quiz 1 fall 2012

2.003 Fall 2012: Dynamics and Control I Quiz 1Massachusetts Institute of Technology Thursday 9/27/2012Department of Mechanical Engineering 9:30-11:00 am Rm. 10-250

Problem 1 (30 points)A box of negligible size is sliding down along a curved path defined by the parabola y = 0.4x2.

When it is at A (xA = 2m, yA = 1.6m), the speed is v = 8m/s and the increase in speed is

dv/dt = 4m/s2.

Figure 1

At this instant, determine the:

(a) two components of a in your choice of coordinate system, and

(b) magnitude of a.

1


Problem 1 - SolutionAs we saw in lecture on September 11, 2012; we will choose the normal-tangent coordinate

system. This is similar to Homework 2 Problem 1.

aA = vAut +v2Aρun

The speed increase is given as dvA/dt = 4m/s2, and then the next step is to findv2Aρ

Since the magnitude of the velocity is given as vA = 8m/s, and the only thing left is to find

the radius of curvature, which is, again from the lecture

ρ =

[1 +

(dydx

)2]3/2∣∣∣ d2ydx2

∣∣∣We take the first and second derivatives of y.

dy

dx=

(ut)y(ut)x

=d

dx(0.4x2) = 0.8x

d2y

dx2=

d

dx(0.8x) = 0.8

At our point of interest A, (x = 2, y = 1.6), we obtain:

dy

dx

∣∣∣∣x=2

= 0.8× 2 = 1.6

d2y

dx2

∣∣∣∣x=2

= 0.8

leading to

ρ|x=2 =[1 + 1.62]

3/2

|0.8|≈ 8.39 [m]

2


Finally we can write the acceleration and its magnitude as:

(a)

aA = vAut +v2Aρun = 4ut +

82

8.39un ≈ 4ut + 7.62un

(b)

|aA| =√

42 + 7.622 ≈ 8.61[m/s2

]

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Problem 2 (40 points)A 20 kg sphere of radius r = 0.15 m rolls down the inclined plane shown in Figure 2 without

slipping. The moment of inertia for a sphere of uniform density about its center of mass is

IG = 25mr2.

g

θ

G

Figure 2

Determine the sphere’s:

(a) free body diagram

(b) equation of motion in terms of its rotation, θ

(c) angular acceleration, α, and

(d) center of mass’s linear acceleration, a.

4


Problem 2 - Solution(a) The first thing to establish, in this problem, is the choice of coordinate system. To

simplify calculations, we will choose x as being aligned with the slope, and y in the

normal direction. Figure 3 shows these axes and the free body diagram.

x

G

θ

f

mg

y

OA

N

f

30

r

Figure 3

(b) The equation of motion is obtained from the kinematic constraint and the sum of the

forces and toques. The relationship resulting from the non-slip condition with the plane

enforces the fact that a rotation of the sphere through an angle θ assumed positive in

the counterclockwise direction (in agreement with our choice of axes) corresponds to

a translation of x = −rθ. Stated differently, xG = θ × rAG. This in turn implies

xG = θ × rAG and xG = θ × rAG.

Summing the forces we obtain

∑F x : f

f−mg sin 30◦i = mxG∑

F y : N −mg cos 30◦ = 0

the first of which reduces to

ff −mg

2= mxG

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And summing the torques about the center of mass

∑τG : rGA × f f = Iθ

which we can simplify

rff k = Iθ

rff = Iθ

Combining these two equations into one by substituting for the frictional force, ff,

yields

ff =Iθ

r= mxG +

mg

2

Along with our kinematic relationship, we have two equations and two unknowns. The

geometric constraint yielded

xG = θ × rAGxGi = θk × rjxGi = −θrixG = −θr

And through substitution, we have our equation of motion:

Iθ

r= −mθr +

mg

2I +mr2

rθ =

mg

2

θ =mrg

2(I +mr2)

θ =mrg

2(25mr2 +mr2

)

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(c) Since θ = α, and x = a,

θ =mrg

2(25mr2 +mr2

)α =

g

2(25

+ 1)r

α =5g

14r

α ≈ 23.4k[rad/s2

](d) And by virtue of aC = α× rAC

aC = −5g

14i ≈ −3.5i

[m/s2

]

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Problem 3 (30 points)The 0.5 kg collar C can slide freely along the smooth rod AB. At a given instant, rod AB

is rotating with an angular velocity of θ = 2 rad/s and has an angular acceleration of θ = 2

rad/s2. Neglect the mass of the rod AB and the size of the collar C. Ignore the effect of

gravity.

Figure 4

At this instant, determine:

(a) the acceleration of the collar C

(b) the normal force of rod AB on the collar C

(c) the radial reaction of the end plate B on the collar C.

8


Problem 3 - Solution

A

BC

i

jur

uθ

L

θ

θ C

uruθ

(a) (b)

Nr

Nθ

Figure 5: (a) kinematic diagram. (b) free-body diagram.

(a) To obtain the acceleration of the collar C we choose to write the position vector rAC in

polar coordinates:

rAC = Lur,

from which we can obtain the velocity and acceleration vectors, vC and aC respectively,

via differentiation:

vC = rAC = Ldurdt

= Lθuθ

aC = Lθuθ + Lθduθdt

= Lθuθ − Lθ2ur

Substituting L = 0.6 m, θ = 2 rad/s and θ = 2 rad/s2 yields:

aC = −2.4ur + 1.2uθ m/s2

(b) The normal force of rod AB on the collar C, Nθ acts in the tangential direction and is

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calculated by summing the forces in uθ:∑Fθ = maθ = Nθ ⇒ Nθ = (0.5 kg)(1.2 m/s2)

Nθ = 0.6 N

(c) The radial reaction of the end plate B on the collar C is similarly obtained by balancing

the forces in ur:∑Fr = mar = −Nr ⇒ Nr = −mar = −(0.5 kg)(−2.4 m/s2)

Nr = 1.2 N

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quiz 1 fall 2012

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