Chapter 9
The Kinetic of Rotation:
In Chapter 4, the kinetic of straight line motion was discussed where ΣF = Ma was the kinetic equation. ΣF was the net force or the cause of motion, M or mass was treated as a measure of resistance toward straight line motion, and a the acceleration caused by ΣF.
Here, in Chapter 9, the kinetic of rotation will be studied where ΣT = Iα is the kinetic equation for rotation. Here, ΣT stands for the net "torque" that is the cause of rotation, I is a measure of resistance of an object toward being rotated, and α is the angular acceleration caused by the net torque ΣT.
Rotational kinetic deals with the cause of rotation that is torque. In order to cause rotation in an object, torque must be applied. If the applied torque causes rotation, the relation between the applied torque and the pace of generated rotation is the basis of rotational kinetic. If rotation does not occur, the applied torque is often referred to as the "moment" or the "bending moment." The first topic to study is "the moment of a force" or the "the bending moment of a force."
Torque or Moment of a Force:
When a force is applied to the handle of a wrench (normally perpendicular to it), the product of the force and the perpendicular distance from the center of the bolt is called the "torque" or the "moment " of that force. Mathematically, torque is the product of a force and a perpendicular distance, or torque is the product of a perpendicular force and a distance. Torque may be either clockwise (cw), or counterclockwise (ccw). By convention, ccw is usually taken to be positive, and therefore cw torque is negative. Because of having direction, torque is a vector quantity.
The SI unit for torque is Nm. In both American systems, torque is expressed in ftlb.
The following figure shows the same force F applied to different points of a wrench handle. It shows how torque varies from maximum to zero if the perpendicular distance varies from maximum to zero. If you were to open the bolt shown, you wouldn't push the wrench handle as shown in (4)! You would naturally do it as shown in (1), correct?
Example 1: In the figure shown, find the torque of force F about point A, the point at which the beam is fixed into the wall.
If more than one force is generating torque on an object, then the sum of torques or the net torque should be calculated.
Example 2: 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.
In this problem the overall torque, bending moment, or tendency of rotation is clockwise as the () in 11.5Nm indicates.
Example 3: 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.
In this problem the torque, bending moment, or tendency of rotation is clockwise as the () in 25Nm indicates.
Example 4: In the figure shown, find the torque of force F = 25N at 54^{o} about point A. Hint: Resolve the force into two components: one along the beam and one perpendicular to the beam.
As you may have noticed, it is possible to draw a perpendicular line from Point A to vector 25N sin54^{o}. The length of the perpendicular line becomes 1.6m. If you decide to draw a perpendicular line from A to vector 25N cos54^{o}, you need to first extend the vector from its tail to the left. Doing this makes the extension to pass through A. Point A falls on the extended vector and makes the perpendicular line to have a length of 0.
Example 5: In the following 4 figures, determine (a) the case for which the torque of the 10N 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_{^}?
Solution: (a) The case in Fig. 1, has the maximum torque because the magnitude of the applied force is the same for all cases, but the perpendicular distance in Fig. 1 is the greatest. That makes the torque in Fig. 1 maximum.
(b) Zero, because the perpendicular distance in Fig. 4 is zero.
(c) To find d_{^} , the line of action of the applied force must be extended, and then from the desired point a line segment be drawn perpendicular to it as shown in Figures 1 through 3.
Example 6: In the figure shown, find the net torque of the forces shown about point A.
Rotational Equilibrium
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 at which the calculation of perpendicular distances is convenient. 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.
Example 7: In the figure shown, find F such that the seesaw is in rotational equilibrium.
Solution: To begin with, the seesaw just by itself (without the loads) is in rotational equilibrium. This is because of the uniformity of the wood plank. The weight of the plank is distributed evenly throughout its length. The weight of its left half equals the weight of its right half. In this case, we may solve the problem without incorporating the weight of the plank. Taking ccw to be positive, we may write:
ΣT = 0 ; + F(2.0m)  (430N)(3.5m) = 0. ; F = 750N.
Example 8: In the figure shown, assume that the beam is weightless and find the unknown force F that brings the seesaw in rotational equilibrium.
