 eθ . (7.56)
Expressing eψ and eθ in terms of i and j
eψ = − sinψi+ cosψj , (7.57)
eθ = − sin [θ + ψ]i+ cos [θ + ψ]j . (7.59)
Substituting (7.59) and (7.57) into (7.56) yields
vC = (−ωR sinψ− (ω+ θ) L
2 sin [θ + ψ])i+(ωR cosψ+(ω+ θ)
L
7 Kinematics and kinetics of planar rigid bodies II 7-13
7.6 Homework A rigid cylinder of radi- us R is moving to the right such that its center C has velocity v. There is no slipping between the cylinder and the bar BD, but there is slipping between the cy- linder and the ground. In the position shown
(a) Determine the angular velocity of the bar BD.
(b) Determine the velocity of the cylinder at the point where it contacts the ground.
Solution
v = vI
I. Choose the reference system Set the reference system as shown below.
II. Set up the variables
III. Angular velocity of the rod The velocity of point C
vC = vCi = vC cos θI + vC sin θJ . (7.61)
Since there is no slip between the cylinder and the bar
vA,cyl = vA,bar . (7.62)
7 Kinematics and kinetics of planar rigid bodies II 7-14
The velocity of point A on the bar is
vA,bar = ωbar × rBA = |rBA|θJ . (7.63)
The velocity of point A on the cylinder can be determined from vC by using velocity transfer formula
vA,cyl = vC,cyl + ωcyl × rCA = vC cos θI + vC sin θJ −RωcylI . (7.64)
Substituting (7.63) and (7.64) into (7.62), we get
|rBA|θJ = vC cos θI + vC sin θJ −RωcylI , (7.65)
0 = (vC cos θ −Rωcyl)I + (vC sin θ − |rBA|θ)J . (7.66)
Equating separately the coefficients of the I-components and the coefficients of the J- components, yields
ωcyl = vC cos θ
rCA rBA
= tan θ
2 . (7.69)
Using all the already expressed variables, the angular velocity of the bar is
ωbar = −θk = −vC sin θ
|rBA| k = −vC
) k . (7.70)
(b) The velocity of the cylinder at the point where it contacts the ground can be deter- mined from vC by using velocity transfer formula
vE,cyl = vC,cyl + ωcyl × rCE =
7 Kinematics and kinetics of planar rigid bodies II 7-15
7.7 Homework A uniform rod of mass M and length 2b is pivoted at a point O, a distance s above the center of mass (CM). The rod is struck with a rapid impulsive force perpendicular to the rod at a point A, a distance a below the center of mass.The magnitude of the impulse is P = Ft. Find the value of a such that there is no horizontal (N) reaction at the pivot, during the impact. (The moment of inertia of a uniform rod with mass M and length L about an axis through its center perpendicular to its longer side is ICM = ML2/12.)
Solution
I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Linear momentum principle Applying linear momentum principle in the x-direction
Px = F ext x ⇒Mx = F −N . (7.72)
IV. Angular momentum principle Applying angular momentum principle
HO + vO × P = M ext O . (7.73)
Since vO = 0, we get
HO = M ext O ⇒ IOθ = (s+ a)F , (7.74)
where the moment of inertia with respect to O can be determined by parallel axis theorem, such as
IO = ICM +Ms2 ⇒ IO = M
[ (2b)2
7 Kinematics and kinetics of planar rigid bodies II 7-16
Now we have two equations and three unknowns x, θ, N . To solve these for N , we need to add a third equation. Considering the geometry and using small-angle approximation 1 we can write
sin θ ≈ θ = x
s ⇒ x = sθ . (7.76)
N = F −Mx = F b2
3 − sa
. (7.77)
Therefore the horizontal normal force will be zero if we set
a = b2
3s . (7.78)
1The small-angle approximation is a useful simplification of the basic trigonometric functions which is approximately true in the limit where the angle approaches zero. They are truncations of the Taylor series for the basic trigonometric functions to a first-order approximation.
7 Kinematics and kinetics of planar rigid bodies II 7-17
7.8 Homework A rigid block of height H, length L, depth D and mass m rests on a rigid cylinder of mass M and radius R, as shown in the sketch. The cylinder rolls on the floor without slip- ping and the block rolls on the cylinder without slipping as well. Determine the kinetic and poten- tial energy of the system.
