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Physics 201 Homework 8
Feb 27, 2013
1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades.
The blades have a total moment of inertia of 0.22 kg-m2 . What is the angular
acceleration of the blades?
8.2 rad/s2
Solution
P
We have
τ = Iα (Newton’s second law). Thus,
P
τ
1.8
α=
=
= 8.2
I
0.22
2. A bicycle wheel has a radius of 0.330 m and a rim whose mass is 1.20 kg. The
wheel has 50 spokes, each with a mass of 0.010 kg. (a) Calculate the moment of
inertia of the rim about the axle. (b) Determine the moment of inertia of any one
spoke, assuming it to be a long, thin rod that can rotate about one end. (c) Find
the total moment of inertia of the wheel, including the rim and all 50 spokes.
(a) 0.131 kg-m2
(b) 3.63 × 10−4 kg-m2
(c) 0.149 kg-m2
Solution
(a) The rim is in the shape of a ring, so I = M R2 . Thus,
Irim = (1.20)(0.330)2 = 0.131
(b) The spoke is in the shape of a rod, rotating at one end, so I = 13 M R2 . Thus,
Ispoke =
1
(0.010)(0.330)2 = 3.63 × 10−4
3
(c) The moment of inertia of the composite is the sum of the individual moments
of inertia. Thus,
Itotal = Irim + 50Ispoke = 0.149
3. A flywheel is a solid disk that rotates about an axis that is perpendicular
to the disk at its center. Rotating flywheels provide a means for storing energy
in the form of rotational kinetic energy and are being considered as a possible
alternative to batteries in electric cars. The gasoline burned in a 300 mile trip
in a typical midsize car produces about 1.2 × 109 J of energy. How fast would a
13 kg flywheel with a radius of 0.30 m have to rotate to store this much energy?
Give your answer in rev/min.
Solution
The formula for the rotational kinetic energy is
KEr = 21 Iω 2
where I is the moment of inertia and ω is the angular speed. We are given KEr , if
we knew I, we could solve for ω. The moment of inertia depends on the shape of
the object, its distribution of mass, and the axis through which it rotates. Since
this requires some calculus to derive from scratch, we must merely look up the
moment of inertia for our objects. For a solid disk rotating through its center,
the moment of inertia is
I = 12 M R2
where M is the mass of the disk and R is its radius. In our case:
I = 12 (13)(0.30)2 = 0.585
Plugging that into the definition above we get:
(1.2 × 109 ) = 12 (0.585)(ω)2 =⇒ ω = 64051 rad/s
1
6.1 × 105 rev/min
But we are asked to quote the answer in rev/min. Since there are $2p radians in
one revolution, we need to covert the units as:
ω=
64051 rad
60 s
1 rev
×
×
= 6.1 × 105
1s
1 min 2π rad
4. A uniform board is leaning against a smooth vertical wall. The board is at
an angle θ above the horizontal ground. The coefficient of static friction between
the ground and the lower end of the board is 0.650. Find the smallest value for
the angle θ, such that the lower end of the board does not slide along the ground.
Solution
This is an equilibrium problem. Let’s begin by cataloging the forces involved.
ˆ The weight of the ladder, pointing down from the center of the block
ˆ The support force from the wall, pointing left from the top of the block
ˆ The support force from the floor, pointing up from the bottom of the block
ˆ The friction from the floor, pointing right from the bottom of the block
There is no friction from the wall because it is “smooth”. The lever arm of all of
these forces will depend on the angle θ. Also on our choice of rotational axis. We
should consider analyzing the torques from one of the places the forces are applied.
It might at first seem good to use the point at the floor since you eliminate two
forces immediately, but notice that we are not given any information about the
weight or mass of the board. This suggests that we should consider using the
center of the ladder as the rotational axis for analysis.
