Download Version 001 – Summer Review #5 Circular Motion, Gravity, Energy

Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20
s
This print-out should have 12 questions.
gR
10. v =
correct
Multiple-choice questions may continue on
µ
the next column or page – find all choices
Explanation:
before answering.
The maximum frictional force due to friction
is fmax = µ N , where N is the inward
Barrel of Fun 04
directed normal force of the wall of the cylin001 (part 1 of 2) 10.0 points
der on the person. To support the person
An amusement park ride consists of a large
vertically, this maximal friction force fsmax
vertical cylinder that spins about its axis fast
must be larger than the force of gravity m g
enough that any person inside is held up
so that the actual force, which is less than
against the wall when the floor drops away
µN , can take on the value m g in the positive
(see figure). The coefficient of static friction
vertical direction. Now, the normal force supbetween the person and the wall is µ and the
v2
radius of the cylinder is R.
plies the centripetal acceleration
on the
R
person, so from Newton’s second law,
R
m v2
N =
.
R
Since
µ m v2
≥ mg,
R
the minimum speed required to keep the person supported is at the limit of this inequality,
which is
fsmax = µ N =
ω
What is the minimum tangential velocity
needed to keep the person from slipping downward?
p
1. v = g R
p
2. v = µ g R
p
3. v = 2 g R
p
4. v = 2 µ g R
p
5. v = µ 2 π g R
1p
gR
µ
1p
7. v =
gR
2
p
8. v = 2 µ g R
p
9. v = µ 2 g R
6. v =
2
µ m vmin
= mg
R
1/2
gR
vmin =
.
µ
002 (part 2 of 2) 10.0 points
Suppose a person whose mass is m is being
held up against the wall with a constant tangential velocity v greater than the minimum
necessary.
Find the magnitude of the frictional force
between the person and the wall.
µ m v2
− mg
R
m v2
2. F =
− µmg
R
µ m v2
3. F = m g +
R
m v2
4. F = µ m g +
R
1. F =
5. F = m g correct
Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20
m v2
6. F =
µR
mg
7. F =
µ
8. F = µ m g
m v2
R
µ m v2
10. F =
R
Explanation:
The vertical friction force must equal m g in
order to balance the force of gravity and not
have any acceleration in the vertical direction.
9. F =
Conceptual 05 11
003 (part 1 of 2) 10.0 points
Calculate the period of a ball tied to a string
of length 1.3 m making 4.4 revolutions every
second.
An air puck of mass 0.031 kg is tied to a string
and allowed to revolve in a circle of radius 1.5
m on a frictionless horizontal surface. The
other end of the string passes through a hole
in the center of the surface, and a mass of
1.4 kg is tied to it, as shown. The suspended
mass remains in equilibrium while the puck
revolves on the surface.
0.031 kg
1.5 m
1.4 kg
What is the magnitude of the force that
maintains circular motion acting on the puck?
The acceleration due to gravity is 9.81 m/s2 .
Correct answer: 13.734 N.
Explanation:
Correct answer: 0.227273 s.
Explanation:
Let :
T =
Let :
f = 4.4 rev/s ,
R = 1.3 m .
and
g = 9.81 m/s2 ,
mp = 0.031 kg ,
m = 1.4 kg , and
r = 1.5 m .
1
1
=
= 0.227273 s .
f
4.4 rev/s
m
r
004 (part 2 of 2) 10.0 points
Calculate the speed of the ball.
Correct answer: 35.9398 m/s.
mp
Explanation:
D
2πR
v=
=
T
T
2 π (1.3 m)
=
0.227273 s
= 35.9398 m/s .
Holt SF 07Rev 52
005 (part 1 of 2) 10.0 points
Fc = Fg = m g = (1.4 kg) (9.81 m/s2 )
= 13.734 N .
006 (part 2 of 2) 10.0 points
What is the linear speed of the puck?
Correct answer: 25.7788 m/s.
Explanation:
Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20
v2
Fc = mp t
s
s r
Fc r
(13.734 N) (1.5 m)
=
vt =
mp
(0.031 kg)
Correct answer: 2778.3 J.
Explanation:
The kinetic energy of the bullet is
1
1
mb vb2 = (56 g) (315 m/s)2
2
2
= 2778.3 J .
Kb =
= 25.7788 m/s .
Firing a Gun
007 (part 1 of 4) 10.0 points
A 30 kg gun is standing on a frictionless surface. The gun fires a 56 g bullet with a muzzle
velocity of 315 m/s.
The positive direction is that of the bullet.
Calculate the momentum of the bullet immediately after the gun was fired.
010 (part 4 of 4) 10.0 points
Calculate the kinetic energy of the gun immediately after the gun was fired.
Correct answer: 17.64 kg · m/s.
The recoil velocity of the gun comes from its
momentum:
pg = mg vg
pg
vg =
mg
Explanation:
Let : mb = 56 g and
vb = 315 m/s .
The momentum of the bullet is
pb = mb vb = (56 g) (315 m/s)
= 17.64 kg · m/s .
008 (part 2 of 4) 10.0 points
Calculate the momentum of the gun immediately after the gun was fired.
Correct answer: 5.18616 J.
Explanation:
Let : mg = 30 kg .
Thus the kinetic energy of the gun is
p2g
1
(−17.64 kg · m/s)2
2
Kg = mg vg =
=
2
2 mg
2 (30 kg)
= 5.18616 J .
Drag Racer 02
011 (part 1 of 2) 10.0 points
A 901 kg drag race car accelerates from rest
to 113 km/h in 1.03 s. What change in momentum does the force produce?
Correct answer: −17.64 kg · m/s.
Explanation:
By conservation of momentum,
pbo + pgo = pbf + pgf
0 + 0 = pb + pg
pg = −pb = −17.64 kg · m/s .
009 (part 3 of 4) 10.0 points
Calculate the kinetic energy of the bullet immediately after the gun was fired.
Correct answer: 28281.4 kg · m/s.
Explanation:
Let :
m = 901 kg and
v = 113 km/h = 31.3889 m/s .
Momentum is defined as
∆p = m ∆v
= (901 kg) (31.3889 m/s − 0 m/s)
= 28281.4 kg · m/s .
Version 001 – Summer Review #5 Circular Motion, Gravity, Energy and Momentum – tubman – (IBII20
012 (part 2 of 2) 10.0 points
What average force is exerted on the car?
Correct answer: 27457.7 N.
Explanation:
Let :
t = 1.03 s .
Momentum and impulse are related by
∆p = F t ,
F =
so
∆p
28281.4 kg · m/s
=
= 27457.7 N .
t
1.03 s