Physics 4A Solutions to Chapter 10 Homework

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Winfred Owen
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1 Physics 4A Solutions to Chapter 0 Homework Chapter 0 Questions: 4, 6, 8 Exercises & Problems 6, 3, 6, 4, 45, 5, 5, 7, 8 Answers to Questions: Q 0-4 (a) positive (b) zero (c) negative (d) negative Q 0-6 F 5, F 4, F, F, F 3 (zero) Q , then 70 and 0 tie Answers to Problems: P 0-6 If we make the units explicit, the function is b g c h c 3 θ = 40. rad / s t 30. rad / s t + 0. rad / s ht but generally we will proceed as shown in the problem letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. 3 (a) Equation 0-6 leads to c h 3. d ω = t t + t = t+ t dt Evaluating this at t = s yields ω = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives ω 4 = 8 rad/s. (c) Consequently, Eq. 0-7 gives (d) And Eq. 0-8 gives α avg ω 4 ω = 4 = rad / s.

2 Evaluating this at t = s produces α = 6.0 rad/s. c h dω d α = = 4 6t+ 3t = 6+ 6t. dt dt (e) Evaluating the expression in part (d) at t = 4 s yields α 4 = 8 rad/s. We note that our answer for α avg does turn out to be the arithmetic average of α and α 4 but point out that this will not always be the case. P 0-3 The wheel has angular velocity ω 0 = +.5 rad/s = rev/s at t = 0, and has constant value of angular acceleration α < 0, which indicates our choice for positive sense of rotation. At t its angular displacement (relative to its orientation at t = 0) is θ = +0 rev, and at t its angular displacement is θ = +40 rev and its angular velocity is ω = 0. (a) We obtain t using Eq. 0-5: which we round off to t s. ( 0 ) (40 re v) θ = ω + ω t t = = 335 s 0.39 rev/s (b) Any equation in Table 0- involving α can be used to find the angular acceleration; we select Eq (40 rev) 4 θ = ωt αt α = = 7. 0 rev/s (335 s) which we convert to α = rad/s. (c) Using θ = ω t + αt (Eq. 0-3) and the quadratic formula, we have 0 ω ± ω + θα (0.39 rev/s) ± (0.39 rev/s) + (0 rev)( 7. 0 rev/s ) α 7. 0 rev/s = = 4 t which yields two positive roots: 98 s and 57 s. Since the question makes sense only if t < t we conclude the correct result is t = 98 s. P 0-6 (a) The angular acceleration is

3 Δω 0 50 rev / min α = = = 4. rev / min. Δt (. h)(60 min / h) (b) Using Eq. 0-3 with t = (.) (60) = 3 min, the number of revolutions is θ = ω ( )( ) 0t+ αt = (50 rev/min)(3 min) +.4 rev/min 3 min = rev. (c) With r = 500 mm, the tangential acceleration is which yields a t = 0.99 mm/s. F π rad at = αr = c 4. rev/minh H G rev (d) The angular speed of the flywheel is I KJ F H G min 60 s I K J (500 mm) ω = (75 rev/min)(π rad/rev)( min/ 60 s) = 7.85 rad/s. With r = 0.50 m, the radial (or centripetal) acceleration is given by Eq. 0-3: ar = ω r = (7.85 rad/s) (0.50 m) 3 m/s which is much bigger than a t. Consequently, the magnitude of the acceleration is a = a + a a = 3 m / s. r t r P 0-4 The particles are treated point-like in the sense that Eq yields their rotational inertia, and the rotational inertia for the rods is figured using Table 0-(e) and the parallel-axis theorem (Eq. 0-36). (a) With subscript standing for the rod nearest the axis and 4 for the particle farthest from it, we have

