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ROTATIONAL MOTION III
Torque, Angular Acceleration and Rotational
Energy
NEWTON’S 2ND LAW & ROTATIONAL
MOTION
The net force on a particle is proportional
to its TANGENTIAL acceleration.
The net torque on a particle is
proportional to its ANGULAR
acceleration.
ROTATIONAL INERTIA

I = ∑mr2

Rot. Inertia = ∑ masses of particles x radius2


Smaller radius 
Lower Rotational Inertia
Larger radius 
Higher Rotational Inertia
EXAMPLE

The motor of an electric saw brings the circular blade up to
the rated angular speed of 80 rev/s in 240 rev. If the
rotational inertia of the blade is 1.41 x 10 -3 kg m2, what net
torque must the motor apply to the blade?
ROTATIONAL WORK

The rotational work WR done by a constant
torque τ in turning an object through an angle θ
is


WR = τθ
Θ must be in radians; Unit – Joule (J)
The rotational work done by a net external
torque equals the change in rotational kinetic
energy
 Therefor the formulas for rotational work and
energy are analogous to the translational
formulas

ROTATIONAL WORK AND KINETIC ENERGY
Work:
Rotational
Translational
W = 
W = Fd = Fx
Kin. Energy: K = ½ I 2
Power:
P = 
Work-Energy Theorem:
W = K = ½ I 2 - ½ I o2
K = ½ mv2
P = Fv
W = ½ mv2 - ½ mvo2
ROLLING BODIES




The kinetic energy of a rolling body (without
slipping) relative to an axis through the contact
point is the sum of the rotational kinetic energy
about an axis through the center of mass and the
translational kinetic energy of the center of mass.
K = ½ ICM 2 + ½ mvCM2
total = rotational + translational
KE
KE
+
KE
If the rolling body experiences a change in height then
potential energy (mgh) must also be included
CONCEPT CHECK
EXAMPLE

A 1 kg cylinder with a rotational inertia of 0.0625 kg m2
rolls without slipping down a one meter high incline. At
the bottom of the incline the cylinder’s translational speed
is 3.13 m/s. What is the cylinder’s angular velocity?
EXAMPLE
A very common problem is to find the velocity of a ball
rolling down an inclined plane. It is important to realize
that you cannot work out this problem they way you
used to. In the past, everything was SLIDING. Now the
object is rolling and thus has MORE energy than
normal. So let’s assume the ball is like a thin spherical
shell and was released from a position 5 m above the
ground. Calculate the velocity at the bottom of the
Ebefore  Eafter
incline.
U g  KT  K R
Ebefore  Eafter
If you HAD NOT
included the
U g  KT
mgh  1 mv 2  1 I 2
2
2
rotational kinetic
2
2
1
2
mgh 
mv energy, you see the
v  R I sphere@ cm 
mR
3
2
answer is very much
2
2
1
v
2
2
mgh 
mv different.
mgh  1 mv  1 ( 2 mR )( 2 )
2
2
2 3
R
1 v2
2
2
gh

1
1
gh 
v  v
2
2
3
v  6 gh  6 (9.8)(5)  7.67 m/s
5
5
v  2 gh  2(9.8)(5)  9.90 m / s