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Phy2048 Spring 2008
Review Exam 2
Q20: Two disks are mounted on low-friction bearings on a common shaft. The first disc
has rotational inertia I and is spinning with angular velocity ω. The second disc has
rotational inertia 2I and is spinning in the same direction as the first disc with angular
velocity 2ω as shown. The two disks are slowly forced toward each other along the
shaft until they couple and have a final common angular velocity of:
a
b
We don’t know the details of how the disks get stuck
together. So all we know is initial and final conditions.
This sounds like a job for .... conservation of angular
momentum
Initially:
1) Li,tot = La + Lb
La = +Iiωi î, Lb = +(2Ii) (2ωi) î
Li,tot = +5 Iiωi î
y
z
x
2) Lf,tot = Li,tot = +5 Iiωi î
3) Lf,tot = Ifωf î
If = Ia + Ib = Ii + 2Ii = 3Ii
(3 Ii)ωf î = +5 Iiωi î
ωf î = +5/3 ωi
Q19: A 2.0 kg block travels around a 0.50 m radius circle with an angular velocity of
12 rad/s. The magnitude of its angular momentum about the center of the circle is:
v
Angular momentum: L⃗ = ⃗
rx⃗
p = r (mv) sinθ
y
1) L = r (mv) sin(90°)
= (0.5 m)(2 kg· v) (1)
r
z
x
2) v = ωr = (12 rad/s)(0.5 m) = 6 m/s
3) L = (0.5 m)(2 kg· 6 m/s) (1) = 6 kg m2/s
Q18: A sphere and cylinder of equal mass and radius are simultaneously released from
the same height, from rest, on the same inclined plane, rolling down the incline without
slipping. On an adjacent incline of the same angle a cube of the same mass is
simultaneously released from rest, from the same height, to slide without friction down
the incline. Then:
Sphere or Cylinder:
ω
y
x
1) Fx,net = +mg sinθ - fs = maCM
2) τCM,net = - Rfs = ICM α
3) aCM = -αR
FN
fs
aCM
Fg
θ
y
4)
5)
6)
7)
-Rfs = ICM (-aCM/R)
fs = ICM aCM/R2
mg sinθ - ICM aCM/R2 = maCM
aCM = + g sinθ / (1 + ICM/MR2)
8) ICM,Sphere = 2/5 MR2, ICM,Cylinder = 1/2 MR2
9) aCM,Sphere = +g sinθ / (1 + (2/5 MR2)/MR2)
= +g sinθ / (1 + 2/5) = -5/7g sinθ
FN
x
aCM
Fg
10)aCM, Cylinder = +g sinθ / (1 + (1/2 MR2)/MR2)
= +g sinθ / (1 + 1/2) = -2/3 g sinθ
Block:
1) Fx,net = +mg sinθ - 0 = maCM.Block
2) aCM,Block = +g sinθ
θ
1) aCM,Block > aCM,Sphere > aCM,Cylinder
Q17: A thin-walled hollow tube rolls without sliding along the floor. The ratio of its
translational kinetic energy to its rotational kinetic energy (about an axis through its
center of mass) is:
ω
For rolling without slipping we have: vCM = ωr
vCM
Translational kinetic energy:
1) KTr = ½ M vCM2
2) KRot = ½ IHoop ω2
IHoop = M r2
ω = vCM/r
KRot = ½ (Mr2) (vCM/r)2 = ½ M vCM2
3) KTr / KRot = 1.0
Note: we did not need to know the mass (M), the radius
(r), or even the linear or angular velocity of the hoop to
solve this problem
Q16: A rod is pivoted about its center. A 5N force is applied 4 m from the pivot and
another 5N force is applied 2 m from the pivot, as shown. The magnitude of the total
torque about the pivot (in N· m) is:
b
a
y
z
x
1) τtot = τa + τb
2) τa = ra x Fa = + Fa ra sin(30°) k̂ = (5N)(4m)(0.5) k̂
3) τb = rb x Fb = + Fb rb sin(30°) k̂ = +(5N)(2m) (0.5) k̂
4) τtot = (10 + 5) k̂ N· m = 15 k̂ N· m
Q15: The rotational inertia of a solid uniform sphere about a diameter is (2/5)MR2,
where M is its mass and R is its radius. If the sphere is pivoted about an axis that is
tangent to its surface, its rotational inertia is:
We know the ICenter of Mass and we want to fins the
rotational inertia around another parallel axis. This
sounds like a job for .... “The parallel axis theorem”
ICenter = 2/5 MR2
Initially:
1) Inew = ICM + Mh2
h=R
2) Inew = (2/5)MR2 + MR2 = 7/5 MR2
IRim = ?
