Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 9
Rotational Kinetics
Younes Sina

2. Rotational kinetics deals with the cause of rotation. In order to cause rotation in
an object, torque must be applied.
When a force is applied to the handle of a wrench (normally perpendicular to it),
the product of the force and the perpendicular distance it has from the center of
the bolt is called the torque or moment of that force.
By convention, CCW is usually taken to be positive, and thus CW is negative;
therefore, torque is a vector quantity.

3. Example : In the figure shown, find the torque of force ( F )
about point A, the point at which the beam is fixed into the wall.
Solution: The perpendicular distance from F to point A is 1.2 m and
the tendency of rotation is clockwise.
dFTA  TA=(25 N).(1.2m)= 30 Nm

4. If more than one force is generating torque on an object,
then the sum of torques or the net torque should be calculated.
Example : In the figure shown, find the net torque of the forces shown about point A,
the point at which the beam is fixed into the wall.

6. Example : In the figure shown, find the net torque of the
forces shown about point B, the point at which the beam
is fixed into the wall.

8. Example : In the figure shown, find the torque of force F about point A.

11. Example :
In the following 4 figures, determine (a) the case for which the torque of the 10-N
force is maximum and explain why. (b) What is the value of torque about point D in
Fig. 4? (c) How do you determine the perpendicular distance ( d) ?

12. Example :
In the figure shown, find the net torque of the forces shown about point A.

14. An object is said to be in rotational equilibrium if the net torque acting on it is zero
that means ΣT = 0. The torque sum may be taken about any single point on the
object or out of it. It is usually taken about a point for which perpendicular
distances from all acting forces are convenient to calculate. ΣT = 0 means that the
sum of CW torques equals the sum of CCW torques; in other words, the total CW
tendency of rotation is equal to the total CCW tendency of rotation.
Rotational Equilibrium
Example : In the figure shown, find F such that the seesaw is rotational equilibrium
that means it is neither rotating cw nor ccw.

16. Example : In the figure shown, assume that the beam is weightless and find the
unknown force F that brings the seesaw in rotational equilibrium.

18. The Center of Mass of Uniform Objects
The center of mass of an object is the point that all of its mass can be assumed to
be concentrated at. For geometrically symmetric objects, such as rectangular
boxes, spheres, cylinders, etc., geometric center is easily determined. Geometric
center is the same thing as the center of symmetry. If the material of an object is
uniformly distributed throughout its volume to where it has the same mass
everywhere, the geometric center and the mass center are the same point in that
object. For geometrically symmetric and constant density objects, the geometric
center is the same point as the mass center is.
For a uniform and rectangular plank of wood, the geometric center or mass
center is at its midpoint. It means that all of the mass of such plank can be
assumed to be concentrated at its geometric center or its middle.

19. Example : In the figure shown, an 8.0-m long 550-N heavy uniform plank of wood
is pivoted 2.0m off its middle at P to form an unbalanced seesaw. It is then loaded
with a 420-N force as shown. Find the magnitude of F that keeps the plank in
rotational equilibrium.

21. Example : In the figure shown, a force of 300.N is applied on the lever at A.
Find the reaction force F that the crate exerts on the lever at B.

23. Torque-Angular Motion Relation:
Newton's 2nd Law for straight line motion is ΣF = Ma
M = mass of the object
a= linear acceleration
Newton's 2nd Law for rotational motion is ΣT = I α
I= mass moment of inertia
α = angular acceleration
M is a measure of the resistance an object has toward being
linearly accelerated.
I is a measure of the resistance an object has toward being
angularly accelerated about an axis.
I = MR2

25. Moment of Inertia of a Thin Ring:
The mass moment of inertia of a thin ring about its centroidal axis is given by
I = MR2
The "centroidal axis" is an axis that is perpendicular to the plane of a ring
and passes through its center. The vertical rods in the above figures are
centriodal axes.

26. Moment of Inertia of a Solid Disk:
A solid disk of radius R may be thought as a combination of infinite number of
thin rings which radii range from zero to R. As we go from the inner rings to the
outer ones, each ring has more mass as well as a greater radius that make its
corresponding (I) increasingly greater. To find the total (I) for all rings, calculus
must be employed and integration performed.
The result is: I = (1/2) MR2 .
Again, this is good only for finding I about the centriodal axis of a disk.

27. Example : A 1.50-kg thin ring is connected to weightless spokes of
length 45.0 cm and then attached to a vertical rod (axis) that is free
to rotate as shown in following Figure. The apparatus is initially at
rest. The vertical axis is then given a twist by a constant net torque
of 6.08Nm for a period of 2.00 seconds. Find
(a) the angular acceleration of the spinning mass
(b) the angle it travels within the 2.00-sec. period.
Assume frictionless rotation.

28. Solution:
(a) The mass moment of inertia ( I ), or the resistance toward rotation is
I = MR2
I = MR2
I = (1.50 kg)(0.45 m)2
I = 0.304 kg m2
ΣT = I a
6.08Nm = (0.304kg m2 ) a
a = 20.0 rd/s2
(b) q = (1/2) a t 2 + wit
q = (1/2) a t 2
q = 40.0 rd

29. Example : A 255-kg solid disk of radius 0.632 m is free to spin about a
frictionless axle-bearing system in a vertical plane as shown. A force
of 756 N is applied tangent to its outer edge for 1.86 s that puts it in
rotation starting from rest. Calculate
(a) the mass moment of inertia of the disk,
(b) the torque applied as a result of the applied force,
(c) the angular acceleration of the disk, and
(d) its angular speed and the linear speed of points on its edge at the
end of the 1.86-sec. period.

30. Solution:
(a)
I = (1/2)MR2 = (0.5)(255kg)(0.632m)2 = 50.9kgm2
(b)
TC = F R = (756N)(0.632m) = 478 Nm
(Torque of F about the axle at C )
(c)
ΣTC = I α
478 Nm = (50.9 kgm2)(α)
α = 9.39 rd/s2
(d) α =(ωf -ωi)/t
ωf =α t
ωf = (9.39 rd/s2)(1.86s) = 17.5 rd/s
vf = Rωf
vf = (0.632m )( 17.5 rd/s2) = 11.1 m/s

31. Homework :
problems 1 through 4 chapter 9