Download lecture notes on Newton`s laws`s applications

PHYS1110H, 2011 Fall.
Shijie Zhong
2. Applications of Newton’s Laws - continued.
.
Contact force: Tension of a string, and friction
Example 5. Block and String
A block of a mass M on a friction-free surface is attached to a string with
mass m. A force F is applied to the string to pull the block. What’s force that
the string is applied to the block?
Applying Newton’s 2nd law to both block and string:
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(2.14)
F1 = Ma M ,
(2.15)
F − F1 = ma m ,
where aM and am are the acceleration for the block and string, respectively,
and are equal to each other . Combining these two equations, we obtain:
a M = a m = F /(M + m) , (2.16)
M
F1 =
F.
(2.17)
(M + m)
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If m is zero or the mass of the string is ignored, F1=F. Although we often
assume this in solving mechanics problems, as we see here, it is only an
assumption or approximation to the reality.
Example 6, Tension on a whirling rope.
A uniform rope of mass M and length L is pivoted at one end and whirls
with uniform angular velocity ω. Assuming that gravity can be ignored,
what is the tension in the rope at distance r from the pivot?
Consider a small segment of the rope between r and r+Δr. The segment has a
length of Δr, and a mass of m=MΔr/L. Now let’s treat this segment as a
particle and consider the forces and Newton’s law on this segment. The
inward tensional force from the rope to the segment is T(r), and the outward
force to the segment is T(r+Δr). This segment has an acceleration rω2
pointing towards the center or pivot. Suppose that the positive r direction is
outward, then the acceleration is negative, and the NT’s 2nd law leads to:
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(2.18)
T(r + Δr) − T(r) = −mrω 2 ,
M
T(r + Δr) − T(r) = − Δr ⋅ rω 2 ,
(2.19)
L
T(r + Δr) − T(r)
Mrω 2
=−
.
(2.20)
Δr
L
Let Δr approach to zero, and use the definition of derivative, (2.20) becomes
dT
Mrω 2
=−
,
(2.21)
dr
L
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which is a first order ODE that can be solved for how T depends on position
r of the rope. Re-arrange (2.21) and integrate it.
Mω 2 2
T =−
r + C,
(2.22)
2L
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where C is an integration constant and can be determined by considering the
constraint that tensional force T is zero at the tip of the rope, that is, T=0 at
r=L, which leads to
Mω 2
C=
L,
2
Mω 2 2 2
T=
(L − r ) .
(2.23)
2L
(2.23) describes how tensional force varies with r with the maximum tension
at the pivot r=0, and zero tension at r=L.
Example 7. Force on a pulley from a string.
A string with constant tension T is deflected through angle 2θ0 by a smooth
fixed pulley. What’s the force on the pulley?
Due to the symmetry, the y component of the force on the pulley is zero, and
we do not need to consider it. Consider a small segment of the deflected
string that starts at θ and ends at θ+Δθ (see the figure). There are two
tensional forces pulling on the segment from each end, but they are not
parallel. There is also a force pushing the segment outward from the pulley
ΔF. At the equilibrium, these three forces cancel. In the tangential direction,
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the two tensional forces cancel each other. In the radial direction, or the
direction of force ΔF, the force balance is (i.e., when acceleration is zero):
ΔF − 2T sin(Δθ /2) = 0 ,
(2.24)
For small x, sin(x) ≈ x . Therefore, sin(Δθ /2) = Δθ /2 , because Δθ/2 is small.
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ΔF = 2TΔθ /2 = TΔθ .
(2.25)
€ x component of this €
The
force ΔF on the pulley is
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ΔFx = ΔF cos θ = TΔθ cos θ .
(2.26)
The x component of the total force on the pulley is the sum or integral of
ΔFx for all the small segments or for θ to vary from -θ0 to θ0.
θ0
Fx = ∫ ΔFx = ∫ T cos θdθ = 2T sin θ 0 .
(2.27)
−θ 0
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We do not need to compute y component of the force on the pulley, because
they all cancel.
Friction.
Friction arises when the surface of one body moves or tries to move along
the surface of another body.
For many surfaces, the maximum friction is equal to µsN, where µs is
coefficient of static friction, and N is the normal force against the surface.
When there is a relative motion between two bodies with a contact surface,
then the friction force has a magnitude of µkN and a direction opposite to the
motion, where µk is the coefficient of kinetic friction. For bodies not in
relative motion, the frictional force is between 0 and µsN. For example,
when a force F is applied to a body but is not sufficient to overcome the
maximum frictional force µsN (i.e., F< µsN), then the frictional force is
equal to F in magnitude and is opposite in direction (see the figure).
Static friction coefficient µs is larger than kinetic friction coefficient µk.
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Example 8. The spinning terror.
A large vertical cylinder has a coefficient of friction µ at its inner wall, and
has a radius of R. How fast does the cylinder must rotate to hold an object to
the wall with no support at its bottom?
Suppose that the angular velocity is ω, the normal force to an object with
mass m from the wall is N = mRω 2 , and points to the center axis of the
cylinder. The maximum frictional force on the object is f=µN, and the
direction of the frictional force is upwards, opposite to the direction of the
gravitational force. To hold the object to the wall and to prevent the object
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from sliding down the wall, the frictional force f must be equal to the weight
of the object, mg.
f = µN = µmRω 2 = mg ,
or
µRω 2 = g .
Therefore, the minimum angular velocity must satisfy:
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ω 2 ≥ g /(µR) .
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(2.28)
The larger the radius R, or the larger the coefficient of friction, the slower
the cylinder can rotate to hold the object onto the wall.
Example 9. Air drag.

v0 , suppose that there is an air
For a cannon ball firing at an
initial
velocity



f
drag force that is given as = −mkv , where v is velocity vector, m is the
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mass of the cannon ball, and k is a constant. What’s velocity as a function of
time during the flight?
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

Fnet = −mgeˆ y − mkv = −(mg + mkv y )eˆ y − mkv x eˆ x ,
(2.29)
dv y
dv
−kv x eˆ x − (mg + mkv y )eˆ y = m( x eˆ x +
eˆ y ) ,
(2.30)
dt
dt
(2.30) is the NT’s 2nd law. Treat each component separately and we have
dv
−kmv x = m x ,
(2.31)
dt
dv y
−mg − mkv y = m
,
(2.32)
dt
We have solved these two equations (2.31) and (2.32) in our homework set
2, Q3. The solutions are given as follows..
g
g
(2.33)
v x = v x 0 e−kt , v y = (v y 0 + )e−kt − .
k
k
The air drag causes the horizontal velocity of the ball to decrease with time,
and also a terminal velocity for vertical component, -g/k.
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Note
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