Download Quiz3 Solutions

Quiz 3 SOLUTIONS — PHYS 121, Summer 2010
There were 35 points possible on this quiz.
1. A child runs across the grass, and then makes a running jump into a wagon (initially at rest).
The child-carrying wagon rolls for a distance across the grass before coming to a stop. If
the child has a mass of 20 kg and the wagon has a mass of 15 kg, the coefficient of rolling
friction is 0.1, and the child is moving at 5 m/s when she jumps in the wagon, how far does
the wagon roll before coming to a stop? (10 points)
This was a two-part problem. To find the distance the loaded wagon rolled, you needed to find its
initial velocity. That in turn came from the conservation of momentum, because this is an inelastic
collision where a child (initially running at 5 m/s) jumps into a wagon (initially at rest) and the
combination of them rolls off. The momentum equation is
mchild v child = m child wagon v child wagon . We know the masses, and we can get to
mchild
m
20 kg
20 m
v  childwagon = v child
= 5
=
. That gives us the initial
mchild  mwagon
s 20 kg  15 kg
7 s
velocity of the loaded wagon. We need the acceleration of the wagon, which is due to the frictional
force. That in turn is F friction =  n =  m g , and with F = m a we get to
F friction
m g
a =
=
=  g . We know the final velocity is zero, and we want the total
m
m
displacement, so we use v 2f = v 2i  2 a  x , and with algebra (and plugging in the zero final
v 2i
v 2i
20/7 m/s 2
velocity) we get to  x =
=
=
= 4.16 m.
2a
2g
2⋅ 0.1⋅9.8 m/s 2
2. A streetlight can be modeled like two rods at a right angle with a weight
at the end of one of them (see figure at right). The vertical rod is 10 m
long and has a mass of 200 kg. The horizontal rod is 4 m long and has a
mass of 40 kg. Finally, there is a 25 kg lamp mounted at the end of the
horizontal rod. The streetlight is bolted to the ground at its base.
(a) What is the magnitude of the torque exerted on the base of streetlight
by the bolts holding it to the ground? (10 points)
(b) What is the moment of inertia of the streetlight when pivoting around
its base? (5 points)
On part (a), the direct, brute-force way to do this starts with the torque being given by
 = R F sin . Because the streetlight is not moving, the torque exerted by the bolts at the base
of the streetlight is equal in magnitude (and opposite in direction) to the gravitational torque exerted
by the parts of the streetlight. We aren't asked about the direction, so we can simply find the total of
the gravitational torques of the streetlight parts.
The torques due to the individual parts of the streetlight all have the same sign (one is zero); the
ones due to the horizontal rod and the lamp are trying to rotate the streetlight clockwise in the
diagram above.
Remember that the force of gravity exerted by an object ... or part of an object, as here ... acts as if
all the mass is located at the center of gravity for that object. For a symmetric object, the center of
gravity is located at the geometric center of the object.
So, we have three parts to the streetlight: the vertical rod, the horizontal rod, and the lamp at the end
of the horizontal rod. The centers of gravity of the two rods are at the centers of the rods
PHYS 122, Quiz 3 solutions
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themselves.
So we can write total = vertical rod  horizontal rod  lamp . Each individual torque is
 = R F sin  , where R is the distance from the part's center of gravity to the pivot point (here,
the streetlight base) and F is the gravity force on that part, and θ is the angle between the two. For
the vertical rod, θ is 180°, because the vector from the lamp base to the center of gravity of the rod
is directly up, while the gravity vector is straight down, and since sin 180 ° = 0, the torque
exerted by the vertical rod is zero. For the horizontal rod, the center of gravity of the rod is 10
meters up (so y = 10 m) and 2 meters from the vertical rod (so x = 2 m), so
R = 10 m2  2 m2 = 104 m. The angle θ is the inverse tangent (arc tangent) of 2/10, and
the force is F = m g = 40 kg 9.8 N /kg  = 392 N. For the lamp,
R = 10 m2  4 m2 =  116 m , F = 25 kg 9.8 N /kg  = 245 N , and θ is the inverse
tangent of 4/10. So now we can sum up the parts:
total = 0   104 m 392 N  sin  tan−1 0.2   116 m 245 N  sin  tan−1 0.4 and when
you grind through the arithmetic, that is a total of 1784 N m.
NOTE: An alternative way to do this one – much shorter, but
you have to understand the geometry and see what R ⊥
really is – is to use  = R ⊥ F. Extend the direction of the
force vector (here, that's straight down) past the pivot point,
and construct a perpendicular to this extended force vector
line through the pivot point. Then R ⊥ is the distance
along that perpendicular between the pivot point and the
extended force direction line.
That sounds horrible when expressed in words, but it's easier
to see in the figure at right. The force vectors point straight
down from the centers of gravity (c.o.g.'s) of the three parts
of the streetlight. The extended line for the vertical rod is at
the center of that rod but is embedded entirely inside the rod;
it passes through the pivot point, so R ⊥ is zero. The one extending downward from the c.o.g. of
the horizontal rod passes through the ground 2 m away from the pivot point, and the ground is
perpendicular to that extended force line, so that 2m is R ⊥ . The same kind of argument holds
for R ⊥ for the lamp being 4 m. So you avoid the unpleasant math of computing R and θ if you
see the geometry. Then you can use  = R⊥ F , using the 2 m and 392 N for the horizontal rod
and 4 m and 245 N for the lamp, and you end up at the same 1784 N m result.
Part (b) is computing the moment of inertia, and for that we use I = ∑ M R 2 , where R is the
distance of the center of mass of the part from the pivot point. The R-values for this part are
(mercifully) the same as the R values for part (a), and since we ducked the location of the center of
gravity of the vertical rod, that is simply the middle of that rod, 5 m above the base. So we can
write I =  200 kg 5 m2   40 kg 104 m 2   25 kg 116 m2  = 1.206 × 10 4 kg m 2 .
3. A bungee cord is stretched by 20 cm beyond its original 50 cm length when 15 kg is hung on
the end of it.
(a) What is the spring constant of the bungee cord? (5 points)
(b) How far is it stretched when it has 22 kg hung on the end of it? (5 points)
Since
F = −k  x , k =
PHYS 122, Quiz 3 solutions
N / kg 
= 735 N⋅m
∣Fx∣ = 15 kg 9.8
0.2 m
for part (a).
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On part (b),  x =
 22 kg 9.8 N /kg 
F
=
= 0.293 m.
k
735 N⋅m
PHYS 122, Quiz 3 solutions
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