Solution
v = vi
I. Choose the reference system Set the inertial reference system as shown below.
II. Set up the variables
Note that point O denotes the center of the cylinder and rOC = xOCi+ yOCj.
III. Angular velocity The angular velocity of the cylinder
ωcyl = −k . (7.79)
ωblock = −θk . (7.80)
7 Kinematics and kinetics of planar rigid bodies II 7-18
IV. Velocity of the block and the cylinder Since there is no slip between the cylinder and the floor
vO = Ri . (7.81)
The velocity of the center of mass is
vC = Ri+ xCOi+ yCOj = (R+ xCO)i+ yCOj . (7.82)
Since there is no slip between the block and the cylinder
vB,block = vB,cyl . (7.83)
The velocity of point B on the cylinder can be determined using velocity transfer formula.
vB,cyl = vO + ωcyl × rOB = Ri+ (−k)× (R sin θi+R cos θj) (7.84)
vB,cyl = R(1 + cos θ)i−R sin θj (7.85)
The velocity of point B on the block can be determined using velocity transfer formula.
vB,block = vC + ωblock × rCB (7.86)
vB,block = (R+ xCO)i+ yCOj + (−θk)× [(R sin θ− xCO)i+ (R cos θ− yCO)j] (7.87)
vB,block = (R+ xCO +Rθ cos θ − θyCO)i+ (yCO −Rθ sin θ + θxCO)j (7.88)
V. Position of the center of mass Substituting (7.85) and (7.88) into (7.83) and equating separately the coefficients of the i-components and the coefficients of the j-components, yields
R+ xCO +Rθ cos θ − θyCO = R(1 + cos θ) , (7.89)
yCO −Rθ sin θ + θxCO = −R sin θ . (7.90)
Multiplying (7.89) by sin θ and (7.90) by cos θ and adding them together yields
xCO sin θ + yCO cos θ − θyCO sin θ + θxCO cos θ = 0 . (7.91)
Now we can realize that (7.91) can be written as
d
Integrating (7.92) with respect to time yields
xCO sin θ + yCO cos θ = constant . (7.93)
Substituting θ = 0, xCO = 0 and yCO = R + H 2
initial conditions into (7.93) yields
xCO sin θ + yCO cos θ = R + H
2 . (7.94)
7 Kinematics and kinetics of planar rigid bodies II 7-19
Multiplying (7.89) by cos θ and (7.90) by sin θ and subtracting them yields
R(− θ) = xCO cos θ − yCO sin θ − θyCO cos θ − θxCO sin θ . (7.95)
Now we can realize that (7.95) can be written as
d
Integrating (7.96) with respect to time yields
R(− θ) + C1 = xCO cos θ − yCO sin θ + C2 . (7.97)
Substituting θ = 0, = 0, xCO = 0 and yCO = R+ H 2
initial conditions into (7.97) yields C1 = C2, so
R(− θ) = xCO cos θ − yCO sin θ . (7.98)
In order to get xCO we multiply (7.94) by sin θ and (7.98) by cos θ and add them together.
xCO =
[ R +
H
2
] sin θ +R(− θ) cos θ . (7.99)
In order to get yCO we multiply (7.94) by cos θ and (7.98) by sin θ and subtract them.
yCO =
[ R +
H
2
] cos θ −R(− θ) sin θ . (7.100)
VI. Energy of the system The potential energy of the system is
V = mgyCO +mgyO = mgyCO , (7.101)
V = mg
) . (7.102)
T = 1
2 Mv2O +
1
2
( 1
7.1 In-class A rigid circular cylinder of radius a and length h has a hole of radius 0.5a cut out. The density of the cylinder is ρ. Assume that the cylinder rolls without slipping on the floor. Compute the ki- netic energy and the potential energy of the cylinder using the generalized coordinate θ shown.