Solution:
ΣT = 0 ; +F(2.0m)
+ (300N)(2.9m)  (730N)(3.5m) = 0
; F = 840N.
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 density everywhere, the geometric center and the mass center are the same point in that object. In the above two examples, this was the case where the geometric center of the seesaw plank was indeed its center of mass as well. For geometrically symmetric and uniform 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. This is specially helpful in the following example:
Example 9: In the figure shown, an 8.0m long 550N uniform plank of wood is pivoted 2.0m off its middle at P to form an unbalanced seesaw. It is then loaded with a 420N force as shown. Find the magnitude of F that keeps the plank in rotational equilibrium.
Solution: Note that the 550N force in the figure is the weight of
the beam that is shown at G, its middle. Using point P as
the reference for summing up the torques, we get:
ΣT = 0 ; F (1.0m) 
(550N)(2.0m)  (420N)( 5.0m) = 0;
F = 3200N.
Example 10: In the figure shown, a force of 300N is applied on the lever at A. Find the reaction force F that the crate exerts on the lever at B.
Solution: As the 300N force pulls end A of the lever to the right, end B wants to move to the left. The crate on the floor stops it and exerts the reaction force F onto it (to the right) as shown. The lever will be in rotational equilibrium if no rotation occurs. Taking ccw to be positive and summing the torques acting on the lever (about point C) and set it equal to zero for equilibrium, we get:
ΣT_{C} = 0 ;
300N(2.0m) + F (0.50m) = 0
; F = 1200N.
Chapter 9 Test Yourself 1:
1) To cause rotation in an object about an axis, (a) a force that goes through that axis must be applied (b) a force that does not go through that axis must be applied (c) a torque must be applied (d) b & c. click here.
2) Torque is defined as the product of (a) a force and a parallel distance (b) a force and a perpendicular distance (c) a force and an angle.
3) When the line of action of a force passes through the point about which torque is to be calculated, (a) the perpendicular distance is zero (b) torque is maximum (c) clockwise rotation occurs. click here.
4) When the line of action of a force passes through the point about which torque is to be calculated, (a) that force does not generate any torque (b) that force either pulls that point or pushes it (c) both a & b.
Problem: A 1/2inch water pipe made of copper has a length of 1.00m and is horizontally fixed into a concrete wall at its left end. The right end of it is free. click here.
5) To easily bend this pipe clockwise about its fixed end, (a) the right (free) end of it must be pushed down (b) the middle of it must be pushed down (c) the left (fixed) end of it, nearest to the wall, must be pushed down.
6) The torque necessary to just bend this pipe clockwise is (a) the same no matter which point of it is pushed down (b) minimum, if the right end of it is pushed down (c) minimum, if the left end of is pushed down (d) both b & c.
7) The force necessary to just bend this pipe clockwise is (a) the same no matter which point of it is pushed down (b) minimum, if the right end of it is pushed down (c) neither a, nor b.
8) It takes a certain amount of bending moment (torque) to bend this pipe. Now, if this torque is provided by applying a downward force at its right end, the force is (a) maximum (b) minimum (c) neither a, nor b. click here.
9) It takes a certain amount of bending moment (torque) to bend this pipe. Now, if this torque is provided by applying a downward force very close to its left end, the force is (a) maximum (b) minimum (c) neither a, nor b.
10) If a downward force is applied exactly at its left end, bending does not occur because (a) the downward force passes through the point about which we want bending to occur and passing through that point means zero perpendicular distance and consequently zero torque (b) applying a downward force at the very left end practically pushes that end down and has no torque arm (c) both a & b. click here.
11) If the necessary torque to bend the pipe is 120Nm and the right end is pushed down, the applied downward force is (a) 6.0N (b) 24N (c) 120N.
12) If the necessary torque to bend the pipe is 120Nm and the middle of the pipe is pushed down, the applied downward force is (a) 60.N (b) 240N (c) 120N.
13) If the necessary torque to bend the pipe is 120Nm and 0.20m from the wall is pushed down, the applied downward force is (a) 600N (b) 150N (c) 1200N. click here.
14) If the necessary torque to bend the pipe is 120Nm and 0.25m from its right end is pushed down, the applied downward force is (a) 720N (b) 480N (c) 160N.
