Version 001 – Circular Motion – tubman – (1818185)
This print-out should have 13 questions.
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before answering.
Barrel of Fun 01
001 10.0 points
As viewed by a bystander, a rider in a
“barrel of fun” at a carnival finds herself stuck
with her back to the wall.
ω
Which diagram correctly shows the forces
acting on her?
1. None of the other choices
1
6.
Explanation:
The normal force exerted by the wall on
the rider provides the centripetal acceleration
necessary to keep her going around in a circle.
The downward force of gravity is equal and
opposite to the upward frictional force on her.
(Since this problem states that it is viewed
by a bystander, we assume that the free-body
diagrams are in an inertial frame.)
Centripetal Acceleration
002 10.0 points
A car rounds a curve while maintaining a
constant speed.
Is there a net force on the car as it rounds
the curve?
1. No – its speed is constant.
2. Yes. correct
2.
3. It depends on the sharpness of the curve
and speed of the car.
Explanation:
Acceleration is a change in the speed and/or
direction of an object. Thus, because its
direction has changed, the car has accelerated
and a force must have been exerted on it.
3.
4.
5.
correct
Serway CP 07 19
003 (part 1 of 2) 10.0 points
A 46.6 kg ice skater is moving at 4.97 m/s
when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then
moves in a circle of radius 0.862 m around the
pole.
Find the force exerted by the rope on her
arms. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 1.33534 kN.
Explanation:
Version 001 – Circular Motion – tubman – (1818185)
2
Correct answer: 7.056 N.
Let :
m = 46.6 kg ,
v = 4.97 m/s ,
r = 0.862 m .
and
Explanation:
Let :
The force exerted by the rope is the centripetal force keeping her moving in a circle,
so
m
m v2
(46.6 kg) (4.97 m/s)2 1 kN
T =
=
·
r
0.862 m
1000 N
= 1.33534 kN .
004 (part 2 of 2) 10.0 points
Find the ratio of this tension to her weight.
Correct answer: 2.92401.
Explanation:
M = 0.72 kg and
g = 9.8 m/s2 .
r
v
Mg
Since the suspended mass is in equilibrium,
the tension is
T = M g = (0.72 kg) 9.8 m/s2
= 7.056 N .
m v2
v2
(4.97 m/s)2
T
= r =
=
W
mg
rg
(0.862 m) (9.8 m/s2 )
= 2.92401 .
006 (part 2 of 3) 10.0 points
What is the horizontal force acting on the
puck?
Correct answer: 7.056 N.
Serway CP 07 25
005 (part 1 of 3) 10.0 points
An air puck of mass 0.382 kg is tied to a string
and allowed to revolve in a circle of radius
0.43 m on a horizontal, frictionless table. The
other end of the string passes through a hole
in the center of the table and a mass of 0.72 kg
is tied to it. The suspended mass remains in
equilibrium while the puck revolves.
0.382 kg
0.43 m
v
Explanation:
The horizontal force acting on the puck is
the tension in the string, so
Fc = T = 7.056 N .
007 (part 3 of 3) 10.0 points
What is the speed of the puck?
Correct answer: 2.81827 m/s.
Explanation:
m v2
r
s
r
Fc r
(7.056 N) (0.43 m)
=
v=
m
0.382 kg
Fc =
0.72 kg
What is the tension in the string? The
acceleration due to gravity is 9.8 m/s2 .
= 2.81827 m/s .
Version 001 – Circular Motion – tubman – (1818185)
Serway CP 07 20
008 10.0 points
A sample of blood is placed in a centrifuge
of radius 14 cm. The mass of a red blood
cell is 3.1 × 10−16 kg, and the magnitude of
the force acting on it as it settles out of the
plasma is 2.1 × 10−11 N.
At how many revolutions per second should
the centrifuge be operated?
Explanation:
Fc = m ac = m g − Wa
Wa = m g − m ac = m(g − ac )
= (43.9 kg) 9.8 m/s2 − 0.034 m/s2
= 428.727 N .
Correct answer: 430.22 N.
