When the weight has fallen a distance D and stopped, the law of conservation of total mechanical energy as applied to the particle-weight pair as a system leads to.. −Mgh =E ' − Mg (h+[r]
Trang 1Solution- Theoretical Question 1
A Swing with a Falling Weight
Part A
(a) Since the length of the string L=s+Rθ is constant, its rate of change must be zero Hence we have
˙s + R ˙θ=0 (A1)*1
(b) Relative to O, Q moves on a circle of radius R with angular velocity θ˙ , so
⃗
v Q=R ˙θ ^t=− ˙s ^t (A2)* (c) Refer to Fig A1 Relative to Q, the displacement of P in a time interval t is
Δ ⃗r '
=(sΔθ)(− ^r)+(Δs)^t=[(s ˙θ)(− ^r )+˙s ^t] Δt It follows
⃗
v '=− s ˙θ ^r+ ˙s ^t (A3)*
(d) The velocity of the particle relative to O is the sum of the two relative velocities
given in Eqs (A2) and (A3) so that
⃗
v =⃗v '
+⃗v Q=(− s ˙θ ^r+ ˙s ^t )+R ˙θ ^t=− s ˙θ ^r (A4)* (e) Refer to Fig A2 The ( − ^t )-component of the velocity change Δ ⃗v is given
by (− ^t )⋅ Δ ⃗v=vΔθ=v ˙θ Δt Therefore, the t^ -component of the
acceleration ⃗a=Δ ⃗v / Δt is given by t^⋅ ^a=− v ˙θ Since the speed v of the
particle is s ˙θ according to Eq (A4), we see that the t^ -component of the
particle’s acceleration at P is given by
⃗
a ⋅ ^t=− v ˙θ=−(s ˙θ) ˙θ=− s ˙θ2
(A5)*
Note that, from Fig A2, the radial component of the acceleration may also be obtained as ⃗a ⋅ ^r=−dv /dt=−d s ˙θ¿/dt ¿
1 An equation marked with an asterisk contains answer to the problem.
tˆ
Q
rˆ
v ⃗
P
v
Figure A2
O
v ⃗
v ⃗ v ⃗
v v
tˆ
Q
rˆ
s
s
P
Figure A1
r ⃗
Trang 2(f) Refer to Fig A3 The gravitational potential energy of the particle is given by
U=− mgh It may be expressed in terms of s and as
U (θ)=− mg[ R(1 −cosθ)+s sin θ] (A6)*
(g) At the lowest point of its trajectory, the particle’s gravitational potential energy
U must assume its minimum value U m If the particle’s mechanical energy E were equal to U m, its kinetic energy would be zero The particle would then remain stationary and be in the static equilibrium state shown in Fig A4 Thus,
the potential energy reaches its minimum value when = /2 or s = L− R /2.
From Fig A4 or Eq (A6), the minimum potential energy is then
U m=U (π
2)=− mg[R+L −(πR/2)] (A7)
Initially, the total mechanical energy E is 0 Since E is conserved, the speed v m
of the particle at the lowest point of its trajectory must satisfy
E=0=1
2mvm
2 +U m (A8)
From Eqs (A7) and (A8), we obtain
v m=√− 2U m/m=√2 g [R+(L − πR/2)] (A9)* Part B
(h) From Eq (A6), the total mechanical energy of the particle may be written as
R
x
O s
A Q
s sin
R−R cos
P
R cos
Figure A3 h
R
x
O s
A
Q
P (at rest)
Trang 32mv
2 +U (θ)=1
2mv
2
− mg[R (1− cos θ)+s sin θ] (B1) From Eq (A4), the speed v is equal to s ˙θ Therefore, Eq (B1) implies
s ˙θ¿2=2 g [R (1− cos θ)+ssin θ]
v2=¿ (B2)
Let T be the tension in the string Then, as Fig B1 shows, the t^ -component of
the net force on the particle is –T + mg sin From Eq (A5), the tangential
acceleration of the particle is (− s ˙θ2
) Thus, by Newton’s second law, we have
m(− s ˙θ2)=−T +mg sin θ (B3)
According to the last two equations, the tension may be expressed as
T =m(s ˙θ2
+g sin θ)=mg
s [2 R(1− cos θ)+3 s sin θ]
2 mgR
s [tan
θ
2−
3
2(θ−
L
R)](sin θ)
2 mgR
s (y1− y2)(sin θ)
(B4)
The functions y1=tan(θ/2) and θ − L/ R y ¿/2
2=3¿ are plotted in Fig B2
From Eq (B4) and Fig B2, we obtain the result shown in Table B1 The angle at
R
x
O s
A Q
P mg
Figure B1
T
mg sin
Figure B2
y 30
20
y1=tanθ
2
y1=tanθ
2
10
θ
θ s 0
-10
y2=3
2(θ −
L
R) -20
-30
5 / 2
/ 2
Trang 4which y2 = y1 is called θ s ( π <θ s<2 π ) and is given by
3
2(θs − L
R)=tan
θ s
2 (B5)
or, equivalently, by
L
R=θ s −2
3tan
θ s
2 (B6)
Since the ratio L/R is known to be given by
L
R=
9 π
8 +
2
3cot
π
16=(π +
π
8)−
2
3tan
1
2(π +
π
8) (B7) one can readily see from the last two equations that θ s=9 π /8 .
