PHY 131
Exam Ch 6 - 9
A cat dozes on a prank cat bed which is at the edge of a stationary
merry-go-round, at a radius of 6.0 m from the center of the ride. The
prank cat bed has a Teflon coated bottom with a coefficient of friction
of 0.04. This merry-go-around has a slight tilt of 2 degrees slanting
toward the center. The operator turns on the ride and brings it up to
its proper turning rate of one complete rotation every T seconds. What
is the minimum value of T that the merry-go-round that will allow the
cat to stay in place, without sliding off the edge?
mgsinθ + µ
11/11
= m v2 / r
FN
11/11
Easy
(max points is 25%)
A cat dozes on a stationary merry-go-round, at a radius of
6.0 m from the center of the ride. The operator turns on
the ride and brings it up to its proper turning rate of one
complete rotation every 7.0 s. What is the least
coefficient of static friction between the cat and the
merry-go-round that will allow the cat to stay in place,
without sliding?
= m v2 / r
µ FN
11/11
mgsinθ + µ mgcosθ = m 4π2r/T2
gsin2 + 0.04gcos2 = 4π2r/T2
T = 23.6 sec
13/13
12/12
µ mg
µ g
µ
= m 4π2r / T2
= 4π26 / 72
= 0.49
(Note, I neglected the component that
increases FN, which is negligible.)
A spring in the vertical position is stretched by 0.1 cm when a 3 kg
package is hung from the spring. Later this spring is compressed 10 cm by
a 3.00 kg package and then is launched and slides across a frictionless
floor. Shortly after launching the package explodes into three parts.
Assume the point of explosion is at the origin of a coordinate system.
Part 1 has mass of 0.500 kg and velocity 18.0 i + 14 j.
Part 2 has mass of 0.750 kg and a speed of 14.0 m/s, and travels at an
angle 30.0° (CCW from the positive direction of the x axis). (a) What is
the velocity of part 3? State your direction as a positive (CCW) angle
from the +x axis.
Workspring = Δ½m vf2
½ k
x2 = ½ m vf2
30000 .12 = 3 vf2
vf = 10 m/s
5/5
F=kx
mg = kx
5/5
30 = k (.001)
k = 30,000 N/m
x plane on floor
MV = m1 v1 + m2 v2
+ m3 v3x
3(10) = ½(18) + ¾(cos30)14 + 1.75 v3x
6/6
7/7
v3x = 6.80 m/s
y plane on floor
0 = m1 v1 + m2 v2
+ m3 v3y
0 = ½(14) + ¾(sin30)14 + 1.75 v3y
v3y = -7.00 m/s
7/7
Or for a max of 30%
A spring gun shoots a 5 gram dart at a cork block. To
load the spring gun (with a spring (un)constant of 300x2
N/m3) you must compress the spring by X cm. After the
dart passed through the 100 gram block, the block is
traveling 10 m/s and the dart is traveling 50 m/s. How
far did you compress the spring? Hint: (Fspring = -k x)
Block collision
1/1
v3 = √(v3x2 + v3y2)
v3 = √(6.82 + -72)
v3 = 9.8 m/s
tanθ = vy / vx
tanθ = -7 / 6.8
θ = -46⁰
3/3
mv + MV = mvf + M Vf
5v + 0 = 5(50) + 100(10)
v = 250 m/s
16/16
dWspring
= ΔKE
F dx = m v dv
k x dx = ΔKE
300x2 x dx = ΔKE
8/8
6/6
300∫x3dx = ΔKE
75 x4
= ½ m vf2
75 x4 = ½ 0.005 2502
x = 1.20 meters
To push a 26.0 kg crate up an incline with a
coef of friction of 0.100, angled at 30.0° to
the horizontal, a worker exerts a force of
300 N along the horizontal plane. As the
crate slides 2.00 m, how much work is done
on the crate by (a) the worker's applied
force, (b) the gravitational force on the
crate, and (c) the normal force exerted by
the incline on the crate? (d) What is the
total work done on the crate? (e) What is
the speed of the crate at 2.00 m?
(a) WorkWorker = cos30(300)(2m)
WorkWorker = 520 Nm 5/5
Or for maximum of 30%
A coated spring with a spring
(un)constant of 150x2 N/m3 is
compressed by 8 cm. If this spring
launches a 50 gram block up a 30
degree incline with a coef of
friction of 0.4, how far does this
block travel up the incline?
Hint: (Fspring = -k x)
Reduced by Friction
Workfric = µ
1/1
(b) WorkGravity = mg
h
WorkGravity = -260 sin30(2m) 5/5
WorkGravity = -260 Nm (or down)
(c) 0 Joules
2/2
5/5
FN
dist
6/6
1/1
Workfric = 0.1 (cos30*260+sin30*300)*2 m
Workfric = 75 Nm (opposing Force)
(d) & (e) Work 3/3
520 – 260 – 75
185
v = 3.8 m/s
= Δ½ m v2
= ½ (26) v2
= 13 v2
Ff
dist
µ m g cosθ L
.4(.05)g cos30 L
0.1732L
dWspring - Ff dist
k x dx – 0.1732L
10/10
5/5
12/12
ΔE
mg Δh
mg sin30 L
0.25 L
= ΔE
= 0.25 L
8/8
150x2 x dx – 0.1732L = 0.25 L
150∫x3dx – 0.1732L = 0.25 L
37.5 (.08)4 – 0.1732L = 0.25 L
L = 0.00363 meters