1 Motion of a single particle - Linear momentum, work and...

1 Motion of a single particle - Linearmomentum, work and energy principle

v0

µx

y m

g

1.1 In-class problem A block of mass m slidesalong a rough horizontal plane with friction coef-ficient µ. The block is given an initial velocity v0at time t0 in the positive x direction.

(a ) Write the principle of linear momentum forthe block.

(b ) At which time instant ts does the block stop?

(c ) Which is the distance traveled by the block from t0 to ts?

(d ) Which is the work done by the friction force?

Solution (a) We set a coordinate system xy as indicated in the figure, identified by theunit vectors i and j, respectively. The principle of linear momentum writes p = F , wherep is the linear momentum and F is the resultant force applied to the block. We needtherefore to determine which forces are acting on the block. For this, we have to draw afree body diagram (FBD):

x

y

F f mg

N

The forces acting to the system are the gravity force mg, the reaction force N and thefriction force F f , which magnitude is |F f | = µ|N |, directed as indicated in the figure. Theprinciple of linear momentum can be then written as

p = mxi+myj = −µNi+ (N −mg)j. (1.1)

Because of the presence of the ground, we have y(t) = 0, ∀t and so y = 0. For eq. (1.1)to be satisfied, it must be

mx=−µN, (1.2)

0 =N −mg. (1.3)

1-1

1 Motion of a single particle - Linear momentum, work and energy principle 1-2

(b) From eq. (1.3) we can immediately know the reaction force: N = mg. The accelerationis then

x(t) = −µg. (1.4)

Note that the acceleration is negative: this agrees with the intuitive fact that the blockwill come to a stop. By integrating eq. (1.4) in time, and using the fact that x(t0) = v0i,we obtain

x(t) = −µgt+ v0. (1.5)

In order to know how much time ts it will take for the block to stop, we simply haveto impose that eq. (1.5) is zero. This gives:

−µgts + v0 = 0→ ts =v0µg

(1.6)

The greater the friction coefficient µ is, the shortest ts becomes1.

(c) The distance travelled by the block can be calculated by integrating eq. (1.5)another time to obtain the position of the block:

x(t) = −1

2µgt2 + v0t+ x0, (1.7)

where the initial position is indicated by x0. Inserting ts in eq. (1.7) we obtain

xs = x(ts) =1

2

v20µg

+ x0, (1.8)

and therefore the total distance travelled is

xs − x0 =1

2

v20µg. (1.9)

Note that the distance travelled by the block is a function of the square of the initialvelocity v0. Suppose that the the block will start with an initial velocity v1 = (1 + a)v0,then the difference ∆x of the two stopping distances relative to v1 and v0 will be

∆x = v20[(1 + a)2 − 1

]= v20a(a+ 2). (1.10)

An increment of the initial velocity of 10% (a = 0.1) will produce an increase of thestopping distance of 21%! This is what speed limits are for...

1It seems from eq. (1.5) that negative velocities, and therefore the mass moving backward, can beobtained for t > ts. This is not true. In fact, the rigorous expression for the friction force is Ff =−sgn(x)µN , where sgn(x) is the sign of the velocity. If x = 0, then Ff = 0 and the block will not moveanymore.


(d) In order to find the work done by the friction force, we simply have to apply thedefinition

W xsx0

=

∫ xs

x0

µN · dr =

∫ xs

x0

−µmgi · dxi = −µmg∫ xs

x0

dx = −µmg1

2

v20µg

= −1

2mv20.

(1.11)

Note that the work done by the friction force is equal to the kinetic energy dissipated.This question could be easily solved by the work-energy principle even without knowingthe expression for the friction force!


v0

µ↵

m

g

x

y

1.2 In-class problem Repeat now problem 1.1,but assume that the block of mass m slides alonga rough inclined plane with friction coefficient µ,where α indicates the inclination angle. The blockis given an initial velocity v0 at time t0 directeddown the incline.

(a ) Write the principle of linear momentum forthe block.

(b ) Is it always possible for the block to stop?

(c ) How much time ts does the block need in order to stop?

(d ) Which is the distance travelled by the block in the time interval ts − t0?

(e ) Is the distance travelled longer than the case relative to the horizontal plane?

(f ) Which is the work done by the forces parallel to the incline? How does it comparewith the work done in the case of the horizontal plane?