Q13: The angular speed of the minute hand of a watch is:
ω = Δθ/Δt = 1 rotation / 1 hr = 2π rad / 3600 s = π/1800 rad / s
Q14: A flywheel is initially rotating at 20 rad/s and has a constant angular acceleration.
After 9.0 s it has rotated through 450 rad. Its angular acceleration is:
θf = θi + ωit + ½ αt2
α = 2[(θf - θi) - ωit] / t2
α = 2[(450 rad) - (20 rad/s)(9 s)] / (9s)2 = 6.67 rad/s2
Q11: A 0.2 kg rubber ball is dropped from the window of a building. It strikes the
sidewalk below at 30 m/s and rebounds up at 20 m/s. The magnitude of the impulse due
to the collision with the sidewalk is:
Impulse: J = Δp = pf - pi = mvf - mvi = (0.2 kg)[20 m/s ĵ - (-30 m/s ĵ)]
= (0.2 kg)[50 m/s ĵ] = 10 ĵ kg· m/s ≡ 10 ĵ N· s
Q10: Two 4.0-kg blocks are tied together with a compressed spring between them.
They are thrown from the ground with an initial velocity of 35 m/s, 45° above the
horizontal. At the highest point of the trajectory they become untied and
spring apart. About how far below the highest point is the center of mass of the twoblock system 2.0 s later, before either fragment has hit the ground?
The trajectory of the center of mass follows the parabola of
a free falling object. We do not know the spring constant and
the direction of how the blocks are tied together or the
details of how they become untied. It does not matter.
y
1) y(t) = y0 + v0yt - ½gt2
vy(t) = v0y - gt
x
Let’s start the problem at the maximum of the trajectory:
2) v0y = 0
3) y(t) = ymax + v0yt - ½gt2
θ
y(2s) = - ½g(2s)2 = 19.6 m
y(t-tmax) = - ½gt2
Q10: Two 4.0-kg blocks are tied together with a compressed spring between them.
They are thrown from the ground with an initial velocity of 35 m/s, 45° above the
horizontal. At the highest point of the trajectory they become untied and
spring apart. About how far below the highest point is the center of mass of the twoblock system 2.0 s later, before either fragment has hit the ground?
The trajectory of the center of mass follows the parabola of a free falling
object. .... The long solution
1) y(t) = y0 + v0yt - ½gt2
vy(t) = v0y - gt
2) tmax = v0y/g
y
x
θ
3) y(tmax+2s) = v0y(v0y/g+2s) - ½g (v0y/g+2s) 2
= (v0sinθ)2/g + (v0sinθ)(2s) - ½g(v0sinθ/g + 2s)2
= (v0sinθ)2/g + (v0sinθ)(2s) - ½g[(v0sinθ/g)2 + (2s)2 + 2(v0sinθ/g)(2s)]
= (v0sinθ)2/g + (v0sinθ)(2s) - ½(v0sinθ)2/g - ½g(2s)2 - (v0sinθ)(2s)
= ½(v0sinθ)2/g - ½g(2s)2 = ½[(v0sinθ)2/g - g(2s)2] = 11.6m
5) y(tmax) = v0y(v0y/g) - ½g (v0y/g) 2
= (v0sinθ)2/g - ½g(v0sinθ/g)2 = ½ (v0sinθ)2/g = 31.2 m
6) y(tmax) - y(tmax +2s) = 31.2 - 11.6 m = 19.6m
Q12: A 75kg man is riding in a 30kg cart at 2.0 m/s. He jumps off in such a way as to land
on the ground with no horizontal velocity. The resulting change in speed of the cart is:
Vi = 2 m/s
Vf,Man = 0 m/s
Initially
y
z
x
Vf,Cart = ? m/s
Finally
We don’t know the details of how the man jump and the forces he exerts on
the cart. So all we know is initial and final conditions. This sounds like a job
for .... conservation of linear momentum