The lever arms are:
ˆ `1 = 0, by design
ˆ `2 = (r/2) sin θ
ˆ `3 = (r/2) sin(90 − θ) = (r/2) cos θ
ˆ `4 = (r/2) sin θ
And the corresponding torques are:
ˆ τ1 = 0, by design
ˆ τ2 = +(F2 )(r/2) sin θ
ˆ τ3 = −(F3 )(r/2) cos θ
ˆ τ4 = +(F4 )(r/2) sin θ
where F4 is the friction force, so F4 = (0.650)(F3 ). Setting the sum of these four
torques to zero yields
(F2 )(r/2) sin θ − (F3 )(r/2) cos θ + (0.650)(F3 )(r/2) sin θ = 0
Notice that the (r/2) cancels out. Unfortuantely, we have two equations with two
unknowns. We need another equation.
We don’t know the weight, but we do know that all the forces cancel out—
including the horizontal ones. In other words,
F2 = F4 = (0.650)(F3 )
2
37.6◦
After subsutition, this makes our torque equation solvable.
(0.650)(F3 )(r/2) sin θ − (F3 )(r/2) cos θ + (0.650)(F3 )(r/2) sin θ = 0
We cancel the (F3 )(r/2) and are left with:
(1.300) sin θ − cos θ = 0
=⇒ (1.300) sin θ = cos θ
=⇒ tan θ = 1/1.300 = 0.7692
=⇒ θ = 37.56◦
5. Figure 1 shows a uniform horizontal beam attached to a vertical wall by a
frictionless hinge and supported from below at an angle θ = 39◦ by a brace that is
attached to a pin. The beam has a weight of 340 newtons. Three additional forces
keep the beam in equilibrium. The brace applies a force P~ to the right end of the
beam that is directed upward at the angle θ with respect to the horizontal. The
hinge applies a force to the left end of the beam that has a horizontal component
~ and a vertical component V
~ . Find the magnitudes of these three forces.
H
Figure 1: Problem 9.21
Solution
(a) Okay. There are four forces here:
ˆ The weight of the beam, located in the center of the beam, pointing down.
ˆ The support force from the brace, located at the end of the beam, pointing
up at 39◦ .
ˆ The horizontal component of the support from the hinge, located at the
start of the beam, pointing left.
ˆ The vertical component of the support from the hinge, located at the start
of the beam, pointing up.
Now we must choose a axis of rotation for the analysis. If we choose the point
where the hinge is, then the last two forces will have no lever arm. This means
that we won’t need to know the magnitude of these forces to calculate the torques.
Now we need to determine the lever arms of these forces to get to the torques:
ˆ The lever arm of the beam’s weight is half the length of the beam, `1 = L/2.
ˆ The lever arm of the brace’s support is `2 = L sin 39◦ .
ˆ The lever arm of the hinge’s horizontal support is zero, `3 = 0.
ˆ The lever arm of the hinge’s vertical support is also zero., `4 = 0.
The only tricky one is `2 . The line of action for the brace’s support is along the
support, so the lever arm is perpendicular to this line. There is a right triangle
here: the hypoteneus is the length of the beam, the adjecent side is along the
brace, and the opposite side is the lever arm itself. So sin θ = `/L.
The next step is to calculate the torques. We use τ = F `, but remember that the
sign of the torque is determined by the direction of the line of force. If the line
of force will generate positive rotation, the torque is positive. Thus,
3
P = 270 newtons. H = 210 newtons. V =
170 newtons
ˆ τ1 = −(340)(L/2).
ˆ τ2 = +(P )(0.62932)(L).
ˆ τ3 = (H)(0) = 0.
ˆ τ4 = (V )(0) = 0.
The sum of these torques must be zero because the system is not moving. Thus,
−(340)(L/2) + (P )(0.62932)(L) = 0 =⇒ P = 270.13
(b) To get the hinge’s horizontal support H, we can use fact that the forces must
balance (otherwise the system would move). The horizontal forces are: H, the
horizontal support from the hinge and Px , the horizontal support from the brace.
Thus,
H + P cos θ = 0
=⇒ H + (270.13)(0.77715) = 0
=⇒ H = −209.93
The negative sign indicates the force H is pointing left.
