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Physics 200
Problem Set 6
Solution
1. (i) What is the moment of inertia I of a propeller with three blades (treated as rods) of mass
CM
◦
m, length L, at 120 relative to each other? (ii) If a torque τ acts on this how long will it take to
reach an angular velocity ω? (iii) How many revolutions will it have made before reaching this ω?
(iv) Get the numerical answers if L = 1.25 m, m = 12 kg, τ = 3000 N·m, ω = 2000 rad/s.
Answer:
(i) We know (if not, look on p. 296) that the moment of inertia of a single rod rotating around
its end is 1mL2. It’s not hard to convince oneself that if there are three of them rotating
3
around the same axis and in the same plane, the moment of inertia is just three times this,
I =mL2.
CM
(ii) Since ω = αt and τ = I α,
CM
2
t = ICMω = mL ω.
τ τ
(iii) From our knowledge of constant acceleration problems,
ω2 ICMω2 mL2ω2
ω2 = 2αθ =⇒ θ = = = .
2α 2τ 2τ
The number of revolutions it made is
θ mL2ω2
N=2π = 4πτ .
(iv)
I =(12 kg)(1.25 m)2 = 19 kg·m2,
CM
t = (12 kg)(1.25 m)2(2000 s−1) = 13 s,
2(3000 N·m)
2 −1 2
N=(12kg)(1.25 m) (2000 s ) =2000.
4π(3000 N·m)
2. Consider I for a rectangle of sides a (along the y-axis) and b (along the x-axis) about the two
symmetry axes. (Rotate the rectangle about one of these axes and think of it as composed of
rods.) Show that about the axis parallel to x, I = 1 Ma2. Going back to the very definition of
12
I, show that if this rectangle is spun around an axis through its CM and perpendicular to its area
1 2 2
the moment of inertia will be I = 12M(a +b ).
A
nswer: To find Ix, the moment of inertia about the symmetry axis parallel to the x-axis, we
think of dividing the rectangle into many thin rods:
1
y
a
x
b
Since each “rod” has length a, it should be obvious that the sum of these contributions is simply
1 Ma2, i.e., the same as if there was one rod of mass M rotating around the axis, but let’s be more
12
explicit. If we divide up the rectangle into n rods (where n is large so they really are rods, though
in the end it doesn’t matter) so that each rod has mass M/n, then with each rod contributing
1 M 2 1 2
( )a and there being n rods, we get I = Ma asexpected. The same logic of course applies
12 n x 12
to the symmetry axis parallel to the y-axis.
From the definition of the moment of inertia,
I = Xr2∆mi.
i
i
Notice from the picture that r2 = (x −b/2)2 +(y −x/2)2, so we can write
i i i
y
(x,y)
r i i
a
(b/2,a/2)
x
b
µ ¶
X 2 X³ ´
b a 2 1 2 2
I = x − ∆m + y − ∆m =I +I =I +I = M(a +b ).
i 2 i i 2 i y x x y 12
i i
It’s easy to see that this kind of argument will work for any flat object, as long as you choose your
axes to be perpendicular to each other.
◦
3. A 4.8 kg block is resting at the top of a 30 slope of height 1 m. It is attached to a cylindrical pulley
of mass 1.7 kg and radius 8 cm by a massless string that unwinds as the block slides downhill. If
the acceleration of the block is 1.9 m/s2 what is µk? Find the velocity at the bottom of the slope
using forces and torques. Repeat using energy ideas. See Fig. 1.
Answer: Numbers are cumbersome, so let’s start by letting m be the mass of the block, θ the
angle, h the height, M the mass of the pulley, R its radius, and a the acceleration felt by the block.
As usual, we decompose the forces on the block into those that are parallel to the slope and those
that are perpendicular to the slope. In the perpendicular direction, the component of gravity in
this direction mgcosθ balances the normal force, so N = mgcosθ. In the parallel direction, we
have
mgsinθ−T −fk =ma
2
Figure 1: The block started at height h.
