Rotational Kinematics and Dynamics
Name :
Date :
Level : Physics I
Teacher : Kim
►Angular Displacement, Velocity, and Acceleration Review
- A rigid object rotating about a fixed axis through O perpendicular to the
plane.
- An arbitrary point P is at a fixed distance r from the origin and rotates
about it in a circle or radius r.
-The relationship between the arc length s, radius r and angle θ is
θ=
- θ is the ratio of an arc length and the radius of the circle with the unit radian, where one radian is the
angle subtended by an arc length equal to the radius of the arc
1rad = 57.3° = 0.159revs
- Angular speed is defined as
w=
and angular acceleration is defined as
α=
When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and
has the same angular speed and the same angular acceleration
►Rotational Kinematics: Rotational Motion with Constant Acceleration
When an object’s rotational motion is under constant angular acceleration, the kinematic relationships can
be expressed as
wf =wi + αt ,
*~Compare with kinematics : vf =vi + at ,
θf = θi + wi t + αt²,
xf = xi + vi t + at²,
wf ²= wi² + 2α( θf – θi )
vf ²= vi ² + 2a(xf – xi )
Q1) Rotating Wheel
A wheel rotates with a constant angular acceleration of 3.5rad/s2. If the angular speed of the wheel is
2rad/s at ti=0, (a) through what angle does the wheel rotate in 2secs? How many rotations does the wheel
make? (b) What is the angular speed at t=2s?
Ans) (a) 630°, 1.75revs (b) 9rad/s
Q2) A wheel starts from rest and rotates with constant angular acceleration and reaches an angular speed
of 12rad/s in 3s. (a) Find the angular acceleration of the wheel and (b) the angle (in radians) through
which it rotates in this time. (c) How many rotations(=rev) does the wheel make through this time?
Ans) 4rad/s2, 18rads , 2.86revs
Q3) An electric motor rotating a grinding wheel at 100rev/min is switched off. Assuming constant
negative angular acceleration of 2rad/s2, (a) how long does it take the wheel to stop? (b) Through how
many rotation does it turn during the time found in part(a)?
Ans) (a) 5.24s , 4.38revs
►Angular and Linear Quantities
- Point P moves in a circle where the linear velocity v is always tangent to
the circular path. Hence, called tangential velocity.
- The tangential speed and angular speed has a relationship as
v = rw
- The tangential acceleration and angular acceleration has a relationship as
=r
at =
= rα
=> at = rα
- Tangential acceleration will only exist if point P is moving with changing
tangential speed
- Furthermore, if an object is moving in circular motion, there is another
acceleration, called radial acceleration(or centripetal acceleration)
ar =
= rw²
- The total acceleration will then be the combination of tangential
acceleration and radial acceleration
- Hence, the total acceleration for point P is
a=
=
=
Q4) A racing car travels on a circular track with a radius of 250m. If the car moves with a constant linear
speed of 45m/s, find (a) its angular speed and (b) the acceleration.
Ans) 0.18rad/s, 8.1m/s2
Q5) A wheel 2m in diameter lies in vertical plane and rotates with a constant angular acceleration of
4rad/s2. The wheel starts at rest at t=0, and the point P on the rim makes an angle of 57.3° with the
horizontal at this time. At t=2s, find the (a) the angular speed of the wheel, (b) the linear speed and the
acceleration of the point P, and (c) the angular position of the point P.
Ans) 8rad/s, (b) 8m/s, a=64.12m/s2 (c) 9rads
►Rotational Energy
- The rotational kinetic energy of a rotating rigid object is expressed as
KR =
, where I = ∑mi ri² : rotational inertia(=moment of inertia)
*~Compare with K =
- The moment of inertia(or rotational inertia) is a measure of the resistance of an object to changes in its
rotational motion, just as mass is a measure of the resistance of an object to changes in its linear motion.
- Note that mass is an intrinsic property of an object, where I depends on the physical arrangement of that
mass
►Calculation of Rotational of Inertia ‘I’ (revisit)
I = ∑mi ri² : moment of inertia
- The moment of inertia for a point particle of mass ‘m’is expressed as
I = mr² (Figure1)
Figure1
- The moment of inertia for a point particle of mass ‘m’is expressed as
I = 2mr² (Figure2)
Figure2
Q7) Four Rotating Mass
Four tiny spheres are fastened to the corners of a
frame of negligible mass lying in the xy plane. We
shall assume that the sphere’s radii are small
compared with the dimensions of the frame. The
mass of M=2kg, m=1kg.
