Download Fall 2013 Physics 172 – Recitation 11

Fall 2013
Physics 172 – Recitation 11
“Real” and “Point-Particle” Systems
Solution
Purpose: The purpose of this recitation is to allow you to gain further experience
working with multi-particle systems and rotational kinetic energy.
Readings: Chapter 9.1-9.5
Learning Objectives:
9.1.1 Recall the momentum principle for multiparticle systems and the concept of
center of mass (from chapter 3)
9.2.1 Keep track of the vibrational and rotational parts of a multiparticle system's
translational kinetic energy
9.3.1 Define rotational kinetic energy in terms of moment of inertia
9.3.2 Calculate the moment of inertia of simple shapes like thin rods, etc.
9.3.4 Calculate the moment of inertia of a rigid body composed of discrete elements
whose individual moments of inertia are known
9.3.5 Account for changes in the moment of inertia when rotation is not about the
center of mass
9.4.1 Demonstrate how to model a real multi-particle system as a point particle system
9.5.1
Combine both point particle and real system models to analyze physical situations
Challenge Problems:
Consider a bicycle wheel with a rim of mass M and an axle of mass m. The masses of
the wheel’s spokes are so small that we will neglect them.
Initially, the wheel is resting on a horizontal surface, and you hold its axle, a hand on
each side of the wheel, so that it doesn’t tip over. At time t = 0 s you exert a horizontal
force of magnitude F on the axle so that the wheel begins to roll without slipping.
• Determine the wheel’s speed when you have pushed the wheel forward a distance
d.
• Determine the magnitude of the friction force that the surface, on which the wheel
rolls, exerts on the wheel while you are pushing it.
Gather:
System: Bicycle Wheel consisting of rim Mass M and axle of mass m.
Interacting with System: The Earth, the horizontal surface, you.
Simplifying assumptions/ approximations: Ignore mass of wheel’s spokes.
Fundamental Principle(s): Energy Principle
Drawing: See problem description.
Organize:
Determine wheel’s speed after travels forward distance d:
1. Determine the rotational kinetic energy of the system.
2. Determine the translational kinetic energy of the system.
3. Determine the total kinetic energy of the system.
4. Determine the amount of work done on the system when it is moved a distance d.
5. Use the Work energy principle to determine the change in internal energy of the
system
6. Determine the wheel speed after the distance d.
Determine the magnitude of the frictional force:
1. Use the work energy principle to determine relationship between the speed of the
wheel; the force, F, and force of friction.
2. Rearrange the relationship to solve for friction.
Analyze:
Determine wheel’s speed after travels forward distance d:
Step 1: Determine the rotational kinetic energy of the system.
Using the definition of the wheel’s internal kinetic energy (and neglecting the tiny
rotational energy of the axle) we find that
1
2
MVCM
2
Step 2: Determine the translational kinetic energy of the system.
Using the definition of the wheel’s translational kinetic energy we find that
K rotational =
1
1
2
2
M sysVCM
= ( M + m)VCM
2
2
Step 3: Determine the total kinetic energy of the system.
The wheel’s total kinetic energy is
1
1
1
2
2
2
K = K rotational + K trans = MVCM
+ ( M + m)VCM
= (2 M + m)VCM
2
2
2
Step 4: Determine the amount of work done on the system when it is moved a distance d.
Since the Force acts in the same direction as the distance moved, we can write the
following:

W ext = F d
Step 5: Use the Work energy principle to determine the change in internal energy of the
system
K trans =
€
We apply the energy principle to the system consisting of the wheel alone because
it relates the final energy of the wheel and, thus, the wheel's final speed, to the
work you do by pushing its axle the distance d,
ΔEsys = K f = K trans, f + K rotational , f = Wext = Fd

1
ΔE sys = ( 2M + m)V f2 = F d
2
In writing this equation we have used the fact that the initial kinetic energy of the
wheel is zero and the very important fact that the (as yet unknown) frictional force
exerted on the wheel by the floor does no work on the wheel because it acts at a
€
point on the wheel
that is at rest (because the wheel rolls without slipping).
Step 6: Determine the wheel speed after the distance d.
Rearranging the equation and solving for v, we have:
1
2Fd
Fd = (2 M + m)V f2 ⇒ V f =
2
2M + m
Determine the magnitude of the frictional force:
Step 1: Use the work energy principle to determine relationship between the speed of the
wheel; the force, F, and force of friction.
If we apply the energy principle to the “point-particle” system corresponding to
the wheel in this situation, we find that it relates the net horizontal force acting on
the wheel (including the friction force exerted on the wheel by the floor) to the
final translational kinetic energy of the wheel. Using the answers to parts a) and
b) we can determine the final translational kinetic energy, so, we can use the
relationship between it and the frictional force to solve the problem. We find that

 
1
M +m
M + m) )V f2 =
Fd = F + F fr ⋅ d = F − F fr d
(
2
2M + m
Step 2: Rearrange the relationship to solve for friction.
This leads to the following equation:
⎛
M +m ⎞
M
⇒ F fr = ⎜ 1−
F=
F
⎟
2M + m
⎝ 2 M + m⎠
ΔK trans = K trans, f =
(
)
(
)
where we’ve used the fact that the frictional force exerted on the wheel by the
floor points in the direction opposite the displacement of the wheel.
As a check, note that the magnitude of the friction force is less than the magnitude
of the force you exert. This is necessary for consistency with the momentum
principle applied to the wheel in this case. The wheel does accelerate in the
direction you are pushing it.
Additional Questions to Answer:
1. How does the speed of the wheel’s rim relative to the wheel’s center of mass
compare to the speed of its hub when the wheel is rolling at speed VCM without
slipping?
Use the definition of rolling without slipping to determine two speeds. Rolling without
slipping implies that there is no relative velocity between the two surfaces (portion of the
wheel and the horizontal surface) in contact, that is, no sliding. Rolling without slipping
occurs when the part of the rim in contact with the surface is moving backward relative to
the wheel’s hub (center of mass) at the same speed, VCM, as the hub is moving forward.
Learn: By solving this problem, you should have learned that we can model a multiparticle, or real world, situation the same as a point particle system.