Download Physics 111 * Test #2 - University of St. Thomas

Physics 111
Test #5 Formula Sheet
UNITS CONVERSIONS
1 mile = 5280 feet = 1.609 km;
1 slug = 14.59 kg;
1 m = 3.281 feet = 39.37 inches;
1 inch = 2.54 cm;
1 year = 365.2425 days;
1 gallon = 3.788 liters = 3788 cm3;
1 ft/s = 0.3048 m/s;
1 pound force (lb) = 4.448 N = 1/14 stones (0.0714 stones);
1 ft·lb = 1.356 J;
1 hp = 746 W;
TRIGONOMETRY FORMULAE
c
sin θ =
cos θ =
tan θ =
c2 = a2
b
θ
a
b/c;
a/c;
sinθ/cosθ = b/a;
+ b2;
“FUNDAMENTAL” CONSTANTS
g = 9.787 m/s2 = 32.11 ft/s2;
G = 6.673 X 10-11 N·m2/kg2;
mE = mass of Earth = 5.97 X 1024 kg;
rE = radius of Earth = 6.38 X 106 m;
MISCELLANEOUS FORMULAE
circumference of circle = 2πr;
If
area of circle = πr2;
surface area of sphere = 4πr2;
volume of sphere = (4/3) πr3;
, then
.
360o = 2π radians;
UNIT 1: Kinematics
UNIT 2: Newton’s Laws
Different forces:
UNIT 3: Work and Energy
net work done = ∆𝐾𝐸
or
energy at start = energy at end
;
UNIT 4: Momentum
UNIT 5: Torque/Rotation
Linear Motion
x
displacement
velocity
acceleration
constant linear
acceleration
equations
v
dx
dt
x  x0  v0 t  12 at 2
x  x0  vt  12 at 2
angular
velocity
angular
acceleration
constant
angular
acceleration
equations


s
r
d vtan 
ds 

 vtan  
dt
r 
dt 

d d 2 atan
 2 
dt
r
dt
   0  0 t  12  t 2
   0   t  12  t 2
v  v0  at
   0  12 (  0 )t
  0   t
v 2  v02  2ax
 2  02  2
x  x0  12 (v  v0 )t
m
force
F
p  mv
momentum
Force Impulse
(2nd Law in
integral form)
angular
displacement
dv d 2 x

dt dt 2
a
mass
Newton’s 2nd
law
Rotational Motion
Fnet 
ext
F
net
ext
dpnet
 ma
dt
dt  pnet
moment of
inertia
(“rotational
mass
distribution”)
n
I   mi ri 2 or
i 1
  r F
torque
angular
momentum
Newton’s 2nd
law –
rotational form
Torque
Impulse
(2nd Law in
integral form)
L  r  p  I
dLnet
 I
dt
 net 
ext

net
ext
dt  Lnet
linear kinetic
energy
Klin  12 mv 2
rotational
kinetic energy
Krot  12 I  2
work done by
a force
WF   F  dr
work done by
a torque
W    d
power
delivered by
force
P
WF
 F v
t
power
delivered by
torque
 r dm
P
W
  
t
2
Physics 111 – Test #5
1 July 2015
Name _____________________________
Directions: This is a CLOSED BOOK test, and you may only use the calculator and cheat
sheet provided. SHOW ALL OF YOUR REASONING! You can only get partial credit
for an incorrect answer if you show your reasoning. You may have as much time as you
like to finish this exam.
1. (5 pts) Is it possible for a spinning thing to have ω = 0 but have α ≠ 0? If not, prove it. If so,
give an example.
2. A wheel is rotating about an axis perpendicular to the plane of the wheel and passing through
the center of the wheel. The angular speed of the wheel is increasing at a constant rate. Point
A is on the rim of the wheel and point B is midway between the rim and center of the wheel
(see diagram). For each of the following quantities, is its magnitude larger at point A, at point
B, or is it the same at both points? Explain.
a) (5 pts) angular speed
b) (5 pts) tangential speed
A
B
3. At t = 5.00 s a point on the rim of a 0.300-m-radius wheel has a tangential speed of
150.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude
45.0 m/s2.
a) (12 pts) Through what angle did the wheel turn between t = 0 and t = 3.00 s?
b) (10 pts) At what time will the radial acceleration of this point on the rim equal g?
45.0 m/s2
0.300 m
150.0 m/s
4. A gate 4.00 m wide and 2.00 m high weighs 500 N. Its centers of gravity and mass are
at its center, and it is hinged at A and B. To relieve the strain on the top hinge, a wire
CD is connected as shown in the figure. The tension in CD is increased until the
horizontal force at hinge A is zero. Determine
a) (8 points) the tension in the wire CD, and
b) (8 points) the horizontal component of the force at hinge B.
C
30o
D
A
2.0 m
B
4.0 m
5. (18 points) A bowling ball (mass mb = 5 kg, radius rb = 0.11 m) is attached to a
lightweight brace so that the ball rolls down the incline on a fixed axis without slipping.
The brace is attached to a rope, which goes over a pulley (mass mp = 1 kg, radius rp =
0.11 m) and is attached to a heavy block (mass M = 10 kg) which sits on a frictional
(µk = 0.40) surface. Use ENERGY considerations to determine the speed of the
bowling ball after it has rolled from rest 2 m down the incline. Careful! There are
several energy terms in your energy equation!
pulley
mp = 1 kg
r = 0.11 m
M = 10 kg
µk = 0.40
bowling ball
mb = 5 kg
r = 0.11 m
2m
θ = 60o
6. (18 pts) An Atwood’s machine is set up as shown in the diagram below. The 50 kg
pulley (R = 0.1m) is fixed in place but is free to rotate about an axis through its center.
It rotates WITH a frictional torque of 4 Nm. Use TORQUES AND FORCES to
determine the tension in BOTH sides of the rope (that’s two different tensions, T1 and
T2). (If you ignore the frictional torque, the best you can get is 16/18.)
M = 50 kg
T1
m1 = 10 kg
T2
m2 = 5 kg
7. Two solid spheres each with mass 3.00 kg move in the same direction. The radius
of each sphere is 10 cm, and they are in deep space, far from any sources of gravity.
The top one catches up to the bottom one, and they make a grazing collision, with
the bottom of the upper one sticking to the top of the lower one, as shown.
a) (6 points) Determine the velocity of the two stuck-together spheres after the
collision (the center of mass velocity).
b) (10 points) Determine the angular velocity () of the two stuck-together
spheres after the collision.
m
10 m/s
m
m
m
4 m/s

m
m
vCM
BONUS BRAIN BUSTER!!!
A lightweight rod (negligible I) connects two identical point masses, m, forming a
dumbbell. It is located in deep space far from any sources of gravity or any other forces.
This spins freely in place (the center of mass of the rod/two-mass system isn’t moving)
with an angular speed ω about an axis perpendicular to the rod and through its center of
mass. Determine an expression for the angular momentum of this dumbbell about the point
indicated in terms of m, d, and ω. If you think about it, the answer might surprise you.
ω
Find
of dumbbell
about this point.
d
2d
Credit will only be given for correct reasoning. If you are able to guess the correct answer but
give no correct reasoning, you will receive no extra credit. However, if you give some correct
reasoning that would (or does) lead to the correct answer, at least some credit will be given.
________________________________________________
By signing my name above, I affirm that this test represents my work only, without aid
from outside sources. In all aspects of this course I perform with honor and integrity.