7 Kinematics and kinetics of planar rigid bodies II

7 Kinematics and kinetics of planar rigidbodies II

7.1 In-class A rigid circular cylinder of radius aand length h has a hole of radius 0.5a cut out. Thedensity of the cylinder is ρ. Assume that the cylinderrolls without slipping on the floor. Compute the ki-netic energy and the potential energy of the cylinderusing the generalized coordinate θ shown.

Solution

0. NotationWe use the following notation for vectors:

v = vi

I. Choose the reference systemSet the reference system as shown below.

II. Moment of inertiaThe mass of the cylinder is

M = ρh[πa2 − π(0.5a)2

]=

3

4πa2ρh . (7.1)

The moment of inertia of a cylinder of radius R without the hole is

IO =π

2R4ρh . (7.2)

7-1

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

2)

πa2 − π(a2)2

=5

6a . (7.4)

The center of mass of the cylinder without the hole and the center of mass of the hole aredenoted by O, O′ respectively.Applying the parallel axis theorem and the additive property of the moment of inertia,we get

IC = ρh

(π

2a4 + πa2|rOC |2 −

[π

2

(a2

)4+ π

(a2

)2|rO′C |2

])=

37

96πa4ρh . (7.5)

III. Potential and kinetic energyThe potential energy of the cylinder is

V = MgyC =3

4πa2ρgh(a− |rOC | cos θ) =

3

4πa2ρgha

(1− cos θ

6

). (7.6)

The kinetic energy of the cylinder is

T =1

2Mv2C +

1

2ICω

2 . (7.7)

The angular velocity is

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)


|vC(t)| = θ(t)

√(a6

)2+ a2 − 2

a

6a cos θ(t) = aθ(t)

√37

36− cos θ(t)

3. (7.11)

Substituting all the known variables into (7.7) yields

T (t) =1

2

(3

4πa2ρh

)[aθ(t)

√37

36− cos θ(t)

3

]2+

1

2

(37

96πa4ρh

)θ(t)2 , (7.12)

T (t) = πρha4θ(t)2(

37

64− cos θ(t)

8

). (7.13)


7.2 In-class A rod of mass m, length2a and centroidal moment of inertia IC =13ma2 is dropped onto the edge of a table

as shown. The rod is horizontal, has zeroangular velocity and has downward veloci-ty v0 at the moment just before touchingthe table.

(a) Determine, in terms of v0, the angularvelocity 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 touchedthe 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 onthe rod.I. Choose the reference systemSet the reference system as shown below.

II. Draw the free-body diagram

III. Draw the configurations


IV. Conservation of energyAssume that the collision occurs at t = t1. A vertical impulse (∆Py) acts on the end ofthe rod (point A′) at t = t1. As a result, velocity of the center of mass would be v1 andangular 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

2m(v0)

2 =1

2m(v1)

2 +1

2IC(ω1)

2 , (7.17)

(v0)2 = (v1)

2 +1

3a2(ω1)

2 . (7.18)

V. Angular momentum principleIn order to get a second relation, we apply angular momentum about point A on the table2

HA + vA × P = M extA . (7.19)

Since vA = 0 and gravity has no time to act, angular momentum is conserved about pointA.

HA(t−1 ) = HA(t+1 ) . (7.20)

VI. Angular momentum transfer formulaApplying 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

3ma2ω1k −mav1k . (7.23)

From (7.23), we get

v0 = v1 +1

3aω1 . (7.24)

1Note that this statement is not valid in general, only if the collision is totally elastic. Elastic collision isdefined 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).


Using (7.24) and (7.18), we get

v1 =v02. (7.25)

The angular velocity of the rod just after the impact

ω1 =3v02a

. (7.26)

(b) VII. Velocity transfer formulaIn order to determine the velocity of point A just after the impact, we use velocity transferformula 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

2j + (−3v0

2ak)× (−ai) =

(−v0

2+

3v02

)j = v0j . (7.29)

We know that just before the impact the velocity of point A was −v0j. Therefore ourresult seems reasonable. Since the magnitude of the velocity of A on the rod is conservedand changes just the direction in the collision. This shows an elastic collision which isexpected when energy is conserved.


7.3 In-class A rigid, uniform flat diskof mass m and radius R is moving in theplane towards a wall with central velocityv0 while rotating with angular velocity ω1,as shown. Assuming that the collision inthe normal direction is elastic and no slipoccurs at the wall, find the velocity of the(center of the) disk after it collides with thewall.