(c) To get the hinge’s vertical support V , we use similar logic. The vertical forces
are: V , W , and Py . Thus,
V − W + P sin θ = 0
=⇒ V − (340) + (270.13)(0.62932) = 0
=⇒ V = 170.00
6. Figure 2 shows a person (weight = 584 newtons) doing push-ups. Find the
normal force exerted by the floor on each hand and each foot, assuming that the
person holds this position.
Figure 2: Problem 9.72
Solution
There are three forces here. The support force from the hands, the feet, and
the person’s weight. We need to choose an axis of rotation. Any axis will work
because the system is at rest, in equilibrium. There are three natural places
to consider: each place where there is a force. Whichever point we choose will
eliminate that force from the torque calculation.
Suppose we choose the center of gravity. Then the lever arm of the feet is 0.840
meters and the lever arm of the hands is 0.410 meters. Both supports point up,
but the direction of the torques are different. The torque from the hands cause
positive rotation and the torque from the feet cause negative rotation.
Let’s agree to call the support force from the hands H and the support force from
feet F . The sum of the torques is therefore,
(H)(0.410) − (F )(0.840) = 0
4
96 newtons from each foot and 196 newtons
from each hand
In order to move forward we need another equation. This will come from the
linear equilibrium of the system. In this case,
H + F − 584 = 0
Two equations, two unknowns. We can use the first to get:
H = (2.0488)(F )
And we can plug this into the second equation to yield:
(2.0488)(F ) + F − 584 = 0 =⇒ F = 191.55
And,
H = (2.0488)(191.55) = 392.45
However, we are asked for the force from each foot and hand. So, divide both of
these numbers by two.
7. A person exerts a horizontal force of 190 newtons in the test apparatus shown
~ that his flexor muscle
in Figure 3. Find the magnitude of the horizontal force M
exerts on his forearm.
1200 newtons
Figure 3: Problem 9.15
Solution
This is an equilibrium problem and there are two forces acting on the forearm:
the apparatus is pulling with 190 newtons of force and the bicep muscle is pulling
with the unknown force. The total torque must be zero—but in this simple case
that simply means that the two torques are equal and opposite.
The torque from the apparatus is:
τ = F ` = (190)(0.34) = 64.6
Similarly, the torque from the muscle is:
τ = F ` =⇒ 64.6 = (M )(0.054) =⇒ M = 1196.3
8. Figure 4 shows a bicycle wheel resting against a small step whose height is
0.120 meters. The weight and radius of the wheel are 25.0 newtons and 0.340
meters, respectively. A horizontal force F~ is applied to the axle of the wheel. As
the magnitude of F~ increases, there comes a time when the wheel just begins to
rise up and loses contact with the ground. What is the magnitude of the force
when this happens?
5
29 newtons
Figure 4: Problem 9.24
Solution
There are two ways to solve this problem. In the first approach we consider the
point of contact between the wheel and the stair as the axis of rotation as the
wheel just begins to lift off the ground. Before the wheel lifts there are three
forces: the pulling force, the weight of the wheel and the support force from the
ground. But as the wheel lifts, there is no longer any support to consider. So
there are only two forces to worry about in this problem. So we will have to
calculate two torques and therefore two lever-arms. For this a drawing might
help:
θ = 49.68◦
0.340
0.220
0.259
0.120
This shows all the key distances and angles in the problem. We know the vertical
height of the triangle because it is the radius of the wheel minus the height of
the step. The angle can be obtained using the cosine against these two lengths
and the horizontal side of the triangle you can get from either the Pythagorean
theorem or using the sine of the angle we just calculated.
Okay—back to calculating lever arms. Fortunately we have already done the
work. The lever arm of the pulling force is the vertical side of the triangle (0.220
meters) and the lever arm of the weight is the horizontal side (0.25923 meters).