Solution (a) It is convenient to align the reference frame xy with the incline, as shownin the figure. As before, in order to write the principle of linear momentum, we first needto determine the forces acting on the block. The FBD looks then as

x

y

F f

N

mg

↵

where the friction force Ff , which magnitude is |Ff | = µ|N |. The principle of linearmomentum can the be written as

p = mxi+myj = (mg sinα− µN)i+ (N −mg cosα)j. (1.12)

(b) Again, the constraint present in the system establishes that y(t) = 0, ∀t and thereforey = 0. This allows us to know the magnitude of the reaction force, which is this timeN = mg cosα, by equating the component j of eq. (1.12) to zero. Substituting this valuein the i component of (1.12), we find the acceleration of the block to be

x(t) = g(sinα− µ cosα). (1.13)


Now, the sign of the acceleration depends on the sign of the term sinα− µ cosα. For theblock to stop, the acceleration must be negative and therefore

sinα− µ cosα < 0→ tanα < µ. (1.14)

We established with eq. (1.14) a relation between the angle of inclination and the frictioncoefficient. The higher the angle α is, the larger must be µ. Note also that, when α = 0,the acceleration is always negative, as found for the previous problem.

(c) The block will stop when its velocity becomes zero. To obtain the velocity, weneed to integrate in time eq. (1.13), which gives

x(t) = g(sinα− µ cosα)t+ v0. (1.15)

We are left now to impose that x(ts) = 0. This yields

x(ts) = g(sinα− µ cosα)ts + v0 = 0→ ts =v0

g(µ cosα− sinα). (1.16)

Note that eq. (1.16) gives a positive value for ts if the stopping condition (1.14) is satisfied.Else, a negative value for ts is physically meaningless.

(d) As before, to calculate the distance travelled by the block we need to computethe position x(t). For this, we integrate eq. (1.15) on more time:

x(t) =

∫ t

0

x(t)dt =1

2g(sinα− µ cosα)t2 + v0t+ x0 (1.17)

where x0 is an arbitrary initial position along the incline. If we insert ts in eq. (1.17) weobtain the total distance xs travelled by the block:

x(ts) = xs =1

2

v20g(µ cosα− sinα)

(1.18)

(e) For an arbitrary inclination angle α ∈ [0; π/2], we can see from eq. (1.18) thatµ cosα − sinα < µ, and therefore, for a given v0, the distance travelled on the inclinedplane is always bigger than on a horizontal plane (see problem 1.1).

(f) In order to calculate the work done by the tangential force F t, we need again toapply the definition and substitute the value of xs found in eq. (1.18):

W xsx0

=

∫ xs

x0

F t·dr = mg(sinα−µ cosα)i·∫ xs

x0

dxi =1

2

mg(sinα− µ cosα)v20g(µ cosα− sinα)

= −1

2mv20.

(1.19)

(f) The work done by the tangential forces is the same as for the horizontal plane! Thisis not surprising: the energy that needs to be dissipated is the kinetic energy of thesystem, which is the same in both cases. The work-energy principle indeed writes thatTts − T0 = W xs

x0, where T0 = 1

2mv20 and Tts = 0.


1.3 In-class problem A block of mass m slidesdown a frictionless incline (see Fig.). The block isreleased at height h above the bottom of the loop,with negligible velocity.

(a ) At what speed vB does the block leave thetrack at point B?

(b ) How far away from point A does the blockland on level ground?

Solution

(a) the direction of the velocity vector vB and the exit point B is known, as it hasto be tangent to the track. We are left to determine the magnitude vB only. It is thenadvisable to check if the system is conservative, and therefore utilize the conservation ofenergy principle. For this, we need to determine whether the forces acting on the blockare conservative or do no work. We have:

• the gravity force F g, of constant magnitude mg, directed in the negative y direction.It is possible then to find a potential V = mgy such that

F g = −∂mgy∂y

j = −mgj; (1.20)

therefore, the gravity force is conservative;

• the other force acting on the particle is the reaction force exerted by the track. Thisforce vector varies in direction and magnitude, which depend on the dynamics ofthe particle. However, it is always oriented normal to the motion and it thereforedoes no work on the block.

We can conclude that the system is conservative, and thus the conservation of energyholds. The interesting locations at which to write the energy of the system are the top ofthe track and the exit point B. The total energy at the top of the track is

Etop = Ttop + Vtop = 0 +mgh. (1.21)

while the total energy at point B:

EB = TB + VB =1

2mv2B +mghB. (1.22)


From the figure, we notice that hB can be written as:

hB = R

(1− 1√

2

). (1.23)

The conservation equation Etot = EB reads

mgh =1

2mv2 +mgR

(1− 1√

2

). (1.24)

Solving equation for v2 gives:

2

[gh− gR

(1− 1√

2

)]= v2, (1.25)

which gives for the magnitude of the exit velocity v

vB =

√2

[gh− gR

(1− 1√

2

)]. (1.26)

(b) We need now to determine the trajectory of the block once it leaves the trackfrom point B with velocity vB = vBxi+ vByj. The direction of the exit velocity is parallelto the track and therefore known. We can, for instance, consider point A as the referencesystem origin.