(we take downhill to be positive), where the kinetic friction term is given by f = µ N =
k k
µkmgcosθ. We now need the tension T, which we find as follows. The acceleration a that the
block feels is converted to angular acceleration in the pulley, where a = Rα. But we also know
that τ = RT = Iα where I = 1MR2 is the moment of inertia of the pulley. Therefore
2
T = Iα = Ia = 1Ma,
R R2 2
and plugging into the above equation we get
µM ¶
fk = mgsinθ−T −ma=mgsinθ− 2 +m a.
Then f mgsinθ−(M/2+m)a
µ = k = .
k mgcosθ mgcosθ
Before plugging in the numbers, note that we can make our lives just a little easier by rearranging
a little bit:
µ = sinθ−(1+M/2m)a/g = sin30◦−[1+((1.7 kg)/2(4.8 kg))(1.9 m/s2)/(9.81 m/s2) = 0.31.
k cosθ cos ◦
30
Since the block travels a distance d = h/sinθ, the velocity at the bottom of the slope, starting
from rest at the top, is
v2 = 2ad = 2ah =⇒ v = r2ah =r2(1.9 m/s2)(1 m) =2.8 m/s.
sinθ sinθ sin30◦
The energy method is somewhat more involved. At the top, the energy of the block is purely
potential energy, mgh. At the bottom, the potential energy is zero, the kinetic energy of the block
is T = 1mv2, and the rotational kinetic energy of the pulley is
2
1Iω2 = 1 µ1MR2¶³v´2 = 1Mv2.
2 2 2 R 4
Someenergywasalsolost to friction, so we must take that into account. The work done by friction
is
f d = µ mgcosθ· h = µkmgh.
k k sinθ tanθ
W
e can finally use the conservation of energy in the form
1 1 µ mgh
mgh= mv2+ Mv2+ k ,
2 4 tanθ
3
which means
s s 2 ◦
v = 2gh(1−µk/tanθ) = 2(9.81 m/s )(1 m)(1−0.31/tan30 ) = 2.8 m/s.
1+M/2m 1+(1.7 kg)/2(4.8 kg)
As one would expect, the two answers agree.
4. Argue that that A · (A × B) = 0. In three dimensions find the expression of A × B in terms of
ˆ ˆ ˆ
vector components and i,j, and k.
Answer: Since A×B is always perpendicular to both A and B, the dot product with A must be
zero. For the second part, we use the basic relations
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i × i = j × j = k × k = 0, i × j = −j ×i = k, j ×k = −k×j=i, k×i=−i×k=j.
Thus when we expand the cross product, we can immediately drop any term that contains the
same unit vector twice:
ˆ ˆ ˆ ˆ ˆ ˆ
A×B=(A i+A j+A k)×(B i+B j+B k)
x y z x y z
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
=A B (i×j)+A B (i×k)+A B (j×i)+A B (j×k)+A B (k×i)+A B (k×j)
x y x z y x y z z x z y
ˆ ˆ ˆ
=(A B −A B )i+(A B −A B )j+(A B −A B )k.
y z z y z x x z x y y x
Wecan also check that
A·(A×B)=A (A B −A B )+A (A B −A B )+A (A B −A B )=0,
x y z z y y z x x z z x y y x
as we argued.
5. A disk of radius R and mass M is spinning at an angular velocity ω0 rad/s. A non-rotating
concentric disk of mass m and radius r drops on it from a negligible height and the two rotate
together. (See Fig. 2). Find the final ω and fraction of initial kinetic energy left.
m, r
M, R
Figure 2: The upper disk, initially at rest, falls with
negligible speed on the lower one which is spinning.
Their centers coincide.
Answer: Let I = 1MR2 be the moment of inertia of the disk of radius R, and I = 1mr2 the
M 2 m 2
momen
t of inertia of the other disk. Initially the angular momentum of the system is I ω , and
M 0
when the two disks are rotating together the angular momentum is (I +I )ω. Since angular
M m
momentum is conserved,
I ω ω
I ω =(I +I )ω =⇒ ω= M 0 = 0 .
M 0 M m I +I 1+mr2/MR2
M m
4
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