Explanation:
At the poles, the rotating velocity is zero,
so
r = 14 cm = 0.14 m ,
m = 3.1 × 10−16 kg , and
Fc = 2.1 × 10−11 N .
Fc = Wa − m g = 0
The centripetal force gives us the angular
velocity:
v2
Fc = m
= m r ω2
r
r
Fc
ω=
sm r
=
Since the centripetal acceleration of a person is inward (toward the Earth’s axis),
010 (part 2 of 2) 10.0 points
What is his apparent weight at the poles?
Correct answer: 110.71 rev/s.
Let :
3
2.1 × 10−11 N
(3.1 × 10−16 kg)(0.14 m)
1 rev
2 π rad
Wa = m g = (43.9 kg) 9.8 m/s2 = 430.22 N .
Car on a Banked Curve 01
011 10.0 points
A curve of radius 51.2 m is banked so that
a car of mass 1.6 Mg traveling with uniform
speed 53 km/hr can round the curve without
relying on friction to keep it from slipping on
the surface.
= 110.71 rev/s .
Serway CP 07 49
009 (part 1 of 2) 10.0 points
Because of Earth’s rotation about its axis,
a point on the Equator experiences a centripetal acceleration of 0.034 m/s2 , while a
point at the poles experiences no centripetal
acceleration.
What is the apparent weight at the equator
of a person with a mass of 43.9 kg? The
acceleration of gravity is 9.8 m/s2 .
Your answer must be within ± 0.1%
Correct answer: 428.727 N.
Explanation:
Let :
1. 6
µ≈
and
0
θ
At what angle is the curve banked? The
acceleration due to gravity is 9.8 m/s2 .
Correct answer: 23.3628 deg.
Explanation:
Let :
ac = 0.034 m/s2
m = 43.9 kg .
Mg
m = 1.6 Mg = 1600 kg ,
v = 53 km/hr ,
r = 51.2 m , and
g = 9.8 m/s2 .
Version 001 – Circular Motion – tubman – (1818185)
4
d
1000 m 1 hr
v = (53 km/hr)
1 km 3600 s
= 14.7222 m/s .
l
θ
Consider the free body diagram for the car.
In the absence of friction the forces acting on
the car are the normal force and the force due
to gravity:
N
θ
N cos θ
y
What is the speed of each seat?
x
N sin θ
Correct answer: 3.47052 m/s.
Explanation:
In the vertical direction we have
mg
To keep an object moving in a circle requires a force directed toward the center of
the circle; its magnitude is
Fc = m ac = m
Fx = N sin θ = m
i
X
where T is the tension in the chain. In the
horizontal direction we have
v2
.
r
T sin θ =
Using the free-body diagram,
X
T cos θ = m g ,
v2
r
m v2
.
r
Since
and
Fy = N cos θ = m g .
i
Dividing,
v2
(14.7222 m/s)2
tan θ =
=
gr
(9.8 m/s2 ) (51.2 m)
= 0.431967
θ = arctan(0.431967) = 23.3628◦ .
Amusement Park Ride
012 (part 1 of 2) 10.0 points
An amusement park ride consists of a rotating
circular platform 10.2 m in diameter from
which 10 kg seats are suspended at the end
of 2.52 m massless chains. When the system
rotates, the chains make an angle of 12.3◦ with
the vertical.
The acceleration of gravity is 9.8 m/s2 .
r = ℓ sin θ +
d
2
= (2.52 m) sin 12.3◦ +
10.2 m
2
= 5.63684 m ,
we have
p
v = g r tan θ
q
= (9.8 m/s2 ) (5.63684 m) tan 12.3◦
= 3.47052 m/s .
013 (part 2 of 2) 10.0 points
If a child of mass 37.4 kg sits in a seat, what is
the tension in the chain (for the same angle)?
Correct answer: 475.433 N.
Explanation:
M = 37.4 kg
Version 001 – Circular Motion – tubman – (1818185)
From the first part we have
T cos θ = (m + M ) g
(m + M ) g
T =
cos θ
(10 kg + 37.4 kg) (9.8 m/s2 )
=
cos 12.3◦
= 475.433 N .
5