Solution
v = vi
I. Choose the reference system Set the reference system as shown below.
II. Moment of inertia The mass of the cylinder is
M = ρh [ πa2 − π(0.5a)2
4 πa2ρh . (7.1)
The moment of inertia of a cylinder of radius R without the hole is
IO = π
7 Kinematics and kinetics of planar rigid bodies II 7-2
Considering a new reference system as shown below,
the center of mass of the cylinder is
XC = 0 because of symmetry , (7.3)
YC = πa2(a)− π(a
2 )2(a+ a
6 a . (7.4)
The center of mass of the cylinder without the hole and the center of mass of the hole are denoted by O, O′ respectively. Applying the parallel axis theorem and the additive property of the moment of inertia, we get
IC = ρh
[ π
2
96 πa4ρh . (7.5)
III. Potential and kinetic energy The potential energy of the cylinder is
V = MgyC = 3
3
T = 1
2 Mv2C +
ω(t) = −θ(t)k . (7.8)
Since point A is the instantaneous center of rotation (vA = 0)
|vC(t)| = θ(t)|rAC | . (7.9)
Note that |rAC | depends on time. Considering the geometry (7.9) becomes
|vC(t)| = θ(t) √ |rOC |2 + |rOA|2 − 2|rOC ||rOA| cos θ(t) , (7.10)
7 Kinematics and kinetics of planar rigid bodies II 7-3
|vC(t)| = θ(t)
√(a 6
√ 37
T (t) = 1
7 Kinematics and kinetics of planar rigid bodies II 7-4
7.2 In-class A rod of mass m, length 2a and centroidal moment of inertia IC = 1 3 ma2 is dropped onto the edge of a table
as shown. The rod is horizontal, has zero angular velocity and has downward veloci- ty v0 at the moment just before touching the table.
(a) Determine, in terms of v0, the angular velocity of the rod just after impact, assuming that energy is conserved in the collision.
(b) Under the same assumptions, determine the velocity of the end of the rod that touched the table just after the impact. Does your result seem reasonable? Explain.
Solution
v = vk
Note that A denotes the contact point on the table and A′ denotes the contact point on the rod. I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Draw the configurations
7 Kinematics and kinetics of planar rigid bodies II 7-5
IV. Conservation of energy Assume that the collision occurs at t = t1. A vertical impulse (Py) acts on the end of the rod (point A′) at t = t1. As a result, velocity of the center of mass would be v1 and angular velocity ω1 just after the impact. Energy is conserved in the collision1
T (t−1 ) + V (t−1 ) = T (t+1 ) + V (t+1 ) . (7.14)
Since gravity does not have enough time to act
V (t−1 ) = V (t+1 ) . (7.15)
Substituting (7.15) into (7.14) yields
T (t−1 ) = T (t+1 ) , (7.16)
1
2 + 1
3 a2(ω1)
2 . (7.18)
V. Angular momentum principle In order to get a second relation, we apply angular momentum about point A on the table 2
HA + vA × P = M ext A . (7.19)
Since vA = 0 and gravity has no time to act, angular momentum is conserved about point A.
HA(t−1 ) = HA(t+1 ) . (7.20)
VI. Angular momentum transfer formula Applying angular momentum transfer formula to point A and C
HA = HC + P × rCA = HC +mvC × rCA . (7.21)
Using (7.20) and (7.21), we get
HC(t−1 ) +mvC(t−1 )× rCA(t−1 ) = HC(t+1 ) +mvC(t+1 )× rCA(t+1 ) , (7.22)
0−mav0k = −1
v0 = v1 + 1
3 aω1 . (7.24)
1Note that this statement is not valid in general, only if the collision is totally elastic. Elastic collision is defined as a collision in which kinetic energy is conserved. In several problems, this is a fair approximation. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.
2Please remember: we can choose any arbitrary point as reference to apply angular momentum prin- ciple. The point is not necessarily on the body (lecture notes page 52).
7 Kinematics and kinetics of planar rigid bodies II 7-6
Using (7.24) and (7.18), we get
v1 = v0 2 . (7.25)
The angular velocity of the rod just after the impact
ω1 = 3v0 2a
. (7.26)
(b) VII. Velocity transfer formula In order to determine the velocity of point A just after the impact, we use velocity transfer formula with respect to point A and C.
vA′(t + 1 ) = v1 + ω1 × rCA . (7.27)
vA′(t + 1 ) = −v1j + (−ω1k)× (−ai) . (7.28)
vA′(t + 1 ) = −v0
) j = v0j . (7.29)
We know that just before the impact the velocity of point A was −v0j. Therefore our result seems reasonable. Since the magnitude of the velocity of A on the rod is conserved and changes just the direction in the collision. This shows an elastic collision which is expected when energy is conserved.