15) If the necessary torque to bend the pipe is 120Nm and 0.15m from its left end is pushed down, the applied downward force is (a) 800N (b) 360N (c) 1600N.
16) If in Example 5, Fig. 1, the angle of the 10N force is 42.0^{o} below horizontal, and the beam's length is 2.00m, then d_{┴} is (a) 1.0m (b) 1.33m (c) 1.49m.
17) If in Example 5, Fig. 3, the angle of the 10N force is 15.0^{o} below horizontal, and the beam's length is 2.00m, then d_{┴} is (a) 0.52m (b) 2.49m (c) 1.93m. click here.
18) In Example 5, Fig. 4, the angle of the 10N force is 0.0^{o} with the horizontal, and the beam's length is 2.00m, then d_{┴} is (a) 0.52m (b) 2.00m (c) 0.
19) For an object to be in rotational equilibrium, which sum acting on it must be zero? of (a) forces (b) torques (c) work.
20) The sum of torques acting on an object may be taken about (a) the center of mass of the object (b) any point on the object (c) any point even if it is not a point of that object (d) a, b & c. click here.
21) For ease of calculation, the sum of torques (moments) may be taken about a point of an object for which perpendicular distances from all forces can either be readily seen or easily determined. (a) True (b) False
22) If force F passes through point A, the torque of F about A is (a) maximum (b) not easy to calculate (c) 0.
23) The center of gravity of a rectangular and uniform sheet of metal that coincides with its geometric center, is at (a) its center (b) the point of intersection of its diagonals (c) both a & b. click here.
24) In Example 8, if the 3.5m is replaced by 2.5m and the 730N by 800N, the value of F becomes: (a) 565N (b)700N (c) 500N. First solve, then look at the solution of the example.
25) In Example 8, if the 300N force acts upward, the 730N is replaced by 800N, and the 0.900m is changed to 1.00m, the value of F becomes: (a) 3000N (b)1850N (c) 5500N. First solve, then look at the solution of the example.
Problem: Neatly redraw the figure of Example 9 and replace the 420N by 500N. Also, let the beam weigh 600N. Answer the following questions: click here.
26) The torque of the weight force (the 600N force) about point P is (a) 1200Nm (b) 1200Nm (c)  2400Nm.
27) The torque of the 500N force about point P is (a) 2500Nm (b) 2500Nm (c)  1500Nm.
28) The torque of F about point P is (a) 1.0F (b) +1.0F (c) 3.0F. click here.
29) Adding the above 3 torques corresponding to the 3 existing forces and, for rotational equilibrium, setting the sum equal to zero, results in an F value of (a) 3700N (b) 2700N (c) 4700N.
30) In Example 10, if distance BC is 0.25m instead of 0.50m, the magnitude of F becomes: (a) 2400N (b) 600N (c) 800N.
TorqueAngular Acceleration Relation:
Newton's 2nd Law for straight line motion is ΣF = Ma where M is the mass of the object. M is a measure of the resistance an object has toward being linearly accelerated.
Newton's 2nd Law for rotational motion is ΣT = Ia where I is the mass moment of inertia. I is a measure of the resistance an object has toward being angularly accelerated about an axis. a is, of course, angular acceleration.
We have so far learned the concepts of torque (T) and angular acceleration (a). What is left to learn here is the mass moment of inertia (I).
Mass moment of Inertia ( I ):
Consider a thin ring of mass M and radius R that is attached to a vertical rod via a number of (say 3) weightless spokes as shown in Fig. A. Also, consider another ring that has the same mass M but a radius of 2R and with weightless spokes as in Fig. B.
