Chapter 9
Rotational Kinetics
Rotational kinetics deals with the cause of rotation. 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 kinetics. If rotation does not occur, the torque applied is often referred to as "moment." The first topic to study is " the 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 it has from the center of the bolt is called the torque or moment of that force. Mathematically, torque is the product of force and perpendicular distance, or torque is the product of perpendicular force and distance. Torque may be either clockwise (CW), or counterclockwise (CCW). By convention, CCW is usually taken to be positive, and thus CW is negative; therefore, torque is a vector quantity.
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 net tendency of rotation is clockwise as the () sign 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 net tendency of rotation is clockwise as the () sign in 25Nm indicates.
Example 4: In the figure shown, find the torque of force F about point A.
As you may have noticed, it is possible to draw a perpendicular line from Point A to vector 25N sin54º. The length of the perpendicular line becomes 1.6m. If you decide to draw a perpendicular line from A to vector 25N cos54º, 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 0.0.
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 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 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.
Example 7: In the figure shown, find F such that the seesaw is rotational equilibrium that means it is neither rotating cw nor ccw.
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.
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 plank was indeed its center of mass as well. 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. This is specially helpful in the following example:
Example 9: In the figure shown, an 8.0m long 550N 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 420N force as shown. Find the magnitude of F that keeps the plank in rotational equilibrium.
Example 10: 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.
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 an 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) maximum, 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) maximum, if the right end of it is pushed down (c) minimum, if the left end of is pushed down (d) neither a, nor b, nor c.
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 42.0º 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 15.0º 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 0.0º 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 300.N 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 a value for F 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 Motion Relation:
The same way a net constant force along a straight path creates constant acceleration, a net constant torque applied to an object creates constant angular acceleration in that object. Newton's Second Law for both translational and rotational motions are shown below:
Newton's 2nd Law for straight line motion is: ΣF = Ma ; where M is the mass of the object.
Newton's 2nd Law for rotational motion is: ΣT = I a ,
where ( I ) is the mass moment of inertia of the object and a is its angular acceleration.
We have learned the concept 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 mass M that is attached to a weightless rod of length R and the rod is connected to a vertical rod at right angle as shown below (Fig. A). Assume that the vertical rod is free to rotate about itself and can be put into rotation by applying torque to the vertical axis. Also, consider exactly the same device but with a rod of length 2R as shown in (Fig. B).
Example 11: A 1.50kg mass is connected to a weightless rod of length 45.0 cm and then attached to a vertical 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 mass, 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 is I = MR^{2}.
I = MR^{2} ; I = (1.50 kg)(0.45 m)^{2} ; I = 0.304 kg m^{2} ( Note the unit of I ).
ΣT = I a ; 6.08Nm = (0.304kg m^{2} ) a ; a = 20.0 rd/s^{2}.
(b) q = (1/2) a t^{ 2} + w_{i} t ; q = (1/2) a t^{ 2} ; q = 40.0 rd.
Moment of Inertia of a Thin Ring:
The mass M at radius R (as shown above) can be distributed uniformly along a thin ring as shown below. Its mass moment of inertia ( I ) will still be given by
I = MR^{2}.
(This is good only for finding I about the axis through the center and normal to the ring's plane).
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 rings, 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 axis through the center and normal to the disc's plane.
Note that for a solid disk, M is much greater than that of a thin ring of the same radius and material.
Example 12: A 255kg solid disk of radius 0.632m is free to spin on 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 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 1.86sec. period.
Solution: (a) I = (1/2)MR^{2 } = (0.5)(255kg)(0.632m)^{2} = 50.9kgm^{2}. (b) T_{C} = F R = (756N)(0.632m) = 478 Nm. ( Torque of F about the axle at C ) (c) ΣT_{C} = Ia ; 478 Nm = (50.9 kgm^{2})( a) ; a = 9.39 rd/s^{2}. (d) α =(ω_{f} ω_{i})/t ; ω_{f} =α t ; ω_{f} = (9.39 rd/s^{2})(1.86s) = 17.5 rd/s. v_{f} = Rω_{f} ; v_{f} = (0.632m )( 17.5 rd/s^{2}) = 11.1 m/s.

Kinetic Energy 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) I w^{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 at constant angular (a) velocity (b) acceleration (c) angular 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 = M a (b) the object rotates at constant angular velocity (c) ΣT = I a. click here.
3) Mass moment of inertia, I of a rigid object, is the resistance that object shows toward (a) motion along a straight line (b) a backandforth or upanddown motion (c) rotation.
4) The mass moment of inertia for a solid disk about the axis through its center and normal to its plane 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 the axis through its center and normal to its plane 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 the axis through its center and normal to its plane 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 the axis through its center and normal to its plane is (a) 1.0kgm^{2} (b) 0.5kgm^{2} (c) 2.0kgm^{2}.
8) Solve Example 12, with 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, will act CW. The net torque is therefore, (a) 576Nm (b) 400Nm (c) 756N.
9) The angular acceleration then is (a) 7.86 rd/s^{2} (b) 11.31 rd/s^{2} (c)  7.86 rd/s^{2}. click here.
10) If the solid disk is accelerated for the 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 of points on its outer edge that are at a radius of 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 weight W_{2} 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.
4) A bicycle wheel (Model it as a ring.) with a mass of 2.00kg and a radius of 35.0cm is initially at rest. Find (a) its mass moment of inertia. 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 will then slow down and eventually come to stop. During the slowing down phase, only the frictional torque is present. 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 turn 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 the axis through its center and normal to its plane (that is the crankshaft), and (b) its rotational K.E. when it is turning at 3600rpm. Note that the translational kinetic energy (Chapter 6) was given by ½Mv^{2}. Here, in Chapter 9, the rotational kinetic energy is of course ½ 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