Solving Problems In Rotational Dynamics PDF

Solving Problems in Rotational Dynamics When working with torque and angular acceleration (Eq. 8–14), it is important to use a consistent set of units, which in SI is: in in and the moment of inertia, I, in kgm2. mN;trads2;a I = ©ta, ©t,a ©mr2 = (©m)R2 = MR2, ©mr2, 210 CHAPTER 8 Rotati...

Solving Problems in Rotational Dynamics When working with torque and angular acceleration (Eq. 8–14), it is important to use a consistent set of units, which in SI is: in in and the moment of inertia, I, in kgm2. mN;trads2;a I = ©ta, ©t,a ©mr2 = (©m)R2 = MR2, ©mr2, 210 CHAPTER 8 Rotational Motion Axis Axis Object Moment of inertia Location of axis Thin hoop, radius R Thin hoop, radius R width w Solid cylinder, radius R Long uniform rod, length l Long uniform rod, length l (a) (b) (c) (f) (g) Through center Through center Hollow cylinder, inner radius R1 outer radius R2 (d) Uniform sphere, radius R (e) Through center Through center Through central diameter Through center Through end Rectangular thin plate, length l, width w (h) Through center MR2 MR2 2 1 Mw2 12 1 MR2 2 1 MR2 5 2 Ml2 12 1 Ml2 3 1 R R Axis Axis Axis Axis l Axis R R w + M(R2 2 1 +1 2R2) M(l2 12 1 + w2) w Axis R2 R1 l l FIGURE 8;20 Moments of inertia for various objects of uniform composition, each with mass M. SECTION 8–6 211 directions of rotation (counterclockwise and clock- wise), and assign the correct sign to each torque. 5. Apply Newton’s second law for rotation, If the moment of inertia is not given, and it is not the unknown sought, you need to determine it first. Use consistent units, which in SI are: in in and I in 6. Also apply Newton’s second law for translation, and other laws or principles as needed. 7. Solve the resulting equation(s) for the unknown(s). 8. Do a rough estimate to determine if your answer is reasonable. ©F B = maB , kgm2.mN;t rads2;a ©t = Ia. P R O B L E M S O L V I N G Rotational Motion 1. As always, draw a clear and complete diagram. 2. Choose the object or objects that will be the system to be studied. 3. Draw a free-body diagram for the object under consideration (or for each object, if more than one), showing all (and only) the forces acting on that object and exactly where they act, so you can deter- mine the torque due to each. Gravity acts at the CM of the object (Section 7–8). 4. Identify the axis of rotation and determine the torques about it. Choose positive and negative A heavy pulley. A 15.0-N force (represented by ) is applied to a cord wrapped around a pulley of mass and radius Fig. 8–21. The pulley accelerates uniformly from rest to an angular speed of in 3.00 s. If there is a frictional torque at the axle, determine the moment of inertia of the pulley. The pulley rotates about its center. APPROACH We follow the steps of the Problem Solving Strategy above. SOLUTION 1. Draw a diagram. The pulley and the attached cord are shown in Fig. 8–21. 2. Choose the system: the pulley. 3. Draw a free-body diagram. The force that the cord exerts on the pulley is shown as in Fig. 8–21. The friction force acts all around the axle, retarding the motion, as suggested by in Fig. 8–21. We are given only its torque, which is what we need. Two other forces could be included in the diagram: the force of gravity mg down and whatever force keeps the axle in place (they balance each other). They do not contribute to the torque (their lever arms are zero) and so we omit them to keep our diagram simple. 4. Determine the torques. The cord exerts a force that acts at the edge of the pulley, so its lever arm is R. The torque exerted by the cord equals and is counterclockwise, which we choose to be positive. The frictional torque is given as it opposes the motion and is negative. 5. Apply Newton’s second law for rotation. The net torque is The angular acceleration is found from the given data that it takes 3.00 s to accelerate the pulley from rest to Newton’s second law, can be solved for I which is the unknown: 6. Other calculations: None needed. 