If
each vertical rod (axis of rotation), Figures A & B, is
held with two fingers and twisted with the intention that they both
accelerate at the same rate, it will be felt that it takes 4
times more torque to twist the system with M at 2R
than it takes to twist the system with M at R. This can be measured (verified) by experiment and proved mathematically. In Fig. B, if a ring of radius 3R is used instead, keeping the same mass, the task of rotation becomes 9 times more difficult; in other words, it would take 9 times more torque to accelerate the ring with M at 3R than it takes to accelerate the system at the same rate with M at R. What this means is that the resistance toward rotation of mass M is proportional to the square of radius (R^{2}). The resistance toward rotation of a thin ring is called the "mass moment of inertia ( I ) of the ring." It is mathematically written as I = MR^{2}. The same way
that, in straight line motion,
the M of an object
is a measure of its resistance toward being linearly accelerated and
ΣF
= Ma, 
in, rotational motion, the I of an object is a measure of its resistance toward being angularly accelerated and ΣT = Ia. 
Mass Moment of Inertia of a Thin Ring:
The mass moment of inertia of a thin ring about its centroidal axis is given by
I = MR^{2}.
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.
Mass 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. Of course, as we go from the inner rings to the outer ones, each ring has more mass as well as a greater radius that makes 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) MR^{2}. Again, this is good only for finding I about the the centriodal axis of a disk.
Note that for a solid disk, M is several times greater than that of a thin ring of the same radius and material.
Example 11: A 1.50kg 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 Fig. A above. 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 ring, and (b) the angle it travels within the 2.00sec. period. Assume frictionless rotation.
Solution: (a) The mass moment of inertia (I), or the resistance toward rotation for a ring is I = MR^{2}.
I = MR^{2} ; I = (1.50 kg)(0.45 m)^{2} ; I = 0.304 kg m^{2} (in SI).
ΣT = I a ; 6.08Nm = (0.304kg m^{2}) a ; a = 20.0 rd/s^{2}.
(b) q = (1/2)at^{ 2} + w_{i}t ; q = (1/2)at^{ 2} ; q = 40.0 rd.
Example 12: A 255kg solid disk of radius 0.632m is free to spin about a frictionless axlebearing system in a vertical plane as shown. A force of 756N is applied tangent to its outer edge for 1.86s 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 1.86s period.
Solution: (a) I = (1/2)MR^{2 }= 0.5(255kg)(0.632m)^{2} = 50.9kgm^{2}. (b) T_{C} = FR = 756N(0.632m) = 478Nm. (Torque of F about the axle at C) (c) ΣT_{C} = Ia ; 478Nm = 50.9kgm^{2}(a) ; a = 9.39rd/s^{2}. (d) α =(ω_{f} ω_{i})/t ; ω_{f} =αt ; ω_{f} = (9.39 rd/s^{2})(1.86s) ; ω_{f} = 17.5 rd/s. v_{f} = Rω_{f} ; v_{f} = 0.632m (17.5 rd/s^{2}) = 11.1 m/s. 

K. E. and Angular Momentum in Rotational Motion:
In straight line motion, K.E. = (1/2)Mv^{2}. In rotational motion, the equivalent is obviously K.E. = (1/2) Iw^{2}.
In straight line motion, linear momentum is Mv. In rotational motion, angular momentum is Iw.
Example 13: In Example 12, find the K.E. and the angular momentum of the solid disk at t = 1.86s.
Solution: At t = 1.86s, ω_{f} = 17.5 rd/s ; therefore,
K.E. = (1/2)Iω^{2} = 0.5(50.9 kgm^{2})(17.5rd/s)^{2} = 7790 J.
Angular Momentum = Iω = (50.9 kgm^{2})(17.5rd/s) = 891 kg m^{2}/s.
Chapter 9 Test Yourself 2:
1) Similar to Newton's Laws for motion along a straight line, Newton's First Law applied to rotation should state that " If the net torque applied to an object (a wheel for example) is zero, the object is either not rotating or if it is rotating, it rotates about its centroidal axis at a constant angular (a) velocity (b) acceleration (c) displacement." click here.
2) Newton's Second Law applied to rotation should state that " A nonzero net torque ΣT acting on an object of mass moment of inertia I, creates an angular acceleration a in it such that (a) ΣT = Ma (b) the object rotates at constant angular velocity (c) ΣT = Ia. click here.
3) Mass moment of inertia I of a rigid object, is a measure of the resistance that the object has toward (a) motion along a straight line (b) a backandforth or upanddown motion (c) rotation about an axis.
