Rotational Dynamics, Moment of Inertia and Angular Momentum Now that we have examined rotational kinematics and torque we will look at applying the concepts of angular motion to Newton s first and second law and how it applies to rigid bodies in terms of inertia, forces, energy and momentum. In previous course work we learned how Newton s 3 Laws can be extrapolated to explain the conservation of momentum. In this lesson we will extrapolate the concepts of torque and rotational dynamics to explain rotational dynamics, moment of inertia and angular momentum. Center of Gravity In many situations in life, masses are suspended away from the center of the lever arm. Take for example the simple act of holding your Physics Textbook. The forces involved are far from simple. There are in fact two centers of gravity involved. The book has a center of gravity that we use to measure its weight (W b ) and the arm has its own center of gravity as it also has its own weight (W a ). As a result this creates a dynamic torque situation where the torques are calculated about the elbow joint. The force that the bicep must apply has to balance the weight of the arm, the book and the force on the elbow. In these situations the weight of the object can be considered to act as a definite point fot the purpose of calculating the torque. This point is called the center of gravity.
Definition of Center of Gravity: The center of gravity (cg) of a rigid body is the point at which its weight can be considered to act when calculating the torque due to the weight. To determine the center of gravity of an object that has multiple masses around the same pivot point follows the relationship: The two values of the net torque must be the same so that: Example: An arm is composed of 3 separate parts: the upper arm (W 1 = 17 N), the lower arm (W 2 = 11 N) and the hand (W 3 = 4.2 N) each with a center of gravity of.13 m, 0.38 m, and 0.61 m respectively. Find the center of gravity with respect to the shoulder joint. 0.13 m 0.38 m 0.61 m W 1 = 17 N W 2 = 11 N W 3 = 4.2 N Calculate the center of gravity using:
Angular versus Tangential Variables Imagine a string with a line of light bulbs attached to it being swung around your head. v T For each individual bulb the vector is drawn at a tangent to the circle as seen above. Of all the light bulbs, the one farthest from the center has to travel the fastest in order to complete a larger swing around a circle with a greater radius and thus circumference. This gives the furthest bulb the greatest tangential speed. The tangential speed of the bulbs can be found using the equation V T = rω (ω is in rad/s) Thus for a given angular speed (ω), the tangential speed (v T ) is directly proportional to the radius (r).
Just as tangential speed can be related to angular speed, so too can tangential acceleration be related to angular acceleration. In linear acceleration we relate the acceleration of an object using a T = v T - v To / t Since we just saw that v T = rω, then the equation can be rewritten in terms of angular speeds. a T = (rω rω o ) / t a T = r (ω ω o ) /t ; and since α = (ω ω o ) Then a T = rα Rolling Motion Rolling objects such as tires, balls, and rolling pins all experience rotational motion about an axis. On a car, when the tires roll, there is a relationship between the angular speed at which the tires rotate and the linear speed at which the car moves. Neglecting slippage, the relationship between linear speed and angular speed is similar to the relationship we derived for tangential speed. v = rω (ω is in rad/s) v = linear speed In the same way, linear accelerations and angular accelerations are related to tangential accelerations. a = rα (α in rad/s 2 ) a = linear acceleration As we will see the shape of the object will affect the rate at which an object rolls. Solid spheres and hollow spheres roll at different rates. To understand why, we must first look at rotational dynamics and understand the concepts of center of gravity and moment of inertia.
Newton s Second Law for Rotational Motion Newton s second law describes the acceleration (a) an object of mass (m) experiences when subjected to a net force (F). Similarly, when a rigid object of mass (m) rotates about a fixed axis, all the particles in the object experience a tangential force (F T ) that is proportional to the tangential acceleration (a T ). F T =ma T Consider a model airplane being flown in a circle: r F T Recall from the torque lesson that angular is directly proportional to torque ( ). The torque produced by the net tangential force is therefore related using r F But recall that the tangential acceleration is related to the angular acceleration according to a T = rα, therefore combining the equations we get (mr 2 )α This indicates that the net torque ( ) is directly proportional to the angular acceleration (α). The constant in the proportionality (mr 2 ) is what is known as the Moment of Inertia represented by the symbol ( ). = mr 2 ( in kgm 2 ) The moment of inertia of an object depends on both the mass of each particle in the object and the distance each particle is from the axis of rotation. The farther a particle is
away from the axis of rotation, the greater it s contribution to the moment of inertia. The moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object. Since is the sum of all the moments we can combine it into one useful equation that represents F = ma. It is known as Newton s Second Law for Bodies Rotating about a Fixed Axis and it is α (α must be in rad/s 2 ) Example: Record players bring a record from rest up to a speed of 33 1 / 3 rev/min in one half of a revolution. The platter of a turntable has a moment of inertia of 0.0500 kgm 2. Neglecting frictional effects, what torque must the motor apply to achieve this performance? θ = π rad ( 1 / 2 rev) ω = 3.49 rad/s (33 1 / 3 rev/min) ω 2 = ω o + 2αθ α = ω 2 - ω o / 2θ α = (3.49 rad/s) 2 / 2(π rad) α = 1.94 rad/s 2 The second law of rotational motion can now be used to find the torque α (0.0500 kgm 2 )(1.94 rad/s 2 ) = 0.0970 Nm
Different Moments of Inertia The shape of an object affects its moment of inertia. Although a rigid objects possesses a unique total mass they do not possess a unique moment of inertia. The moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object. Different objects have different moments of inertia as given by the chart below which shows a few common examples.
