solutions problem set 5 newton’s laws with circular motion; torque...
TRANSCRIPT
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SOLUTIONS PROBLEM SET 5 NEWTON’S LAWS WITH CIRCULAR MOTION; TORQUE
#1
the radius of rotation is 2 21.25 1 0.75R m= − =
1cos 0.81.25
θ = = and 0.75sin 0.61.25
θ = =
T1
T2mg
R
y
xT1
T2
T2cosq
T2sinq
q
T1cosqT1sinq
mg
y
x
Here 1.00cos 0.81.25
θ = =
a) total vertical force should be zero
1
1 2 2cos 60(0.8) 4 9.81cos cos 0 11cos 0.8y
T mgF T T mg T Nθθ θθ− − ⋅
Σ = − − = → = = =
b) the magnitude of the centripetal force:
2
1 2sin sinx cvF T T ma mR
θ θΣ = + = =
1 2( sin sin ) 0.75(60(0.6) 11(0.6)) 2.83 /4
R T Tv m smθ θ+ +
= = =
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#2
m1 m2
R1R2
m1 m2
R1
R2
Top view
a) for m1 :
21
1 1 2 1 1 11
x cvF T T m a mR
Σ = − = = (1)
1 1 1 0y NF F m gΣ = − = (2) for m2 :
22
2 2 2 2 22
x cvF T m a mR
Σ = = = (3)
2 2 2y NF F m gΣ = = (4) if v2 = 3m/s then from (3):
2 22
2 22
33 181.5
vT m NR
= = =
b) 1 1v Rω= and 2 2v Rω= therefore: 1 1 2 112 2 2
3 1 2 /1.5
v R v Rv m sv R R
⋅= → = = =
from (1) 2 2
1 11 2
1
2 2 18 261
m vT T NR
⋅= + = + =
y
x
m1
FN1
m1g
T1
T2
a1
y
x
m2
FN2
m2g
T2
a2
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#3
a) If the speed is low, the car tends to slip down. The vertical force should be zero.
FNFNcosq
FNsinq
mgq
ffs
ffscosqffssinqar q
q
FN
ffs
mg
y
x
cos sin 0y N fsF F f mgθ θΣ = + − = (1) 2
sin cosx N fs cvF F f ma mR
θ θΣ = − = = (2)
the larger the static friction force (ffs), the smaller the speed v. For maximum static friction, v is minimum.
From (2) 2minsin cosN s NvF F mR
θ µ θ− = (3)
From (1) cos sinN s NF F mgθ µ θ+ = (4)
(3)/(4): 2 2min min
minsin cos sin10 0.1cos10 8.58 /cos sin cos10 0.1sin10 9.81 100
s
s
v v v m sgR
θ µ θθ µ θ− −
= = = → =+ + ⋅
b) If the speed is high, the car tends to slip up. The vertical force should be zero.
FNFNcosq
FNsinq
mgq ffs
ffscosqffssinq
arq
q
FN
ffs
mg
y
x
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cos sin 0y N fsF F f mgθ θΣ = − − = (1) 2
sin cosx N fs cvF F f ma mR
θ θΣ = + = = (2)
the larger the static friction force (ffs), the larger the speed v. For maximum static friction, the speed is maximum.