4) The mass moment of inertia for a solid disk about its centriodal axis is (a) I = MR^{2} (b) I = (1/2)MR (c) I = (1/2)MR^{2}.
5) The mass moment of inertia for a ring about its centriodal axis is (a) I = MR^{2} (b) I = πMR (c) I = (1/2)MR^{2}. click here.
6) The mass moment of inertia of a 400kg solid disk of radius 2.00m about its centriodal axis is (a) 1600kgm^{2} (b) 800kgm^{2} (c) 600kgm^{2}.
7) The mass moment of inertia of a 4.00kg metal ring of radius 0.50m about its centriodal axis is (a) 1.0kgm^{2} (b) 0.5kgm^{2} (c) 2.0kgm^{2}. click here.
Problem: Solve Example 12, with the assumption that the axle imposes a frictional torque of 78Nm. Since the externally applied force of 756N to the rim creates a CCW torque to rotate the wheel in CCW direction, the frictional torque that always opposes the direction of rotation, acts CW.
8) The net torque is therefore, (a) 576Nm (b) 400Nm (c) 756N.
9) Its angular acceleration then is (a) 7.86 rd/s^{2} (b) 11.31 rd/s^{2} (c) 0.86 rd/s^{2}.
10) If the solid disk is accelerated for 2.45s, at the end of this period, its angular speed is (a) 19.3 rd/s (b) 19.3 rd/s^{2} (c) 9.3 rd/s. click here.
11) The linear speed v of points on its outer edge at R = 0.632m is (a) 11.2 rd/s (b) 21.2 rd/s^{2} (c) 12.2m/s.
12) The rotational K.E. of the disk at the end of the 2.45s period is (a) 9480 J (b) 7840 watts (c) neither a nor b.
13) The angular momentum of the disk at the end of the 2.45s period is (a) 289 kgm^{2}/s (b) 982 kgm^{2}/s (c) 829 kgm^{2}/s.
Problems: [Apply 3 significant figures to all numbers].
1) In each of the figures shown, assuming CCW torque to be positive, find the net torque of the system of forces about the axis passing through the indicated point:
2) In the figure shown, calculate the position of the the 180 lb force for the equilibrium of the seesaw.
3) A solid disk of mass 200kg and radius 0.800m goes under a constant net torque of 160Nm for 3.00s. Find (a) its mass moment of inertia, (b) its angular acceleration, and (c) its final speed. The friction at its axle is negligible.
4) A bicycle wheel (assumed as a thin ring) with a mass of 2.00kg and a radius of 35.0cm is initially at rest. Find (a) its mass moment of inertia about its centroidal axis. A 16.0N force is applied tangent to its outer edge for 1.50s and then released. Find (b) the resulting applied torque. If the friction at its axle generates a counter torque of 0.100Nm, find (c) its angular acceleration, and (d) its final angular speed. The wheel then slows down and eventually comes to stop due to the frictional torque present at its axle. Calculate (e) the angular deceleration, and (f) the stopping time.
5) The anchor wheel in a car is also called the "fly wheel." This wheel is attached to the crank shaft and one of its important functions is to absorb the pulsations of the pistons and make the crank shaft spin smoothly without vibration. The starter (electric motor) gets engaged with the teeth on the edge of this wheel to crank the engine. The clutch plate also gets attached (by friction) to this wheel when clutch is released. Anyway, the mass of a flywheel (a solid disc) is 18.0kg and has a radius of 19.0cm. Calculate (a) its mass moment of inertia about its centroidal axis (that is its crankshaft), and (b) its rotational K.E. when it is turning at 3600rpm. Note that the translational kinetic energy (Chapter 6) was given by K.E. = (1/2)Mv^{2}. Here, in Chapter 9, the rotational kinetic energy, K.E. = (1/2) Iω^{2}.
Answers: 1) 57.0Nm, 122Nm, 491Nm, 168 ftlb, 64.0Nm 2) 2.20m
3) 64.0kgm^{2}, 2.50rd/s^{2}, 7.50rd/s 4) 0.245kgm^{2}, 5.60Nm, 22.4 rd/s^{2}, 33.7rd/s, 0.408rd/s^{2}, 82.6s
5) 0.325 kgm^{2}, 23100J