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Rotational Mechanics Part III – Dynamics, Energy, Momentum

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Rotational Mechanics Part III – Dynamics, Energy, Momentum Rotational Mechanics Part III – Dynamics Pre AP Physics We have so far discussed rotational kinematics – the description of rotational motion in terms of angle, angular velocity and angular acceleration and rotational equilibrium - Torque. Now our discussion will turn to objects that are rotating because of a net torque NOT equal to zero. You should think of Newton’s 2ND Law for objects that are rotating. ROTATIONAL INERTIA Imagine you are asked to rotate a baseball bat. There are many ways to do this, but some are “easier” than others. Even though the bat’s mass has not changed it seems that rotating the bat about some axes are easier than others. The resistance of an object to changes in its rotational motion is measured by a quantity called the moment of inertia or rotational inertia of the object. The moment of inertia is similar to mass because they are both forms of inertia. However there is an important difference between inertia that resists translational motion (mass) and the inertia that resists changes in rotational motion (moment of inertia). Mass is an intrinsic property of an object, while moment of inertia is not. The moment of inertia depends not only on an objects’ mass but also the distribution of that mass about some axis of rotation. Consider a object moving aT around a circular path. Let the object be an airplane with an engine that causes the plane to have a tangential r acceleration of aT. From Newton’s second law we get: FTangential = m aT. Now the torque produced by this tangential force can be found using the torque equation: T = FT r. Where r is the radial length from the center of the circle to the airplane. aT Also notice that r is also the perpendicular distance from the center of the circle to the tangential r acceleration vector. Therefore the radial distance can be considered to be the torque arm. As a result, the torque T = m aT r. Now remember that the radius relates the tangential acceleration and the angular acceleration. The equation is aT = r, where must be in units of radians/time2. With the proper substitution we can now get: T = m r r or another way of stating the torque equation becomes: T = (mr2) . The “mr2” is sometimes referred to the moment of inertia or rotational inertia of the object in question. The variable “I” is often used to designate this quantity. When all the mass m of an object is concentrated at the same distance r from a rotational axis, then the rotational inertia is I = mr2. When the mass is more spread out, the rotational inertia is less and the formula is different. The units for moment of inertia sometimes referred to as rotational inertia are kg-m2. The equations found on the next few slides were determined using calculus methods. The concept is not hard. All you have to do is to sum the moments of inertia for each particle that makes up the object you are applying the torque to. Here are some examples of different moments of inertia commonly used in problem solving. Thin hoop about symmetry axis or I = mr2 ring a point mass about the axis. R R 2 Isolid disk = ½ mr Disk or cylinder about symmetry axis. 2 Thin spherical I empty sphere = 2/3 mr shell about diameter. I = 2/5 mr2 Solid sphere about solid sphere diameter 2 Thin rod about perpendicular axis Ithin rod = 1/12 ml through center. l 2 Thin rod about perpendicular axis Ithin rod = 1/3 ml through end. l Just as it takes a force to change the linear state of motion of an object, a torque is required to change the rotational state of motion of an object. In the absence of a net torque, a rotating object keeps rotating, while a non-rotating object stays non-rotating. The short pendulum will swing back and forth more frequently than the long pendulum. Rotational inertia depends on the distance of mass from the axis of rotation. Which will roll down an incline with greater acceleration, a hollow cylinder or a solid cylinder of the same mass and radius? Answer The answer is the cylinder with the smaller rotational inertia because the cylinder