Summary Lecture 12

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Summary Lecture 12
Rotational Motion 10.8 Torque 10.9 Newton 2 for
rotation 10.10 Work and Power 11.2 Rolling
motion
ProblemsChap. 10 21, 28 , 33, 39, 49, 54, 67
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Some Rotational Inertia
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Parallel-axis Theorem
The rotational inertia of a body about any
parallel axis, is equal to its
R.I. about an axis through its CM,
PLUS
R.I. of its CM about a parallel axis through the
point of rotation
I ICM Mh2
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Proof of Parallel-axis Theorem
One rotation about yellow axis involves one
rotation of CM about this axis plus one
rotation of body about CM.
I Icm Mh2
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Example
R
RI of CM about suspension point, distance R away
is MR2. So total RI is 2MR2
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The Story so far...
?, ?, ? relation to linear variables vector
nature
Rotational Variables
Rotational kinematics with const. ?
Analogue equations to linear motion
Rotation and Kinetic Energy
Rotational Inertia
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torque
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Torque
is the turning ability of a force
Where would you put the door knob?
The magnitude of the torque is Fr,
and this is greater here!
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Torque
Axis
If F and r are perpendicular ??? ?r??F?
(Unit N m)
The same F at larger r has bigger turning effect.
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In General
Torque is a vector
? r x F
??? r F sin? ??? ?r? x perp component
of F
Direction of ? Perpendicular to r and
F Sense Right-hand screw rule (out of screen)
(Hint Which way would it accelerate the body?
Sense is same as change in ang vel. ?.)
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Newton 2 for Rotational motion
For Translational motion we had
For Rotational motion we expect
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Example
? I? (Newton 2) ? I ?/? ? I
6400/1.2 ? I 5334 kg m2
? r x F 6400 N m
What is I for rotating object?
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Work
Work done by a torque ? which rotates a body
through an angle Dq Dw ?.Dq
If ? is constant w ? q
Power
P ?.w
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Example
A car engine has a power of 100 HP at 1800 RPM
What is the torque provided by the engine?
1 HP 746 W ?P 74600 W
P ?.? ? ? P/ ?

• ? 74600/ 188
• ? 397 N m

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Rolling motion
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Rolling Motion
2?R
vcm t 2?R But in turning one revolution (2?
radian) in time t, ? 2?/t So that t 2?/?
vcm ?R
vcm 2?/? 2?R
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Rolling Motion
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Cons. Energy says PEinitial KEfinal mgh
Ktrans Krot ½ mvcm2 ½ Iw2
remember vcm Rw or
w vcm/R
so mgh ½ mv2 ½ Iv2/R2
total energy (KE) ½ mv2 ½ v2bmR2/R2
½ mv2 ½ mv2b
The kinetic energy is shared between
translational and rotational motion.
½ mv2(1 b)
b is the coefficient in the expression for the
Rotational Inertia I I bmR2
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mgh Ktotal Ktrans KRot ½
mvcm2 ½ mvcm2b
h
½ mvcm2(1 b)
The kinetic energy is shared between
translational and rotational motion.
The larger b, the more of the available energy
goes into rotational energy, and the smaller the
centre of mass velocity
The fraction of KE that is translational is

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That's all folks