7 Kinematics and kinetics of planar rigid bodies II 7-7
7.3 In-class A rigid, uniform flat disk of mass m and radius R is moving in the plane towards a wall with central velocity v0 while rotating with angular velocity ω1, as shown. Assuming that the collision in the normal direction is elastic and no slip occurs at the wall, find the velocity of the (center of the) disk after it collides with the wall.
Solution
ω = ωk
Note that B denotes the contact point on the wall and B′ denotes the contact point on the disk. I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Draw the configurations
7 Kinematics and kinetics of planar rigid bodies II 7-8
IV. Velocity transfer formula Assume the disk collides with the wall at point B, where an impulse (Px,Py) acts on it at t = t1. Collision in the normal direction (y) is elastic so the magnitude of the velocity in the normal direction is conserved
|(vC)y|(t+1 ) = |(vC)y|(t−1 ) , (7.30)
|(vC)y| = (|v1)y| = |v0| cos θ . (7.31)
Since no slip occurs at the wall
(vB′)x(t + 1 ) = 0 . (7.32)
Using the velocity transfer formula, we get
(vB′)x(t + 1 ) = (vC)x(t
+ 1 ) + (ω1 × rCB)x = 0 , (7.33)
(vC)x(t + 1 ) = (v1)x = Rω1 . (7.34)
V. Angular momentum principle Applying angular momentum principle about point B is
HB + vB × P = M ext B . (7.35)
Since there is no force which produces external torque on B and vB = 0,the angular momentum is conserved about B
HB = 0⇒ HB(t−1 ) = HB(t+1 ) . (7.36)
Using the angular momentum transfer formula (7.36) becomes
HB = 0⇒ HC(t−1 ) + P (t−1 )× rCB(t−1 ) = HC(t+1 ) + P (t+1 )× rCB(t−1 ) . (7.37)
lim t−1 →t1
|rCB| = R .
1
2 mR2ω1k +mR(v1)xk . (7.38)
Using (7.38) and (7.34) the x-component of the velocity of the disk after the impact is
(v1)x = 2
3 Rω . (7.39)
Using (7.31) and (7.39) the velocity of the disk after the collision is
|v1| =
√( 2
7 Kinematics and kinetics of planar rigid bodies II 7-9
7.4 Homework A cube with sides of length 2a and a mass M is moving with an initial speed v0 along a frictionless ta- ble. When the cube reaches the end of the table it is caught abruptly by a short lip and begins to rotate. What is the minimum speed v0 such that the cube falls off the ta- ble? (The collision is not elastic.)
Solution
0. Notation Note that B denotes the contact point on the lip.
I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram at t = t1
III. Angular momentum principle Applying angular momentum principle about point B
HB + vB × P = M ext B . (7.41)
Three external forces (N , F , mg) act on the cube while only N and mg produce torque about point B. Since forces N and mg are not impulsive, we get
M ext B = rBC × (N +mg)⇒
t+∫ t−
M ext B dt = 0 . (7.42)
Since vB = 0 angular momentum with respect to B is conserved so
HB(t−1 ) = HB(t+1 ) , (7.43)
Ma|v0| = IB|ω| . (7.44)
7 Kinematics and kinetics of planar rigid bodies II 7-10
From which the angular velocity of the cube just after the collision
|ω| = Ma|v0| IB
. (7.45)
For the block to tip over the lip, its center of mass must end up a distance a( √
2 − 1) above its original position.
The energy of the rotational motion 1 (just after the impact) has to be large enough to raise the center of mass with a(
√ 2− 1).
√ 2− 1) (7.46)
√ 2− 1)
Ma . (7.48)
The centroid moment of inertia of a cube which has mass m and edge k is
I = 1
IC = 2
v0 >
√ 28Ma2
2− 1) . (7.52)
1Note that the energy of the rotational motion transforms to potential energy, therefore ω continuously decreases till the center of mass reaches its maximum height.
7 Kinematics and kinetics of planar rigid bodies II 7-11
7.5 Homework A pendulum consists of a rod of length L with a frictionless pivot at one end. The pendu- lum is suspended from a flywheel of radius R which rotates with fixed angular velocity ω, as shown be- low.
(a) Determine the angular velocity of the rod in terms of ω and the generalized coordinate θ indicated in the sketch.