with the greater rotational inertia requires more time to get rolling. A heavy iron cylinder and a light wooden cylinder, similar in shape, roll down an incline. Which will have more acceleration? Answer The cylinders have different masses, but the same rotational inertia per mass, so both will accelerate equally down the incline. Their different masses make no difference, just as the acceleration of free fall is not affected by different masses. All objects of the same shape have the same “laziness per mass” ratio. Example # 1 Determine the moment of inertia for a bowling ball with radius 10 cm and mass 4.5 kg. Given: radius = 10 cm = 0.10 m; mass = 4.5 kg 2 ISOLID SPHERE = (2/5) mr 2 ISOLID SPHERE = (2/5) (4.5 kg) (0.10 m) 2 ISOLID SPHERE = 0.018 kg-m Example #2 Brian is twirling a rubber stopper, mass = 50 grams, which is attached to an almost mass less, string 75 cm long. By what amount does the rubber stopper resist changes in its rotational motion? Given: radius = 75 cm = .75 m; mass = 50 g = 0.050 kg 2 IPOINT MASS = mr 2 IPOINT MASS = (0.050kg) (0.75 m) 2 IPOINT MASS = 0.0281 kg-m Newton’s 2nd Law for rotational motion about a fixed axis — Dynamics — Now changes in angular velocity “” imply angular acceleration “”. The angular acceleration is directly proportional to the torque applied to the object and inversely proportional to the rotational inertia of the object. T I and, using some algebra, we get the expression T = I If you’ll notice this looks a lot like “F = ma”. In fact the equation relating the torque on an object to its angular motion is sometimes referred to as the rotational analog of Newton’s second law of motion. Note that “I” is always some form of “mr2”. Example #3 Most turntables can bring a record from rest up to the rated angular speed of 33 1/3 rpm in one-half revolution. The platter of one turntable has a rotational inertia of 0.05 kg-m2 (including the effect of the record). Neglecting frictional effects, what net torque, assumed constant, must the turntable motor apply to the platter to achieve this performance? Given: i = 0; f = 33 1/3 rev/min= 10/9 rads/s; = ½ revolution = rads. I = 0.05 kg-m2 2 T = I 2210 2 fi 9 rads/sec 0 2 2 2 rads = 1.94 rads/sec T = (0.05 kg-m2) 1.94 rads/sec2 T = 0.0970 kg-m2/sec2 or 0.0970 N-m. Example #4 A clay vase on a potter’s wheel experiences an angular acceleration of 8.0 rads/sec2 due to the application of a 10.0 N-m net torque. Find the rotational inertia of the vase and potter’s wheel. Given: T = 10.0 N-m; = 8.0 rads/s2 T = I 10 N-m = I (8.0 rads/sec2) I 10Nm 8.0rads /sec2 I = 1.25 N – m – s2 I1.25 Kg m m s2 s2 I = 1.25 Kg – m2 Example # 5 One day Kirstin gave Courtney a ride on a merry-go-round at a park when the temperature was 85. The merry-go-round had a diameter of four meters, and Courtney along with the merry-go- round weighed 980 Newtons. If the merry-go-round was initially at rest, and one minute later Kirstin pushed it to a rotational speed of 25 rpm: (A) What angular acceleration in radians/sec/sec did the merry-go-round undergo? (B) what force did Kirstin apply in Newtons? Given: radius = 2.0-m; i = 0; t = 1 min = 60 sec; f = 25 rpm = 5/6 rads/s; Weight = 980 N m = 100 kg A merry-go-round can be considered to be a solid disk I = ½ mr2. 5 0 (A) fi 6 = /72 rads/s2 t 60 (B) T = I and T = F r I x = F x r I 1 mr2 F F2 1 mr r r 2 F = (1/2) (100 kg)(2 m)(/72 rads/s2) F = 4.36 Newtons Example # 6 A cord is wrapped around the rim of a flywheel 0.500 m in radius, and a steady pull of 30.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel is 4.00 kg- m2. Compute the angular acceleration of the wheel. T 2 Given: radius = 0.500-m; F = 30.0 N; Iflywheel = 4.00 kg-m ; α = ? I Fr I 30.0Nm 0.500 T = I ∙ α 4.00kg m2 T = F ∙ r 2 F ∙ r = I ∙ α α = 3.75 rads/s Example # 7 A bicycle wheel of 1.25 kg mass and radius 30.0 cm is rotating with an angular speed of 10 rads/s. What frictional force, tangent to the tire will bring it to a stop in 5.00 seconds? Given: radius = 0.300-m; m = 1.25 kg; i = 10; f = 0; t = 5.00 s; F = ? T I F I r F = −2.36 N 2 T = I ∙ α Iwheel = Iring = mr T = F ∙ r 0 10 rads / s = −2 rads/s2 5.00s F ∙ r = I ∙ α 1.25kg 0.3 m2 2 rads / s2 F 0.3m.
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