Download PHY131 Physics Exam Ch 7

PHY131
Physics Exam Ch 7-10
Name
A 20 gram bead slides w/o friction down a ramp into a
circular loop-the-loop. The bead is released from a height
h = 5R. What is the normal force provided by the track at
the top, bottom, and side of the loop?
mghtop + ½ mvtop2 = mghlooptop + ½ mvlooptop2
mg5R + 0
= mg2R
+ ½ mvlooptop2
g3R
=
+ ½ vlooptop2
vlooptop2
= 6gR
11/11
2/2
2/2
( vloopside2 = 8gR ;
vloopbottom2 = 10gR )
FN + mg = m v2 / r
FN
= m6gR / r - mg
FN
= 5mg = 1N
Top
10/10
FN = m v2 /r
FN = m8gR / r
FN = 8mg = 1.6N
Side
4/4
FN – mg = m v2 / r
FN
= m10gR/ r + mg
FN
= 11mg = 2.2 N
Bottom 4/4
A pulley of radius 0.200 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about
the axis is 0.0600 kg·m2. A massless cord wrapped around the wheel is attached to a 5.00 kg block that slides on a
horizontal frictionless surface. If a horizontal force of magnitude P = 16.0 N is applied to the block with a 1.00 kg
block hanging down from the other side of the pulley, what is the angular
acceleration of the wheel?
τ
=
I
α
2
2
[16 - (m1kgg)]r = ([.06 + 5(.2 ) + 1(.2 )] α
6/5
6/6
2/2
2
6/6
6/6
6/6
1/1
α = 4.00 rad/s
A 10.0 gram bullet moving directly upward at 2000 m/s strikes and passes
through the center of mass of a 8.00 kg block initially at rest. The bullet
emerges from the block moving directly upward at 400 m/s. To what
maximum height does the block then rise above its initial position if the
block rests inside a box full of compressible green ooze? (The force applied
by the green ooze is given as F = a + b(y), where a = 20 N and b = 400 N/m.)
SI units
m v = M V + m vf
.01(2000) = 8(V) + .01(400)
Total energy before block rises = Total energy of block at top of path
M g h + ½ M v2 + F dy
0 + ½ 8 22 + 0
= M g hf + ½ M vf2 + F dy
= 8(10) hf + 0 + 20dy + 400ydy
V = 2.0 m/s
16
yf = 0.1275 meters
= 80 yf
5/5
5/5
5/5
4/4
4/4
10/10
+
+ 20yf + 200yf2
Note: hf = yf
PHY131
Physics Exam Ch 7-10
Name
A disk rotates about its central axis with an angular velocity of 2t2 - 7. At the time the disk is
rotating at 11 rad/sec (a) what is its angular position; (b) the angular acceleration (rad/s2).
(c) What is the angular distance travelled during the time the angular velocity has increased
from 11 to 43 rad/s.
θ = 2/3 t3 – 7t
α = d ω / dt
ω = 2t2 – 7
θ = ω dt
8/8
ω = 2t2 – 7
θ = 2/3 53 – 7(5)
α = 4t
11 = 2t2 – 7
θ = 2/3 t3 – 7t
43 = 2t2 – 7
θ = 48.7 radians
2
3
Δθ = 48.7 – -3
t = 3 sec
θ = 2/3 3 – 7(3) α = 12 rad/s
t = 5 sec
8/8
3/3
3/3
Δθ = 51.7 rad 8/8
θ = -3 radians
A force F = (3x2 i + 4y3 j - 2z k) Newtons
acts on an object as the object moves in the x
dir from the origin to x = 5.00 m. Find the
work done on the object.
Work = F dx (ignore y and z components)
Work = 3∫x2 dx 18/18 integrate
Work = x3 from 0 to 5 m
8/8 dx
Work = 125 Joules
In the figure, two 8.00 kg blocks are connected by a massless string over
a frictionless pulley of radius 10.0 cm and rotational inertia, I. When this
system is released from rest, the blocks drop 1.00 meters before
impacting with a spring. The spring (k = 50 N/cm) compressed by 0.100
meters. What is the inertia of the pulley?
vave = Δy / Δt
12/12
½ (0+2.5) = 1 / t
½ k
x2 = ½ mv2
t = 0.800 sec
½ 5000(.1)2 = ½ 8v2
6/6
8/8
3/3
4/4
4/4
3/3
2
v = 2.50 m/s
d=½at
τ =
I
α
2
2
1 = ½ a (.8)2
(m8kgg)r = ([ I +8(.1 )+8(.1 )] 3.125/.1
a = 3.125 m/s2
I = 0.096 kg m2
4/4