Download Tue Sep 21

Phys 2101
Gabriela González
2
Pure rotation of a reference line in a solid body around
an axis is described with an angle as a function of time.
θ=s/r
ccw: positive
cw: negative
radians
rad/sec
rad/sec2
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ω = ω 0 + αt
v = v0 + at
x = x0 + v0t + at
1
2
2
θ = θ 0 + ω 0 t + αt
1
2
2
y
x
θ
x
4
•  Trajectory is a circle (“circular” motion)
•  Speed is constant (“uniform motion”)…
•  … but velocity is not constant!?

x (t ) = R cos(ωt ) iˆ + R sin(ωt ) ˆj

v (t ) = − Rω sin(ωt ) i + Rω cos(ωt ) ˆj

a (t ) = − Rω 2 cos(ωt ) iˆ − Rω 2 sin(ωt ) ˆj
2
= − ω x (t )
v2
= ω2R
Derive the following formulae: s = θR; v = ωR; a =
R
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s = θr
Assume a constant radius: circular motion
ds dθ
=
r ⇒ v =ωr
dt dt
dv dω
=
r ⇒ at = α r
dt dt
ar = ω 2 r
If speed is not constant, there is angular and tangential acceleration,
apart from radial acceleration.
If speed is constant, there is no angular acceleration, but there still is
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radial acceleration.
A wheel A of radius rA = 10 cm is coupled by belt B to
wheel C of radius rC = 25 cm. The angular speed of
wheel A is increased from rest at a constant rate of 1.6
rad/s2. Find the time for wheel C to reach a rotational
speed of 100 rev/min, assuming the belt does not slip.
(Hint: If the belt does not slip, the linear speeds at the
rims of the two wheels must be equal.)
7
Kinetic energy:
K = 12 m1v12 + 12 m2 v22 + 12 m3v32 + 12 m4 v42
= 12 m1r12ω 2 + 12 m2 r22ω 2 + 12 m3 r32ω 2 + 12 m4 r42ω 2
=
1
2
(m r
2
1 1
)
+ m2 r22 + 12 m3 r32 + 12 m4 r42 ω 2
= 12 Iω 2
8
9
Moment of inertia with respect to an arbitrary rotation axis:
I = I com + Mh 2
Four identical particles of mass 0.50 kg each are placed at the vertices of a 2.0 m 2.0 m square
and held there by four massless rods, which form the sides of the square. What is the rotational
inertia of this rigid body about an axis that
(a) passes through the center of mass, the midpoints of opposite sides and lies in the plane of
the square,
(b) passes through two masses on one side?
10
The figure shows an overhead view of a meter stick that
can pivot about the point indicated, which is to the left of
the stick's midpoint. Two horizontal forces, F1 and F2,
are applied to the stick. Only F1 is shown. Force 2 is
perpendicular to the stick and is applied at the right end.
If the stick is not to turn,
(a) what should be the direction of F2, and
(b) should F2 be greater than, less than, or equal to F1?
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How effective is a force to produce rotation?
1
6
2
3
4
5
Torque = Force x distance x sin(angle)
 
τ = r × F = rF sin φ
Positive if the force produces a ccw rotation;
negative if the force produces a cw rotation.
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€
The body in Fig. 11-38 is pivoted at O. Three forces act
on it in the directions shown: FA = 10 N at point A, 8.0 m
from O; FB = 16 N at point B, 4.0 m from O; and FC = 19
N at point C, 3.0 m from O. What is the net torque about
O?
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

F = ma ⇒
F r = mar
F cos φ − T = mω 2 r
F t = mat
F sin φ = mα r
τ = Fr sin φ = mα r 2
τ = m r2 α
τ =Iα
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We know all of that already!!
15
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?
1
y = at 2 ⇒ a = 2y /t 2 = 2 × 0.75m /25s2 = 0.06m /s2
2
What is the tension in the part of the cord that supports
(b) the heavier block and
€
Mg − T1 = Ma ⇒ T1 = M(a + g) = 0.5kg × 9.86m /s2 = 4.93N
(c) the lighter block?
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?
α = a /R = 0.06m /s2 /0.05m = 1.2rad /s2
(e) What is its rotational inertia?
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:
1
1
1
ML2 + ML2 = ML2
12
4
3
1
L
MgL
g
Ibotω 2 − 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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€