Right as the wheel lifts these two torque are in equilibrium: they are equal and
opposite. Well, we know the torque from the weight around this pivot:
τ = F`
= (25.0)(0.25923)
= 6.4808
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Since the magnitude of the torque from the pulling force must be the same:
τ = F`
=⇒ 6.4808 = (F )(0.220)
=⇒ F = 29.458
I said earlier there was a second way to do this problem. You don’t even need
to consider torque (though you still need to do that trig). The net force is the
combination of the weight and the pulling force. The wheel will lift when that
net force pulls the wheel above and over the pivot. The angle of this net force is:
P Fy
θ = tan−1 P
Fx
P
P
But in this case
Fy is just the weight and
Fx is just the pulling force. The
only remaining trick is to remember to use the “canonical angle”. In this case
that is 270◦ plus the angle in the triangle. Thus,
tan(319.68
◦
) = -25.0 F
=⇒ F =29.458
9. A stationary bicycle is raised off the ground, and its front wheel (mass = 1.3
kilograms) is rotating at an angular velocity of 13.1 rad/s (see Figure 5). The
front brake is then applied for 3.0 seconds, and the wheel slows down to 3.7 rad/s.
Assume that all the mass of the wheel is concentrated in the rim, the radius of
which is 0.33 meters. The coefficient of kinetic friction between the brake pad
and the rim is µk = 0.85. What is the magnitude of the normal force that each
brake pad applies to the rim?
Figure 5: Problem 9.45
Solution
The friction from the brake pads provides the torque to decelerate the wheel. We
know that the friction force is given by F = µk N . So by determining this force
we can answer the question. On the other hand, clearly we can determine the
angular acceleration from the data given:
ω0 = 13.1
t = 3.0
ω = 3.7
α=?
7
0.79 newtons
We can use ω = ω0 + αt...
(3.7) = (13.1) + (α)(3.0) =⇒ α = −3.1333
The torque required to produce this deceleration is given by tau = Iα. For a ring,
the moment of inertia is simply mr2 . In this case:
I = (1.3)(0.33)2 = 0.14157
So the torque needed is
τ = (0.14157)(3.1333) = 0.44358
I’ve dropped the sign because the direction doesn’t matter in this question. Well,
the torque comes from the friction force. But since there are two brake pads,
there are two forces applied. The lever arm is simply the radius of the wheel. So
we can plug all this into the definition of torque to yield:
τ = F`
=⇒ 0.44358 = (2)(0.85)(N )(0.33)
=⇒ N = 0.79070
10. A bowling ball encounters a 0.760 meter vertical rise on the way back to
the ball rack, as Figure 6 illustrates. Ignore frictional losses and assume that the
mass of the ball is distributed uniformly. The translational speed of the ball is
3.50 m/s at the bottom of the rise. Find the translational speed at the top.
Figure 6: Problem 9.57
Solution
Hopefully this looks a lot like those inclined plane problems. Like those problems,
we can use energy to solve this one. In that case we need to catalog our energy
before and after the slope.
Before, the height is zero and so is the potential energy. The linear kinetic energy
is simply 12 mv 2 . We are not given the mass, but we are used to that cancelling
out in these energy problems.
The only difficulty is the rotational kinetic energy 12 Iω 2 . Assuming that the ball
is rolling (we must or we can’t solve this problem), we can say v = rω, but we
don’t know the radius of the ball. This also is an issue in calculating the moment
of inertia I = 52 mr2 . But notice that we don’t really need them individually. In
fact, the radius cancels out:
KErot = 12 Iω 2
= ( 21 )( 25 mr2 )(v/r)2
= 15 mv 2
8
1.3 m/s
This is the formula for the rotational kinetic energy of a rolling sphere. Now we
can proceed with our catalog:
PE = 0
KElin = ( 12 )(m)(3.50)2
KErot = ( 15 )(m)(3.5)2
= (6.125)(m)
= (2.450)(m)
This yields a total initial energy of (8.575)(m) joules. Since we ignore frictional
losses, the final energy is the same. But what do we know about the final energy?
PE = mgh = (m)(9.8)(0.760) = (7.448)(m)
KElin = ( 12 )(m)(v)2 = (0.500)(m)(v)2
KErot = ( 15 )(m)(v)2 = (0.200)(m)(v)2
Adding these up must give us the total final energy:
(8.575)(m) = (7.448)(m) + (0.700)(m)(v)2
=⇒ v 2 = 1.61
=⇒ v = 1.2689
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