(b) Calculate the velocity of the mid point C of the rod
Solution
v = ve1
I. Choose the reference system Set the reference system as shown below.
II. Draw the reference and displaced configuration
III. Angular velocity of the rod To find the angular velocity of the rod, compare orientation of AB to A’B’
ωrod = [θ + ]k = [θ + ω]k . (7.53)
7 Kinematics and kinetics of planar rigid bodies II 7-12
IV. Draw the displaced configuration
V. Velocity transfer formula The velocity transfer formula
vC = vA + ωrodk × rAC , (7.54)
where vA is
Now the velocity of point C is
vC = ωReψ + (ω + θ) L
2 eθ . (7.56)
Expressing eψ and eθ in terms of i and j
eψ = − sinψi+ cosψj , (7.57)
eθ = − sin [θ + ψ]i+ cos [θ + ψ]j . (7.59)
Substituting (7.59) and (7.57) into (7.56) yields
vC = (−ωR sinψ− (ω+ θ) L
2 sin [θ + ψ])i+(ωR cosψ+(ω+ θ)
L
7 Kinematics and kinetics of planar rigid bodies II 7-13
7.6 Homework A rigid cylinder of radi- us R is moving to the right such that its center C has velocity v. There is no slipping between the cylinder and the bar BD, but there is slipping between the cy- linder and the ground. In the position shown
(a) Determine the angular velocity of the bar BD.
(b) Determine the velocity of the cylinder at the point where it contacts the ground.
Solution
v = vI
I. Choose the reference system Set the reference system as shown below.
II. Set up the variables
III. Angular velocity of the rod The velocity of point C
vC = vCi = vC cos θI + vC sin θJ . (7.61)
Since there is no slip between the cylinder and the bar
vA,cyl = vA,bar . (7.62)
7 Kinematics and kinetics of planar rigid bodies II 7-14
The velocity of point A on the bar is
vA,bar = ωbar × rBA = |rBA|θJ . (7.63)
The velocity of point A on the cylinder can be determined from vC by using velocity transfer formula
vA,cyl = vC,cyl + ωcyl × rCA = vC cos θI + vC sin θJ −RωcylI . (7.64)
Substituting (7.63) and (7.64) into (7.62), we get
|rBA|θJ = vC cos θI + vC sin θJ −RωcylI , (7.65)
0 = (vC cos θ −Rωcyl)I + (vC sin θ − |rBA|θ)J . (7.66)
Equating separately the coefficients of the I-components and the coefficients of the J- components, yields
ωcyl = vC cos θ
rCA rBA
= tan θ
2 . (7.69)
Using all the already expressed variables, the angular velocity of the bar is
ωbar = −θk = −vC sin θ
|rBA| k = −vC
) k . (7.70)
(b) The velocity of the cylinder at the point where it contacts the ground can be deter- mined from vC by using velocity transfer formula
vE,cyl = vC,cyl + ωcyl × rCE =
7 Kinematics and kinetics of planar rigid bodies II 7-15
7.7 Homework A uniform rod of mass M and length 2b is pivoted at a point O, a distance s above the center of mass (CM). The rod is struck with a rapid impulsive force perpendicular to the rod at a point A, a distance a below the center of mass.The magnitude of the impulse is P = Ft. Find the value of a such that there is no horizontal (N) reaction at the pivot, during the impact. (The moment of inertia of a uniform rod with mass M and length L about an axis through its center perpendicular to its longer side is ICM = ML2/12.)
Solution
I. Choose the reference system Set the reference system as shown below.
II. Draw the free-body diagram
III. Linear momentum principle Applying linear momentum principle in the x-direction
Px = F ext x ⇒Mx = F −N . (7.72)
IV. Angular momentum principle Applying angular momentum principle
HO + vO × P = M ext O . (7.73)
Since vO = 0, we get
HO = M ext O ⇒ IOθ = (s+ a)F , (7.74)
where the moment of inertia with respect to O can be determined by parallel axis theorem, such as
IO = ICM +Ms2 ⇒ IO = M
[ (2b)2
7 Kinematics and kinetics of planar rigid bodies II 7-16
Now we have two equations and three unknowns x, θ, N . To solve these for N , we need to add a third equation. Considering the geometry and using small-angle approximation 1 we can write
sin θ ≈ θ = x
s ⇒ x = sθ . (7.76)
N = F −Mx = F b2
3 − sa
. (7.77)
Therefore the horizontal normal force will be zero if we set
a = b2
3s . (7.78)
1The small-angle approximation is a useful simplification of the basic trigonometric functions which is approximately true in the limit where the angle approaches zero. They are truncations of the Taylor series for the basic trigonometric functions to a first-order approximation.