From (2) 2maxsin cosN s NvF F mR
θ µ θ+ = (3)
From (1) cos sinN s NF F mgθ µ θ− = (4)
(3)/(4): 2 2min min
minsin cos sin10 0.1cos10 16.5 /cos sin cos10 0.1sin10 9.81 100
s
s
v v v m sgR
θ µ θθ µ θ+ +
= = = → =− − ⋅
the range of speeds the car can have without slipping up or down: 8.58 / 16.5 /m s v m s≤ ≤
#4
2
2
(1)
net r c
N
N
F ma ma
vmg F mr
vF m gr
= =
+ =
⎛ ⎞= −⎜ ⎟
⎝ ⎠
Top of loop:
(force of seat on pilot is the normal force
2
2
(2)
net r c
N
N
F ma ma
vF mg mr
vF m gr
= =
− =
⎛ ⎞= +⎜ ⎟
⎝ ⎠
Bottom of loop:
mg FNar
mg
FNar
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2 (3)
net r c
N
F ma mavF mr
= =
= Side of loop:
Comparing (1), (2) and (3) the normal force is greatest at the bottom of the loop. The least force is at the top of the loop. #5
9545 0.45N m
L cm mτ = ⋅
= =
with r F r Lτ = × =rr r
magnitude of torque is sin 95rFτ θ= = θ is the angle between rr and F
r
95 229.3sin sin 0.45 sin 67
N mF Nr L mτ τθ θ
⋅= = = =
⋅ o
#6
balance torques at the right end of the board and balance forces. Clockwise torque is negative.
mg
FNar
F
67o
L
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support
support
support
00
500 3 280 1.5 1 0 0
500 3 280 1.5 19201
net
girl board end
N m N m F m
F N
τ
τ τ τ τ
=
+ + + =
− ⋅ − ⋅ + ⋅ + =
⋅ + ⋅= =
rr r r r
support
0
0
ˆ1920 280 500 1140
y
end board girl
end
F
F F F F
F Nj
Σ =
+ + + =
= − − = −
r r r r
r
#7
22.7700 700(2 ) / 60 73.3 /I kg mrpm rad sπ
= ⋅
= =
use Iτ α= but what is α ?
2 2
2 2 22
2 2
1( ) ( )2
(73.3 / ) 0 17.11 /2( ) 2(2 (25) 0)
2.7 17.11 / 46.2
o o
o
o
rad s rad s
kg m rad s N m
α θ θ ω ω
ω ωα
θ θ π
τ
− = −
− −= = =
− −
= ⋅ ⋅ = ⋅
#8
m2
T2
ff2
FN2
m2gcosq
m2gsinq
m1T1
FN1
m1g
ff1
m2g
T1T2
R
y
x x +
consider m1:
1 1 1
1 1 1 1
1 1 1 1 1 1
0x f
y N N
F T f m aF F m g F m gT m g m a T m a m gµ µ
Σ = − =
Σ = − = → =
− = → = +
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7
consider m2: 2 2 2 2
2 2 2 2
2 2 2 2 2 2 2 2
sin 0
cos30 0 cos30
cos30 sin 0 sin30 cos30
x f
y N N
F T f m g m aF F m g F m gT m g m g m a T m a m g m gµ µ
Σ = − − − 3 =
Σ = − = → =
− − + 3 = → = − + −
consider pulley:
2 tan1 2
1 2
1 2
( is clockwise, so negative)12
12
12
IaI MRR
T R T R MRa
T T Ma
τ α α
τ τ α
Σ =
⎛ ⎞− = = −⎜ ⎟⎝ ⎠
− = −
− = −
r
substitute expression for T1 and T2
1 1 2 2 2
1 2 2 1 2
2 22 1 2
1 2
2
1( ) ( sin 30 cos30 )2
130 cos302
sin30 cos30 6 9.81 / sin 30 0.360 9.81 / (2 6 cos30)0.5 (2 6 0.5 10)
0.309 /
m a m g m g m g m a Ma
m g m gsin m g M m m a
m g m g m g kg m s m s kg kgaM m m kg
a m s
µ µ
µ µ
µ µ
+ − − − = −
⎛ ⎞− + = − − −⎜ ⎟⎝ ⎠
− − ⋅ ⋅ − ⋅ ⋅ += = =
+ + + + ⋅
= then find T1 and T2 1 1
2 21 1
2 2
2 2 22 2
( )
2 (0.309 / 0.360 9.81 / ) 7.67
( sin30 cos30 )
6 (9.81 / sin30 0.360 9.81 / cos30 0.309 / ) 9.21
T m a g
T kg m s m s N T
T m g g a
T kg m s m s m s N T
µ
µ
= +
= ⋅ + ⋅ = =
= − −
= ⋅ = ⋅ − = =