(a) If the system rotates about the y-axis with an
angular speed w=3rads, find the rotational of inertia
and the rotational kinetic energy about this axis.
(b) Suppose the system rotates in the xy plane
about an axis through O(z-axis) with the same
angular speed. Calculate the rotational of inertia
and the rotational kinetic energy about this axis.
y
m
3
M
M
O
x
-4
4
-3
m
Ans) (a) I=64kg·m2, KR=288J (b) I=82kg·m2, KR=369J
►Torque Revisit
∑F = F1 + F2 + ∙ ∙ ∙ ∙FN = ma
 ∑τ = τ1 + τ2 + ∙ ∙ ∙ ∙ τN = Iα
1. A meter-stick is place on a fulcrum at the 50cm mark. m2 > m1
10cm
50cm
m1
90cm
4kg
m2
∑τ = τL + τR = Iα => ∑τ = – rL×FgL + rR ×FgR = Iα
Q8) Atwood’s machine revisit.
The pulley is free to rotate on a horizontal axis through
its center. There is no slippage between the cord and the
pulley. Assume the rotational inertial of the pulley is I =
8kg·m2 and the radius R=2m. The mass of m1 is 2kg and
m2 is 4kg. Find the acceleration of both blocks and the
tension in each side of the cord.
R=2m
T2
T1
m1
m2
Ans) 2.45m/s2, T1=29.4N, T2=24.5N
*~notice that T1> T2. This causes the pulley turn clockwise~*
Q9) Two block system revisit.
The frictional force between the block and table is 20N.
The mass m1 is 5kg and m2 is 2.5kg. If the rotational
inertia of the wheel is 4kg·m2 and the radius r=0.3m, find
the acceleration of the blocks. Assume no slippage
between the rope and wheel.
m1
T2
r
T1
m2
Ans) 0.087m/s2
Q10) A flywheel is a solid disk that rotates about an axis that is
perpendicular to the disk at its center. Rotating flywheels provide means
for storing energy in the form of rotational kinetic energy and are being
considered as a possible alternative to batteries in electric cars. The
gasoline burned in a 300-mile trip in a midsize car produces about
1.2×109J of energy. How fast would a 13kg flywheel with a radius of 0.3m
have to rotate to store this much energy? Answer in rev/s
Ans: 1.02×104rev/s
Q11) Rotating Rod
A uniform rod of length L=4m and mass M=1.5kg is attached at
one end to a frictionless pivot and is free to rotate about the pivot
in the vertical plane. The rod is released from rest in the
horizontal position.
(a) What is the initial angular acceleration of the rod and initial
linear acceleration of its right end? (b) What is the angular speed
when the rod reaches its lowest position? (c) Determine the
linear speed of the center of mass and the linear speed of the end
of the rod when it reaches its lower position
Ans) (a) α=3.68rad/s2, at =14.7m/s2 (b) 2.71rad/s (c) vcm = 5.42m/s, vend=10.84m/s
The rotational of inertia for homogenous rigid bodies are expressed as below
Formula Summary
● If a particle rotates in a circle of radius r through an angle of θ(measured in radians), the arc
length it moves through is s= rθ.
s= r θ
● Angular speed is defined as
w=
or θ = s / r ,
1rad = 57.3° = 0.159revs
and angular acceleration is defined as
α=
● When an object’s rotational motion is under constant angular acceleration, the kinematic
relationships can be expressed as
wf =wi + αt ,
θf = θi + wi t + αt²,
wf ²= wi² + 2α( θf – θi )
● When a rigid object rotates about a fixed axis, the angular position, angular speed, and angular
acceleration are related to the linear position, linear speed, and linear acceleration through the
relationships
s= r θ
,
v=rw
, at = rα
● The rotational inertia of a system of particles
I=
i
i
2
● The rotational kinetic energy of a rotating rigid object with an angular speed of w can be expressed as
KR =
● If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object
undergoes an angular acceleration α, where
∑τ = Iα
● Total linear acceleration will then the combination of tangential acceleration and radial acceleration
a = at + ar
where at is the tangential acceleration and ar is the centripetal(=radial) acceleration (ar = v2/r)