Table B1 shows that the tension T must be positive (or the string must be taut and straight) in the angular range 0< <s Once reaches s, the tension
T becomes zero and the part of the string not in contact with the rod will not be straight afterwards The shortest possible value smin for the length s of the line segment QP therefore occurs at θ=θ s and is given by
smin=L− Rθ s=R( 9 π
8 +
2
3cot
π
16 −
9 π
8 )=
2 R
3 cot
π
16=3 352 R (B8) When θ=θ s , we have T = 0 and Eqs (B2) and (B3) then leads to
v2=− gs sin θ Hence the speed v s is
v s=√−gsminsin θ s=√2 gR3 cot
π
16 sin
π
8=√4 gR3 cos
π
16
(i) When θ ≥θ s , the particle moves like a projectile under gravity As shown in Fig B3, it is projected with an initial speed v s from the position P=(x s , y s)
in a direction making an angle φ=(3 π /2 −θ s) with the y-axis.
The speed v H of the particle at the highest point of its parabolic trajectory
is equal to the y-component of its initial velocity when projected Thus,
v H=v s sin (θ s − π )=√4 gR
π
16 sin
π
The horizontal distance H traveled by the particle from point P to the point of
Table B1 (y1− y2) sin θ tension T
π <θ<θ s negative negative positive
θ s<θ <2 π positive negative negative
Trang 5maximum height is
H= v s
2
sin 2(θ s − π )
v s2
2 gsin
9 π
4 =0 4535 R (B11)
The coordinates of the particle when θ=θ s are given by
x s=R cos θ s − sminsin θ s=− R cosπ
8+sminsinπ
8=0 358 R (B12)
y s=R sin θ s+smincos θ s=− R sin π
8− smincosπ
8=− 3 478 R (B13) Evidently, we have ¿y s∨¿(R+H ) Therefore the particle can indeed reach its maximum height without striking the surface of the rod
Part C
(j) Assume the weight is initially lower than O by h as shown in Fig C1.
When the weight has fallen a distance D and stopped, the law of conservation of
total mechanical energy as applied to the particle-weight pair as a system leads to
−Mgh =E ' − Mg (h+D) (C1) where E is the total mechanical energy of the particle when the weight has stopped It follows
E '=MgD (C2) Let be the total length of the string Then, its value at = 0 must be the same
A
h
M
m
L
R
Figure C1
x
O
m
R
L
s
min
s
x
O
s
s
v
H
v
s
Figure B3
y
Q
) , ( xs ys
P
H
Trang 6as at any other angular displacement Thus we must have
Λ=L+ π
2 R+h=s+R(θ +
π
2)+(h+D) (C3)
Noting that D = L and introducing ℓ = L−D, we may write
ℓ=L − D=(1− α) L (C4)
From the last two equations, we obtain
s=L − D − Rθ=ℓ − Rθ (C5)
After the weight has stopped, the total mechanical energy of the particle must be conserved According to Eq (C2), we now have, instead of Eq (B1), the following equation:
E '=MgD=1
2mv
2
−mg [R(1 − cos θ)+s sin θ] (C6)
The square of the particle’s speed is accordingly given by
s ˙θ¿2=2 MgD
m +2 gR[(1 − cos θ)+
s
R sinθ ]
v2=¿
(C7)
Since Eq (B3) stills applies, the tension T of the string is given by
−T +mg sin θ=m(− s ˙θ2) (C8) From the last two equations, it follows
T =m(s ˙θ2+g sin θ)
mg
s [
2 M
m D+2 R(1 −cos θ)+3 s sin θ]
2 mgR
s [
MD
mR +(1− cos θ)+
3
2(
ℓ
R −θ)sin θ]
(C9)
where Eq (C5) has been used to obtain the last equality
We now introduce the function
f (θ)=1− cos θ+3
2(
ℓ
R −θ)sin θ (C10) From the fact ℓ = (L−D) >> R, we may write
f (θ)≈ 1+3
2
ℓ
R sin θ −cos θ=1+ A sin(θ − φ) (C11)
where we have introduced
3 2
ℓ
R¿
2
1+¿
A=√¿
,
3 ℓ
2 R¿
2
¿
1+¿
√¿
φ=tan − 1
3 ℓ
2 R
¿
(C12)
From Eq (C11), the minimum value of f() is seen to be given by
Trang 73 2
ℓ
R¿
2
1+¿
fmin=1 − A=1 −√¿
(C13)
Since the tension T remains nonnegative as the particle swings around the rod,
we have from Eq (C9) the inequality
3 ℓ
2 R¿
2
¿
1+¿
MD
mR +fmin=M (L − ℓ)
(C14)
or
3 ℓ
2 R¿
2
¿
1+¿
(ML
mR)+1≥ (
Mℓ
mR )+√¿
(C15)
From Eq (C4), Eq (C15) may be written as
(ML
mR)+1≥ [(
ML
mR )+(
3 L
2 R)](1 − α) (C16) Neglecting terms of the order (R/L) or higher, the last inequality leads to
α ≥ 1−
(ML
mR)+1 (ML
mR)+(
3 L
2 R)
=
(3 L
2 R)−1
(ML
mR)+(
3 L
2 R)
=
1− 2 R
3 L
2 M
3 m+1
1+2 M
3 m
(C17)
The critical value for the ratio D/L is therefore
α c= 1 (1+2 M
3 m)
(C18)*