2. β’ Considers the kinetic energy of a rigid body rotating about an axis
β’ It is composed of other small bodies of mass ππ and at separation
distance ππ from axis of rotation and the kinetic energy of a rigid body
is expressed as:
πΎ = π=1
π 1
2
πππ£π
2
= π=1
π 1
2
ππ πππ 2
= π=1
π 1
2
πΌππ2, (1)
where π is the angular velocity for all bodies constituting a rotating
rigid body and πΌ = π=1
π
ππππ
2
is rotational inertia/moment of inertia
which is analogous to mass in linear motion equations. Therefore πΌπ can
be thought of as a resistant of a body to undergo rotational motion.
πΎ = π=1
π 1
2
πΌππ2 = 1
2
πΌπ2(radian).
Rotational Kinetic Energy
4. Torque and Newtonβs 2nd Law
Figure T. (a) Schematic showing a rigid body cause to rotate about point O when force πΉ is applied at position π from point O; these vectors are
oriented at an angle β to each other. (b) force πΉ has a perpendicular component πΉβ₯ to π or tangential component (πΉπ‘) to point P. (c) Position
vector π has a perpendicular component (πβ₯) to πΉ whose interaction causes rotation about point O.
5. Torque
β’ When opening a door you need to apply a force F at an angle close to right-angle and far
enough (separation displacement r ) from hinges, otherwise more force would be
needed if the you pushed close to hinges or at an angle very different from 900.
β’ As shown for Figure T (b) β (c ) even though πΉ and π were oriented at angle π which was
different from 900
they both have perpendicular components relative to each vector
enabling easy rigid body rotation
β’ The interaction of πΉ and π can be visualized using a vector or cross product,
π = π Γ πΉ, and this result in another vector known as torque whose direction can be
determined by right-hand rule and the magnitude is,
π = πβ₯πΉ = ππ πππ πΉ = ππΉπ‘ = π πΉπ πππ = ππΉπ πππ.
β’ Unit of torque π is π. π, but do not confused it with Work which has the same SI units
6. Newtonβs 2nd Law
β’ Having previously shown the analogy of linear and rotational motion, we can further
show that Newtonβs 2nd law applies to rotational motion
πΉπ‘ = πππ‘ = ππΌπ (1)
Multiplying both side of (1) by π yields:
π = ππΉπ‘ = ππππ‘ = ππ2 πΌ = πΌπΌ (2)
Therefore Newtonβs second law for rotational motion is
ππππ‘ = πΌπΌ
7. Worked Examples
1. Figure Ex1 shows a uniform disk, with mass π = 2.5 ππ and radius π = 20 ππ, mounted on a
fixed horizontal axle. A block with mass π = 1.2 ππ hangs from a massless cord that is
wrapped around the rim of the disk. Find the acceleration of the falling block, the angular
acceleration of the disk, and the tension in the cord. The cord does not slip, and there is no
friction at the axle.
Figure Ex1.
8. Worked Problemsβ¦.
For the block with mass π = 1.2 ππ:
π β ππ = π(βπ) (1)
For a solid disk with mass π = 2.5 ππ:
Since ππππ‘ = πΌπΌ
πΌ =
1
2
ππ 2
(moment of inertia of solid disk), and
βπ π =
1
2
ππ 2(βπΌ), and dividing both sides by π and solve for π would yield:
π =
1
2
ππΌ =
1
2
πππ‘ =
1
2
ππ (2)
π = 4.8 π/π 2 and T can be calculated from (2) and π = 6.0 π
πΌ =
π
π
=
4.8
0.20
= 24 πππ/π 2
9. Worked Problemsβ¦.
2. Let the disk in Figure Ex1 start from rest at time π‘ = 0 and also let the tension in the
massless cord be 6.0 π and the angular acceleration of the disk be β24 πππ/π 2. What is
its rotational kinetic energy πΎ at π‘ = 2.5 π ?
Solution:
You already calculated πΌ =
1
2
ππ 2
and the acceleration is a constant (β24 πππ/π 2
)
and therefore equations of motion can be used.
πΎ =
1
2
πΌπ2
, while π0 = 0 πππ/π at time π‘ = 0 π , what is π at time π‘ = 2.5 π ?
Using equations of motions one can work out π,
π = π0 + πΌπ‘,
π = 0 β 24 2.5 = β60 πππ/π , and πΎ =
1
2
πΌπ2
=
1
2
1
2
2.5 0.20 2
β60 2
= 90 J