Pointparticle Motion - PowerPoint PPT Presentation

Title: Pointparticle Motion


1
Point-particle Motion
Rigid-body Motion
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Summary of Dynamics of Rigid Bodies
If O is fixed
Gravitational Potential Energy
Kinetic Energy
Linear Momentum
Angular Momentum
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Work-Energy Relation
The same as for point particles
Applied couples (moments)
Applied forces
Dont double count forces either a force or
part of a couple
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Energy methods useful to relate initial position,
velocity to final position, velocity
No friction Reaction forces do no work
Initially starts at rest, with yG0 ? Initial
energy is zero
At any angle ?, we then have
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From last lecture Alternative approach use
moment equation for fixed-axis rotation
Reaction forces (two of the unknowns) do not
appear in the moment equation
One EOM for ?(t), and all the rest follows from
other equations
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EOM angular acceleration is a function only of
angle
A bit of trickier using differentiation
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To get reaction forces, recall from last lecture
(see previous slides) that we had
At bottom position, ?3?/2, we thus have
As found via analyzing full MATLAB solution
and
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Fy
Reaction Forces (N)
Fx
Time (s)
Bottom position
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Loop-the-Loop the role of rotational kinetic
energy
At what height must the ball start so as not to
lose contact with the loop?
First, loss of contact
Contact with the loop lost first at the top of
the loop.
To maintain contact, the normal force must be Ngt0.
For motion along the circular loop
Draw FBD and work out N
For N just 0, the ball must have
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Loop-the-Loop the role of rotational kinetic
energy
Need
Apply work-energy to find initial height required
to attain final velocity
Rolling Static friction neglect air resistance
? Energy is Conserved
Neglect ball radius compared to loop radius, rltltR
Final height is yGf2R
For a solid uniform sphere, IG(2/5)mr2
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Angular Momentum

• Recall two things
• Angular momentum is conserved if there are no
  moments acting on the body over time
• The point O (for a rigid body in plane motion, an
  axis of rotation through O) at which we compute
  angular momentum is arbitrary. So, we can choose
  it to eliminate moments and obtain conservation.

For plane motion, if no moments acting along axis
of rotation, then angular momentum along this
axis is conserved
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Angular Momentum
M70kg m 5 kg R0.25m, L0.7m
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z
R
Rotate wheel vertically (applying internal
moments to change hx and hy)