Rotational Work and Energy Similar to linear work, rotational work takes advantage of applying a force through a distance. Rotational work occurs when a force (F) is used in rotating an object through the angle θ. It is done by a constant torque ( ). * It requires that θ be expressed in radians. ** The units of rotational work are expressed in Joules (J) Example: An electric drill is turned on and the chuck exerts a constant torque of 4.0 x 10-4 Nm to turn the drill bit through 9.0 revolutions. Find the rotational work done by the chuck. First find the angular displacement in radians: θ = (9.0 rev) x (2π rad / 1 rev) = 57 rad Then find rotational work using: W R = θ = (4.0 x 10-4 Nm)(57 rad) = 2.3 x 10-2 J
Rotational Kinetic Energy The kinetic energy of a rotating object is analogous to linear kinetic energy and can be expressed in terms of the moment of inertia and angular velocity. The total kinetic energy of an extended object can be expressed as the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass. For a fixed axis of rotation, the rotational kinetic energy can be expressed in the form * It requires that ω be expressed in rad/s. ** The units of rotational kinetic energy are expressed in Joules (J) Example: A hollow cylinder and a solid cylinder start from rest at the top of an incline (h o ). Ignoring friction, which cylinder will have the greatest translational speed and reach the bottom first? Solution: The total mechanical energy (E) at any height (h) must equal the sum of all the energies including the gravitational potential, translational kinetic and rotational kinetic. E = ½ mv 2 + ½ Iω 2 + mgh E f = E o ½ mv f 2 + ½ Iω f 2 + mgh f = ½ mv o 2 + ½ Iω o 2 + mgh o ½ mv f 2 + ½ Iω f 2 = mgh o Since ω f = v f /r, we get v f = 2mgh o / (m + I / r 2 ) If for a hollow cylinder I = mr 2, then v f = gh o If for a solid cylinder I = ½ mr 2, then v f = 4 / 3 gh o = 1.15 gh o Thus, the solid cylinder having the greatest translational speed, arrives at bottom of the ramp first.
Angular Momentum In lessons #1 and #2, we describe linear momentum (p) as the product of the mass (m) and linear velocity (v); (p = mv). For rotational motion, the analogous concept to linear momentum is known as angular momentum (L), and is described instead in terms of the product between the moment of inertia (I), and the angular velocity (ω). Unit = kg m 2 /s L = I ω (ω must be measured in rad/s) Similar to linear momentum, angular momentum also obeys the laws of conservation. Since the angular momentum is the product of the moment of inertia (I), and the angular velocity (ω), any change in one of these two variables must be accompanied by a change in the other to keep the product constant. That is, if you change the moment of inertia (I), the angular velocity (ω) must also change to keep the angular momentum constant. A great example of this can be seen in figure skating. A skater spins slowly on one skate with both arms and one leg outstretched. But as she pulls her arms and legs in toward the rotational axis, her moment of inertia (I) decreases, and the angular speed (ω) increases. Example: A skater is spinning at an angular velocity of ω o = 9.0 rad/s. By pulling her arms inwards she reduces her moment of inertia to 30% of its initial value. What was the final angular velocity of her spin? I f = 0.30 I o I f ω f = I o ω o ; therefore (0.30 I o )ω f = I o ω o Thus ω f = ω o / 0.30 = 30 rad/s
Hand In Assignment 1. Explain how it is possible for a large force to produce only a small (or even zero) torque and how a small force can produce a large amount of torque. 2. Two children are sitting on a seesaw of a uniform board with a fulcrum on its center of mass. One child (m=30kg) is 2.0m to the left of the fulcrum. Where should the second child (m=50kg) sit so that they are in equilibrium? (1.2 m right) 3. A jet transport has a weight of 1.00 x 10 6 N and is a rest on a runway. The two rear wheels are 15.0 m behind the front wheel, and the planes center of gravity is 12.6 m behind the front wheel. Determine the normal force exerted on the front wheel and each of the two rear wheels.(1.6 x 10 5 N front, 4.20 x 10 5 N each rear) 4. How many revolutions per minute would a 15.0-m diameter Ferris wheel need to for the passengers to feel weightless at the topmost point of the trip? (7.70rev/min)
5. A rifle is a long gun whose barrel has been grooved or "rifled" on the inside with spiral channels. (For comparison, a long gun with a smooth bore is called a musket.) Bullets fired from a rifled barrel spin. This gives them greater stability in flight and thus greater accuracy when fired. Since 1964, the standard infantry weapon in the US Army has been the.22 caliber M16 rifle. Due to rifling, a bullet fired from an M16 rotates two and a half times on its journey from the breech to the muzzle. Given a barrel length of 510 mm and a muzzle velocity of 950 m/s, determine Looking out from the inside of a rifled barrel. The opening of every James Bond film. a. the average translational acceleration (8.8 x 10 5 m/s 2 ) b. the average rotational acceleration (in rad/s 2 ) (2.7 x 10 7 rad/s 2 ) c. the final angular velocity (in rotations per second) (4700 rot/s) 6. Consider a motor that exerts a constant torque of 25.0 N m to a horizontal platform whose moment of inertia is 50.0 kg m 2. Assume that the platform is initially at rest and the torque is applied for 12.0 rotations. How much work does the motor do on the platform during this process? (1885 J) 7. In a race down an incline between a solid and a hollow sphere of the same mass (m), which will win? Prove this using moment of inertia and rotational kinetic energy. 8. The mass of the earth is 6.0 x 10 24 kg, and its radius is 6.4 x 10 6 m. If the earth has a rotational speed of 1 rev/day, what is its rotational kinetic energy (in Joules)? (2.6 x 10 29 J) 9. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater's final moment of inertia to his initial moment of inertia. (0.581) 10. To prepare homemade ice cream, a crank must be turned with a torque of 3.95 N m. How much work is required for each complete turn of the crank? (24.8 J)