7 Kinematics and kinetics of planar rigid bodies II 7-17
7.8 Homework A rigid block of height H, length L, depth D and mass m rests on a rigid cylinder of mass M and radius R, as shown in the sketch. The cylinder rolls on the floor without slip- ping and the block rolls on the cylinder without slipping as well. Determine the kinetic and poten- tial energy of the system.
Solution
v = vi
I. Choose the reference system Set the inertial reference system as shown below.
II. Set up the variables
Note that point O denotes the center of the cylinder and rOC = xOCi+ yOCj.
III. Angular velocity The angular velocity of the cylinder
ωcyl = −k . (7.79)
ωblock = −θk . (7.80)
7 Kinematics and kinetics of planar rigid bodies II 7-18
IV. Velocity of the block and the cylinder Since there is no slip between the cylinder and the floor
vO = Ri . (7.81)
The velocity of the center of mass is
vC = Ri+ xCOi+ yCOj = (R+ xCO)i+ yCOj . (7.82)
Since there is no slip between the block and the cylinder
vB,block = vB,cyl . (7.83)
The velocity of point B on the cylinder can be determined using velocity transfer formula.
vB,cyl = vO + ωcyl × rOB = Ri+ (−k)× (R sin θi+R cos θj) (7.84)
vB,cyl = R(1 + cos θ)i−R sin θj (7.85)
The velocity of point B on the block can be determined using velocity transfer formula.
vB,block = vC + ωblock × rCB (7.86)
vB,block = (R+ xCO)i+ yCOj + (−θk)× [(R sin θ− xCO)i+ (R cos θ− yCO)j] (7.87)
vB,block = (R+ xCO +Rθ cos θ − θyCO)i+ (yCO −Rθ sin θ + θxCO)j (7.88)
V. Position of the center of mass Substituting (7.85) and (7.88) into (7.83) and equating separately the coefficients of the i-components and the coefficients of the j-components, yields
R+ xCO +Rθ cos θ − θyCO = R(1 + cos θ) , (7.89)
yCO −Rθ sin θ + θxCO = −R sin θ . (7.90)
Multiplying (7.89) by sin θ and (7.90) by cos θ and adding them together yields
xCO sin θ + yCO cos θ − θyCO sin θ + θxCO cos θ = 0 . (7.91)
Now we can realize that (7.91) can be written as
d
Integrating (7.92) with respect to time yields
xCO sin θ + yCO cos θ = constant . (7.93)
Substituting θ = 0, xCO = 0 and yCO = R + H 2
initial conditions into (7.93) yields
xCO sin θ + yCO cos θ = R + H
2 . (7.94)
7 Kinematics and kinetics of planar rigid bodies II 7-19
Multiplying (7.89) by cos θ and (7.90) by sin θ and subtracting them yields
R(− θ) = xCO cos θ − yCO sin θ − θyCO cos θ − θxCO sin θ . (7.95)
Now we can realize that (7.95) can be written as
d
Integrating (7.96) with respect to time yields
R(− θ) + C1 = xCO cos θ − yCO sin θ + C2 . (7.97)
Substituting θ = 0, = 0, xCO = 0 and yCO = R+ H 2
initial conditions into (7.97) yields C1 = C2, so
R(− θ) = xCO cos θ − yCO sin θ . (7.98)
In order to get xCO we multiply (7.94) by sin θ and (7.98) by cos θ and add them together.
xCO =
[ R +
H
2
] sin θ +R(− θ) cos θ . (7.99)
In order to get yCO we multiply (7.94) by cos θ and (7.98) by sin θ and subtract them.
yCO =
[ R +
H
2
] cos θ −R(− θ) sin θ . (7.100)
VI. Energy of the system The potential energy of the system is
V = mgyCO +mgyO = mgyCO , (7.101)
V = mg
) . (7.102)
T = 1
2 Mv2O +
1
2
( 1
