Phys 2101
Gabriela González
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F = ma ⇒
F r = mar
F t = mat
F cos φ − T = mω 2 r
F sin φ = mα r
τ = Fr sin φ = mα r 2
τ = m r2 α
τ =Iα
2
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We know all of that already!!
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The figure shows the massive shield door at a neutron test facility
at Lawrence Livermore Laboratory; this is the world's heaviest
hinged door. The door has a mass of 44,000 kg, a rotational
inertia about a vertical axis through its huge hinges of 8.7x104
kg·m2, and a (front) face width of 2.4 m. Neglecting friction, what
steady force, applied at its outer edge and perpendicular to the
plane of the door, can move it from rest through an angle of 90° in
30 s?
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In the figure, a block has mass M = 500 g, the other has mass m = 460 g, and
the pulley, which is mounted in horizontal frictionless bearings, has a radius of
5.00 cm. When released from rest, the heavier block falls 75.0 cm in 5.00 s
(without the cord slipping on the pulley).
(a) What is the magnitude of the blocks' acceleration?
What is the tension in the part of the cord that supports
(b) the heavier block and
(c) the lighter block?
(d) What is the magnitude of the pulley's angular acceleration?
(e) What is its rotational inertia?
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(a)  What is the magnitude of the blocks' acceleration?
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y = at 2 ⇒ a = 2y /t 2 = 2 × 0.75m /25s2 = 0.06m /s2
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What is the tension in the part of the cord that supports
(b) the heavier block and
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Mg − T1 = Ma ⇒ T1 = M(a + g) = 0.5kg × 9.86m /s2 = 4.93N
(c) the lighter block?
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T2 − mg = ma ⇒ T2 = m(a − g) = 0.496kg × 9.794m /s2 = 4.86N
(Tensions are not equal any more!!!)
(d) What is the magnitude of the pulley's angular acceleration?
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α = a /R = 0.06m /s2 /0.05m = 1.2rad /s2
(e) What is its rotational inertia?
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τ = −T1R + T2 R = Iα
I = (T2 − T1 )R /α = 0.07N × 0.05m /1.2rad /s2 = 0.0029kgm 2
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A meter stick is held vertically with one end on the floor and is
then allowed to fall. Assuming that the end on the floor does not
slip…
(a) What forces are acting on the stick?
(b) What forces are doing work?
(c) Find the speed of the other end when it hits the floor.
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A meter stick is held vertically with one end on the floor and is then allowed to
fall. Assuming that the end on the floor does not slip…
(a) What forces are acting on the stick?
Gravity (at the center of mass), static friction and normal force (at bottom end)
(b) What forces are doing work?
Only gravity does (positive) work.
(c) Find the speed of the other end when it hits the floor.
The motion is pure rotation about the lower end.
Use conservation of energy:
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1
1
ML2 + ML2 = ML2
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3
1
L
MgL
g
2
Ibotω − Mg = 0 ⇒ ω =
= 3
2
2
Ibot
L
Ibot = Icom + M(L /2) 2 =
v top = ωL = 3gL = 5.4 m/s
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Assume a 50 kg woman is in a rotor, with
µs=0.5, R=2m.
•  What’s the minimum angular velocity that
will let the woman be stuck to the wall
without a floor?
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Assume a 50 kg woman is in a rotor, with
µs=0.5, R=2m.
What’s the minimum angular velocity that will let the
woman be stuck to the wall without a floor?
We need enough static friction to cancel gravity.
Maximum value of static friction is µsN. The
normal force is the centripetal force, or mω2R.
mg ≤ µsN = µsmω 2 R ⇒ ω ≥ g /(µsR) = 3.13 rad/s ~ 30 rpm
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Answer is independent of the mass, but very much dependent on
coefficient of friction!
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It depends on the reference frame!
If we sit at the center of the wheel
If we sit on the road,
following the bike
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2
KE = 12 I comω 2 + 12 Mvcom
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A solid ball starts from rest at the upper end of the track
shown in the figure, and rolls without slipping until it rolls
off the right-hand end. If H = 6.0 m and h = 2.0 m and
the track is horizontal at the right-hand end, how far
horizontally from point A does the ball land on the floor?
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A solid ball starts from rest at the upper end of the track shown in the figure, and rolls without
slipping until it rolls off the right-hand end. If H = 6.0 m and h = 2.0 m and the track is horizontal
at the right-hand end, how far horizontally from point A does the ball land on the floor?
From H to h: use conservation of energy to get speed.
Notice that kinetic energy has both translational and
rotational components:
ΔK + ΔU = 0
1 2
1
mv com + Icomω 2 − mg(H − h) = 0
2
2
 v 2
1 2
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mv com +
mR 2  com  = mg(H − h)
 R 
2
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2
mv com = mg(H − h)
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v com = (10 /7)g(H − h) = 7.48 m/s
Notice that this is a smaller velocity than a point mass: rolling bodies fall slower than point masses, and the
larger the rotational inertia, the slower they fall). The velocity of the center of mass at h is horizontal.
€ (and the ball keeps rotating, but with a constant angular
From h on, com motion is a projectile motion
velocity, since gravity does not produce a torque)
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h = gt 2 ⇒ t = 2h /g = 0.64 s
2
x = vt = 4.78 m
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•  No slipping! Friction is static; and does no work.
•  There is a net force, so the center of mass is accelerated.
•  If the center of mass is accelerated, there is also angular
acceleration.
Fnet = m acom
τ net = I com α
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acom =
Mgsin θ − f s = Macom
R f s = Icomα = Icom acom /R
gsin θ
1+ Icom /R 2
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A constant horizontal force of magnitude 10 N is applied to a
wheel of mass 10 kg and radius 0.30 m as shown in the figure.
The wheel rolls smoothly on the horizontal surface, and the
acceleration of its center of mass has magnitude 0.60 m/s2.
(a) What are the magnitude and direction of the frictional force on
the wheel?
(b) What is the rotational inertia of the wheel about the rotation
axis through its center of mass?
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A constant horizontal force of magnitude 10 N is applied to a wheel of mass 10 kg and radius 0.30
m as shown in the figure. The wheel rolls smoothly on the horizontal surface, and the acceleration
of its center of mass has magnitude 0.60 m/s2.
(a) What are the magnitude and direction of the frictional force on the wheel?
Fnet = ma ⇒ F − f s = macom ⇒ f s = F − macom = 4N
(b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?
τ net = Iα ⇒ f sR = Iacom /R ⇒ I = f sR 2 /acom = 0.6kgm2
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