Solution


ω = ωk

Note that B denotes the contact point on the wall and B′ denotes the contact point onthe disk.I. Choose the reference systemSet the reference system as shown below.


III. Draw the configurations


IV. Velocity transfer formulaAssume the disk collides with the wall at point B, where an impulse (∆Px,∆Py) acts onit at t = t1. Collision in the normal direction (y) is elastic so the magnitude of the velocityin 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 principleApplying angular momentum principle about point B is

HB + vB × P = M extB . (7.35)

Since there is no force which produces external torque on B and vB = 0,the angularmomentum 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)

limt−1 →t1

|rCB| = limt+1→t1

|rCB| = R .

1

2mR2ωk +mv0R sin θk =

1

2mR2ω1k +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

3v0 sin θ +

1

3Rω . (7.39)

Using (7.31) and (7.39) the velocity of the disk after the collision is

|v1| =

√(2

3v0 sin θ +

1

3Rω

)2

+ (v0 cos θ)2 . (7.40)


7.4 Homework A cube with sides oflength 2a and a mass M is moving withan initial speed v0 along a frictionless ta-ble. When the cube reaches the end of thetable it is caught abruptly by a short lipand begins to rotate. What is the minimumspeed v0 such that the cube falls off the ta-ble? (The collision is not elastic.)

Solution

0. NotationNote that B denotes the contact point on the lip.


II. Draw the free-body diagram at t = t1

III. Angular momentum principleApplying angular momentum principle about point B

HB + vB × P = M extB . (7.41)

Three external forces (N , F , mg) act on the cube while only N and mg produce torqueabout point B. Since forces N and mg are not impulsive, we get

M extB = rBC × (N +mg)⇒

t+∫t−

M extB 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)


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 toraise the center of mass with a(

√2− 1).

1

2IB|ω|2 > Mga(

√2− 1) (7.46)

Substituting (7.45) into (7.46)

1

2IBM2a2v20I2B

> Mga(√

2− 1) . (7.47)

v0 >

√2IBg(

√2− 1)

Ma. (7.48)

The centroid moment of inertia of a cube which has mass m and edge k is

I =1

6mk2 . (7.49)

Substituting mass M and edge 2a, we get

IC =2

3Ma2 . (7.50)

Using parallel axis theorem

IB =2

3Ma2 +M(a

√2)2 =

8Ma2

3. (7.51)

Substituting IB into (7.48) yields

v0 >

√28Ma2

3g(√

2− 1)

Ma⇒ v0 >

√16

3ag(√

2− 1) . (7.52)

1Note that the energy of the rotational motion transforms to potential energy, therefore ω continuouslydecreases till the center of mass reaches its maximum height.


7.5 Homework A pendulum consists of a rod of lengthL with a frictionless pivot at one end. The pendu-lum is suspended from a flywheel of radius R whichrotates with fixed angular velocity ω, as shown be-low.

(a) Determine the angular velocity of the rod in terms ofω and the generalized coordinate θ indicated in the sketch.

(b) Calculate the velocity of the mid point C of the rod

Solution


v = ve1


II. Draw the reference and displaced configuration

III. Angular velocity of the rodTo find the angular velocity of the rod, compare orientation of AB to A’B’

ωrod = [θ + ϕ]k = [θ + ω]k . (7.53)


IV. Draw the displaced configuration

V. Velocity transfer formulaThe velocity transfer formula

vC = vA + ωrodk × rAC , (7.54)

where vA is

vA = ωrod × rOA = ωReψ . (7.55)

Now the velocity of point C is

vC = ωReψ + (ω + θ)L

2eθ . (7.56)

Expressing eψ and eθ in terms of i and j

eψ = − sinψi+ cosψj , (7.57)

eθ = − cos[θ −

(π2− ψ

)]i− sin

[θ −

(π2− ψ

)]j , (7.58)

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

2sin [θ + ψ])i+(ωR cosψ+(ω+ θ)

L

2cos [θ + ψ])j . (7.60)


7.6 Homework A rigid cylinder of radi-us R is moving to the right such thatits center C has velocity v. There is noslipping between the cylinder and the barBD, but there is slipping between the cy-linder and the ground. In the positionshown

(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


II. Set up the variables

III. Angular velocity of the rodThe 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)


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 velocitytransfer 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

R, (7.67)

θ =vC sin θ

|rBA|. (7.68)

From the geometry, we know that

rCArBA

= tanθ

2⇒ rBA = R cot

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

Rsin θ tan

(θ

2

)k = −2vC

Rsin2

(θ

2

)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 =

[vC +

vC cos θ

RR

]i = vC(1 + cos θ)i . (7.71)


7.7 Homework A uniform rod of mass M and length 2b ispivoted at a point O, a distance s above the center of mass (CM).The rod is struck with a rapid impulsive force perpendicular tothe rod at a point A, a distance a below the center of mass.Themagnitude of the impulse is P = F∆t. Find the value of a suchthat there is no horizontal (N) reaction at the pivot, during theimpact. (The moment of inertia of a uniform rod with mass Mand length L about an axis through its center perpendicular toits longer side is ICM = ML2/12.)