Initially:
1) Pi,tot = Mtot vi = (Mman + Mcart) vi
2) Pf,tot = Mman vf,man + Mcart vf,cart = Mman 0 + Mcart vf,cart = Mcart vf,cart
3) Pf,tot = Pi,tot
Mcart vf,cart = (Mman + Mcart) vi
vf,cart = 7 m/s. Δv = vf - vi = 5 m/s
vf,cart = vi(Mman + Mcart)/Mcart
Q9: Block A, with a mass of 4 kg, is moving with a speed of 2.0 m/s while block B, with a
mass of 8 kg, is moving in the opposite direction with a speed of 3 m/s. The center of
mass of the two block-system is moving with the velocity of:
PCM = PA + PB = mAvA + mBvB = (4kg)(2m/s) + (8kg)(-3m/s) = -16 kg· m/s
PCM = (mA+mB)vCM = (-16 kg· m/s) vCM = (-16 kg· m/s)/(mA+mB) = -1.33 m/s
1.33 m/s in the same direction as B
Q8: A 25g ball is released from rest 80m above the surface of the Earth. During the fall
the total thermal energy of the ball and air increases by 15J. Just before it hits the
surface its speed is:
Wtot = ΔK = Kf - Ki = Kf - 0 = ½mvf2
Wtot = Wgrav + Wdrag = +Fg· d - 15J = (0.025kg· 9.81m/s2)(80m) - 15J = 4.62J
vf = √[2(4.62 J)/0.025kg] = 19.2 m/s
Q7: The potential energy of a body of mass m is given by U = −mgx + ½kx2. The
corresponding force is:
F = -dU/dx = -d(-mgx + ½kx2)/dx = -[-mg + kx] = mg - kx
Q5: A 0.20 kg particle moves along the x axis under the influence of a conservative
force. The potential energy is given by U(x) = (8.0 J/m2)x2 + (2.0 J/m4)x4, where x is the
coordinate of the particle (in meters). If the particle has a speed of 5.0 m/s when it is at
x = 1.0 m, its speed when it is at the origin is:
K1 + U1 = K2 + U2
½mv12 + U(x1 = 1m) = ½mv22 + U(x2 = 0)
½mv12 + (8.0 J/m2)(1m)2 + (2.0 J/m4)(1m)4 = ½mv22 + (8.0 J/m2)(0m)2 + (2.0 J/m4)(0m)4
½mv12 + (8.0 J) + (2.0 J) = ½mv22 + 0 J ½mv22 = ½mv12 + 10 J
½mv22 = ½(0.2 kg)(5 m/s)2 + 10 J = 12.5 J
v2 = 11.2 m/s
Q4: At time t = 0 a 2kg particle has a velocity in m/s of (4 m/s) î −(3 m/s) ĵ. At t = 3 s its
velocity is (2 m/s) î + (3 m/s) ĵ. During this time the work done on it was:
Wtot = ΔK = Kf - Ki = ½Mvf2 - ½Mvi2
= ½M[(2 m/s)2 + (3 m/s)2] - ½M[(4 m/s)2 + (3 m/s)2]
= ½M[13 (m/s)2] - ½m[25 (m/s)2] = ½M[-12 (m/s)2]
= -½(2 kg)[12 (m/s)2] = - 12 J
Q3: A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An
external force of 3.0 N, always tangent to the track, causes the object to speed up as it
goes around. The work done by the external force as the mass makes one revolution is:
W = F· x = F x sinθ = F x sin(90°) = (3N)(2π 2.5m)(1) = 47 J
Q2: An object is constrained by a cord to move in a circular path of radius 0.5m on a
horizontal frictionless surface. The cord will break if its tension exceeds 16N. The
maximum kinetic energy of the object can have is:
F = T = mv2/r
K = ½mv2
½mv2 = ½T r = ½(16 N)(0.5 m) = 4J
Q1: A 8000-N car is traveling at 12 m/s along a horizontal road when the brakes are applied.
The car skids to a stop in 4.0s. How much kinetic energy does the car lose in this time?
ΔK = Kf - Ki
Ki = ½mvi2, m = 8000N / g =
Kf = 0
ΔK = 0 - ½(8000N / 9.81 m/s2)(12 m/s)2 = 58715 J = 5.9x104 J