The only force acting on the block is the gravity force, as shown in the FBD here below:

We can then simply apply the principle of linear momentum

p = mxi+myj = −mgj (1.27)

In components, the principle of linear momentum writes

mx= 0, (1.28)

my=−mg. (1.29)

Eqs. (1.28) and (1.29) are two second order differential equations that require initialconditions, namely xB, yB, vBx vBy in position and velocity to be solved. By integratingtwice with respect to time we get the equation of motion for the projectile:

x = xB + vBxt , (1.30)


y = yB + vByt−1

2gt2 . (1.31)

The initial conditions are known:

xB = R1√2

; yB = R

(1− 1√

2

); vBx = vBy =

vB√2

(1.32)

Substituting the initial conditions (1.32) into (1.30) and (1.31) we obtain:

x =R√

2+vB√

2t , (1.33)

y = yB +vB√

2t− 1

2gt2 . (1.34)

The block reaches the ground at time ts when y(ts) = 0. We can then solve eq. (1.34) forts

gt2s − 2yB −√

2vBts = 0 ,

ts =

√2vB ±

√2v2B + 8gyB

2g. (1.35)

We discard the negative root since it gives a negative time and substitute into eq. (1.33):

xs =R√

2+vB√

2

(√2vB +

√2v2B + 8gyB

2g

). (1.36)

Now inserting the previous expressions for vB and hB yields

xs = (√

2− 1)R + h+

√h2 − 3

2R2 +

√2R2 . (1.37)


1.4 Homework A skier of mass m = 65 kgstarts from rest at A. Assuming that the friction,air resistance and the size of the skier can beneglected:

(a ) Determine his speed at point B as a functi-on of hAB.

(b ) Determine the distance l and the time twhen he lands in C.

(c ) Calculate the distance l and the time t when he lands in C, if α = 30◦, hAB = 15 mand hBD = 4.5 m .

Solution

I. Choose the reference systemConsider the reference system as shown in the figure of the assignment.

II. Draw the free-body diagramFree-body diagram at point B:

III. Apply the appropriate principle(s)(a) Since all forces are either potential (G = mg) or do no work (N),the system is conservative, thus, we can use conservation of energy principle(vA = 0 and yB = y0).

1

2mvA

2 +mg(yA − y0) =1

2mvB

2 +mg(yB − y0)⇒ vB =√

2g(yA − yB) =√

2ghAB

(1.38)

(b) Kinematics.By considering the motion in the x direction:

xB(t) = xB(t0) + vBxt⇒ xB(t) =√

2ghAB cos(α)t , (1.39)

and from the geometry

∆x = lx = l cosα ,


thus,

l cosα =√

2ghABt cosα⇒ l =√

2ghABt . (1.40)

By considering the motion in the y direction

yB(t) = yB(t0) + vByt+1

2aByt

2 =√

2ghABt sinα + 0.5(−g)t2 , (1.41)

and from the geometry

|∆y| = hBD + ly = hBD +√

2ghABt sinα . (1.42)

Equating the geometrical quantity to the kinematic one (i.e. the vertical position of theskier with respect the chosen coordinate frame) we get

yB(t) = −|∆y| , (1.43)

thus,

0 = hBD + 2√

2ghABt sinα− 0.5gt2 . (1.44)

Therefore the solution for t is:

t =−2 sinα

√2ghAB −

√(2 sinα

√2ghAB)2 + 2hBDg

−g. (1.45)

(c) Applying (1.45) and (1.40):t = 3.743 s and l = 64.2 m .


1.5 Homework A car of mass m = 150 kg is atta-ched to a rod of negligible mass and of length l = 4m, which pivots around one of its tips. A momentL = 30t2 Nm is applied to the rod and the engineof the car supplies a time dependent traction force ofF = 15t N to the wheels, where t is in seconds. Ne-glect the contribution of friction and assume that thecar is at rest a time t = 0.Determine the velocity of the car at t = 5 s.