Solution



III. Linear momentum principleApplying linear momentum principle in the x-direction

Px = F extx ⇒Mx = F −N . (7.72)

IV. Angular momentum principleApplying angular momentum principle

HO + vO × P = M extO . (7.73)

Since vO = 0, we get

HO = M extO ⇒ 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

12+ s2

]= M

[b2

3+ s2

]. (7.75)


Now we have two equations and three unknowns x, θ, N . To solve these for N , we needto add a third equation. Considering the geometry and using small-angle approximation1 we can write

sin θ ≈ θ =x

s⇒ x = sθ . (7.76)

Using (7.72), (7.74), (7.75) and (7.76) yields

N = F −Mx = Fb2

3− sa

b2

3+ s2

. (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 whichis approximately true in the limit where the angle approaches zero. They are truncations of the Taylorseries for the basic trigonometric functions to a first-order approximation.


7.8 Homework A rigid block of height H,length L, depth D and mass m rests on a rigidcylinder of mass M and radius R, as shown in thesketch. The cylinder rolls on the floor without slip-ping and the block rolls on the cylinder withoutslipping as well. Determine the kinetic and poten-tial energy of the system.

Solution


v = vi

I. Choose the reference systemSet 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 velocityThe angular velocity of the cylinder

ωcyl = −ϕk . (7.79)

The angular velocity of the block

ωblock = −θk . (7.80)


IV. Velocity of the block and the cylinderSince there is no slip between the cylinder and the floor

vO = Rϕi . (7.81)

The velocity of the center of mass is

vC = Rϕi+ xCOi+ yCOj = (Rϕ+ 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 = Rϕi+ (−ϕk)× (R sin θi+R cos θj) (7.84)

vB,cyl = Rϕ(1 + cos θ)i−Rϕ 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 = (Rϕ+ xCO)i+ yCOj + (−θk)× [(R sin θ− xCO)i+ (R cos θ− yCO)j] (7.87)

vB,block = (Rϕ+ xCO +Rθ cos θ − θyCO)i+ (yCO −Rθ sin θ + θxCO)j (7.88)

V. Position of the center of massSubstituting (7.85) and (7.88) into (7.83) and equating separately the coefficients of thei-components and the coefficients of the j-components, yields

Rϕ+ xCO +Rθ cos θ − θyCO = Rϕ(1 + cos θ) , (7.89)

yCO −Rθ sin θ + θxCO = −Rϕ 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

dt(xCO sin θ + yCO cos θ) = 0 . (7.92)

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 + H2

initial conditions into (7.93) yields

xCO sin θ + yCO cos θ = R +H

2. (7.94)


Multiplying (7.89) by cos θ and (7.90) by sin θ and subtracting them yields

R(ϕ− 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

dt(R(ϕ− θ)) =

d

dt(xCO cos θ − yCO sin θ) , (7.96)

Integrating (7.96) with respect to time yields

R(ϕ− θ) + C1 = xCO cos θ − yCO sin θ + C2 . (7.97)

Substituting θ = 0, ϕ = 0, xCO = 0 and yCO = R+ H2

initial conditions into (7.97) yieldsC1 = C2, so

R(ϕ− 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(ϕ− 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(ϕ− θ) sin θ . (7.100)

VI. Energy of the systemThe potential energy of the system is

V = mgyCO +mgyO = mgyCO , (7.101)

V = mg

([R +

H

2

]cos θ −R(ϕ− θ) sin θ

). (7.102)

The kinetic energy of the system is

T =1

2Mv2O +

1

2Icylω

2cyl +

1

2Mv2C +

1

2Iblockω

2block , (7.103)

T =1

2M(Rϕ)2+

1

2

(1

2MR2

)ϕ2+

1

2M [(Rϕ+ xCO)2+ y2CO]+

1

2

(1

12m(H2 + L2)

)θ2 .

(7.104)

7 Kinematics and kinetics of planar rigid bodies II

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