7. Solve for unknowns. From Newton’s second law, 8. Do a rough estimate. We can do a rough estimate of the moment of inertia by assuming the pulley is a uniform cylinder and using Fig. 8–20c: This is the same order of magnitude as our result, but numerically somewhat less. This makes sense, though, because a pulley is not usually a uniform cylinder but instead has more of its mass concentrated toward the outside edge. Such a pulley would be expected to have a greater moment of inertia than a solid cylinder of equal mass. A thin hoop, Fig. 8–20a, ought to have a greater I than our pulley, and indeed it does: I = MR2 = 0.436 kgm2. I L 1 2 MR2 = 1 2 (4.00 kg)(0.330 m)2 = 0.218 kgm2. I = ©t a = 3.85 mN 10.0 rads 2 = 0.385 kgm2. I = ©ta. ©t = Ia, a = ¢v ¢t = 30.0 rads - 0 3.00 s = 10.0 rads2. v = 30.0 rads: a ©t = RFT - tfr = (0.330 m)(15.0 N) - 1.10 mN = 3.85 mN. tfr = 1.10 mN; RFT F B T F B fr F B T tfr = 1.10 mN30.0 rads R = 33.0 cm,M = 4.00 kg F B TEXAMPLE 8;10 P R O B L E M S O L V I N G Usefulness and power of rough estimates T fr R 33.0 cm F B F B FIGURE 8;21 Example 8–10. Additional Example—a bit more challenging Pulley and bucket. Consider again the pulley in Example 8–10. But instead of a constant 15.0-N force being exerted on the cord, we now have a bucket of weight (mass ) hanging from the cord. See Fig. 8–22a. We assume the cord has negligible mass and does not stretch or slip on the pulley. Calculate the angular acceleration of the pulley and the linear accel- eration a of the bucket. Assume the same frictional torque acts. APPROACH This situation looks a lot like Example 8–10, Fig. 8–21. But there is a big difference: the tension in the cord is now an unknown, and it is no longer equal to the weight of the bucket if the bucket accelerates. Our system has two parts: the bucket, which can undergo translational motion (Fig. 8–22b is its free- body diagram); and the pulley. The pulley does not translate, but it can rotate. We apply the rotational version of Newton’s second law to the pulley, and the linear version to the bucket, SOLUTION Let be the tension in the cord. Then a force acts at the edge of the pulley, and we apply Newton’s second law, Eq. 8–14, for the rotation of the pulley: [pulley] Next we look at the (linear) motion of the bucket of mass m. Figure 8–22b, the free-body diagram for the bucket, shows that two forces act on the bucket: the force of gravity mg acts downward, and the tension of the cord pulls upward. Applying Newton’s second law, for the bucket, we have (taking downward as positive): [bucket] Note that the tension which is the force exerted on the edge of the pulley, is not equal to the weight of the bucket There must be a net force on the bucket if it is accelerating, so We can also see this from the last equation above, To obtain we note that the tangential acceleration of a point on the edge of the pulley is the same as the acceleration of the bucket if the cord doesn’t stretch or slip. Hence we can use Eq. 8–5, Substituting into the first equation above (Newton’s second law for rotation of the pulley), we obtain The unknown appears on the left and in the second term on the far right, so we bring that term to the left side and solve for The numerator is the net torque, and the denominator is the total rotational inertia of the system. With and, from Example 8–10, and , then The angular acceleration is somewhat less in this case than the of Example 8–10. Why? Because is less than the 15.0-N force in Example 8–10. The linear acceleration of the bucket is NOTE The tension in the cord is less than mg because the bucket accelerates. 8–7 Rotational Kinetic Energy The quantity is the kinetic energy of an object undergoing translational motion. An object rotating about an axis is said to have rotational kinetic energy. By analogy with translational kinetic energy, we might expect this to be given by the expression where I is the moment of inertia of the object and is its angular velocity. We can indeed show that this is true. v1 2 Iv2, 1 2 mv2 FT a = Ra = (0.330 m)A6.98 rads2B = 2.30 ms2. FT (= mg - ma = 15.0 N - ma) 10.0 rads2 a = (15.0 N)(0.330 m) - 1.10 mN 0.385 kgm2 + (1.53 kg)(0.330 m)2 = 6.98 rads2. tfr = 1.10 mNI = 0.385 kgm2 (m = 1.53 kg)mg = 15.0 N AI + mR2BAmgR - tfrB a = mgR - tfr I + mR2. a: a Ia = ©t = RFT - tfr = R(mg - mRa) - tfr = mgR - mR2a - tfr. FT = mg - ma = mg - mRa atan = a = Ra. a, FT = mg - ma. FT 6 mg. (= mg = 15.0 N). FT , mg - FT = ma. ©F = ma, FT Ia = ©t = RFT - tfr. FTFT ©F = ma. ©t = Ia, tfr = 1.10 mN a m = wg = 1.53 kgw = 15.0 N EXAMPLE 8;11 212 CHAPTER 8 T fr m (a) (b) R TF B F B F B gB FIGURE 8;22 Example 8–11. (a) Pulley and falling bucket of mass m. This is also the free-body diagram for the pulley. (b) Free- body diagram for the bucket.