Solution

I. Choose the reference systemConsider the reference system as shown in the assignment.

II. Draw the free-body diagramFree body diagram

III. Apply the appropriate principle(s)At the pivot point, there is a reaction moment MS which lies on the xy plane, andtherefore has no component about the z axis, and a reaction force F S. Furthermore, theline of action of mg and N are parallel to the z axis, and thus these forces produce notorques about the z axis. Since all the mentioned forces and moments produce no momentabout the pivot point O, it is wise to write the principle of angular momentum w.r.t. tothis point:

Ho + vo × P =∑

M o . (1.46)

Integrating both sides respect to time from t0 = 0 to tf = 5 and considering vo = 0

Ho|tft0 =

∑ tf∫t0

M odt⇒ Ho(tf ) =∑ tf∫

t0

M odt+Ho(t0) . (1.47)

Since the car starts from rest Ho(t0) = 0, we get

r × (mv(tf )) =

tf∫t0

Lodt+

tf∫t0

r × Fdt . (1.48)


Since r is always perpendicular to v and F , we get

m|r||v(tf )| =tf∫

t0

(30t2 Nm)dt+

tf∫t0

|r|(15tN)dt . (1.49)

Performing the integrations and substituting the remaining values, we find the speed atthe final time tf = 5:

|v(tf )| = 3.33 m/s . (1.50)


1.6 Homework The simple pendulum of mass m andlength l is released from rest at θ = 0◦.(a) Using only the principle of angular momentum, deter-mine the expression for θ in terms of θ and the velocity vof the pendulum at θ = 90◦.(b) Compare this approach with a solution by the work-energy principle. (Hint:

∫αdα =

∫αdα.2)

Solution

I. Choose the reference system(a) Put the origin of the reference system as shown on the free body diagram.

II. Draw the free-body diagramFree body diagram:

III. Apply the appropriate principle(s)Applying angular momentum principle:

Ho =∑

M o + vo × P . (1.51)

Since vo = 0, we get

d

dt(r × P ) = r × F ⇒ d

dt(|r||mlθ| sin 90◦) = |r|m|g| sin(90◦ − θ). (1.52)

d

dt(ml2θ)) = lm|g| cos θ. (1.53)

From (2.26)

θ =|g|l

cos θ. (1.54)

In order to get the velocity we integrate both sides with respect to θ∫θdθ =

∫ |g|l

cos θdθ. (1.55)

2∫αdα =

∫d2αdt2 dα =

∫ddtdαdt dα =

∫dαdt d

dαdt =

∫αdα


Using the following integration identity∫θdθ =

∫θdθ , (1.56)

we get

θ2

2=|g|l

sin θ ⇒ θ =

√2|g|l

sin θ. (1.57)

So the magnitude of the velocity at θ = 90◦

|v| = lθ =√

2|g|l . (1.58)

(b) Since all forces are either potential (m|g|) or do no work (N), the system is conserva-tive, thus, we can use conservation of energy principle to compute |v|.

T0◦ + V0◦ = T90◦ + V90◦ ⇒ 0 =1

2m|v|2 −m|g|l . (1.59)

From 2.32 we get

|v| =√

2|g|l. (1.60)


1.7 Exam-type problem A particle of mass m isattached to point O by an inextensible string of lengthl. Initially the string is slack when m is moving tothe left with a speed v0 in the position shown. Cal-culate the speed of m just after the string becomestaut.

(1) v = v0(2) v = d2

l2v0

(3) v = dlv0

(4) v = clv0

Solution

I. Choose the reference systemThe reference system is already given as it is shown above.

III. Apply the appropriate principle(s)Average force in the string over the small time interval ∆t required to change the directionof the particle’s motion. Initially the force exerted on m by the slack string is zero. Whenthe string becomes taut, the force exerted on m passes through O. Therefore, the momentof the net force on m about O remains zero and the angular momentum is conservedabout point O.

Ho|t0 = Ho|tf (1.61)

Initially, the position and velocity of the particle are

r0 = ci+ dj , (1.62)

v0 = −v0i . (1.63)

The angular momentum with respect to point O at the initial state is

Ho|t0 = r0 ×mv0 = mv0dk . (1.64)

Just after the string becomes taut the position and the velocity of the particle are

rf = −√l2 − d2i+ dj , (1.65)

vf = vxi+ vyj . (1.66)

and the angular momentum is

Ho|tf = rf ×mvf = −mvxd−mvy√l2 − d2k . (1.67)

Substituting (1.64) and (1.67) into (1.61), yields

mv0d = −mvxd−mvy√l2 − d2 . (1.68)


The string is inextensible, so the component of the velocity of m along the string mustbe zero, thus vf · rf = 0. Substituting vf and rf from the previous equations and solvingfor vy we get

vy = vx

√d

l− 1 . (1.69)

Solving (1.68) and (1.69) for vx and vy leads to

vx = −d2

l2v0 , (1.70)

vy = −√

1− d2

l2d

lv0 . (1.71)

Thus, the speed, v =√v2x + v2y after the string becomes taut is

v =d

lv0 . (1.72)

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