4 3 I = I+ I + I3 + I4 = Md + M d + md + Md + M d + m( d) 8 8 = Md + 5 md = (. kg)(0.056 m) +5(0.85 kg)(0.056 m) 3 3 =0.03 kg m. (b) Using Eq. 0-34, we have K = Iω = M+ m d ω = (. kg) + (0.85 kg) (0.056 m) (0.30 rad/s) 3 3 = 3. 0 J. P 0-45 We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r F sin θ is associated with F and a negative torque of magnitude r F sin θ is associated with F. The net torque is consequently Substituting the given values, we obtain τ = rfsinθ rf sin θ. τ = (.30 m)(4.0 N) sin 75 (.5 m)(4.90 N) sin 60 = 3.85 N m. P 0-5 (a) We use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block m, then its coordinate is given by y at, so y ( m) a = = = m/s. t (. 500s) Block has an acceleration of m/s upward. (b) Newton s second law for block is mg T = ma, where m is its mass and T is the tension force on the block. Thus, = ( ) T m ( g a = ) = (0.500 kg) 9.8 m/s m/s = 4.87 N.

5 (c) Newton s second law for block is mg, T = ma where T is the tension force on the block. Thus, ( ) T m ( g a = + ) = (0.460 kg) 9.8 m/s m/s = 4.54 N. (d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the pulley must be the same as the acceleration of the blocks, so a m/s α = = R m = 0. rad / s. (e) The net torque acting on the pulley is τ = ( T T) R. Equating this to Iα we solve for the rotational inertia: ( T T ) R ( )( ) 4.87 N 4.54 N m.38 0 I = = = kg m. α.0 rad/s P 0-5 According to the sign conventions used in the book, the magnitude of the net torque exerted on the cylinder of mass m and radius R is τ net = FR FR Fr 3 = (6.0 N)(0. m) (4.0 N)(0. m) (.0 N)(0.050 m) = 7N m. (a) The resulting angular acceleration of the cylinder (with I = MR according to Table 0- (c)) is τ net 7N m α = = = 9.7rad/s. I (.0kg)(0. m) (b) The direction is counterclockwise (which is the positive sense of rotation). P 0-7 We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a = a = Rα (for simplicity, we denote this as a). Thus, we choose rightward positive for m = M (the block on the table), downward positive for m = M (the block at the end of the string) and (somewhat unconventionally) clockwise for positive sense of disk rotation. This means that we interpret θ given in the problem as a positivevalued quantity. Applying Newton s second law to m, m and (in the form of Eq. 0-45) to M, respectively, we arrive at the following three equations (where we allow for the possibility of friction f acting on m ).

6 mg T = ma T f = m a TR TR= Iα (a) From Eq. 0-3 (with ω 0 = 0) we find θ (0.30 rad) θ = ω0t+ αt α = = = 3.4 rad/s. t (0.090 s) (b) From the fact that a = Rα (noted above), we obtain Rθ (0.04 m)(0.30 rad) a = = = t (0.090 s) m/s. (c) From the first of the above equations, we find ( ) Rθ (6.0 kg) 9.80 m/s (0.04 m)(0.30 rad) T = m g a = M g = t (0.090 s) = 56. N. (d) From the last of the above equations, we obtain the second tension: T 4 Iα ( kg m )(3.4 rad/s ) = T = 56. N = 55. N. R 0.04 m P 0-8 L The center of mass is initially at height h = sin 40 when the system is released (where L =.0 m). The corresponding potential energy Mgh (where M =.5 kg) becomes rotational kinetic energy Iω as it passes the horizontal position (where I is the rotational inertia about the pin). Using Table 0- (e) and the parallel axis theorem, we find I = ML + M( L/ ) = 3 ML. Therefore, L 3gsin40 Mg sin 40 = ML ω ω = = 3. rad/s. 3 L

a +3bt 2 4ct 3) =6bt 12ct 2. dt 2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s. Thus,

a +3bt 2 4ct 3) =6bt 12ct 2. dt 2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s. Thus, 1. a) Eq. 11-6 leads to ω d at + bt 3 ct 4) a +3bt 4ct 3. dt b) And Eq. 11-8 gives α d a +3bt 4ct 3) 6bt 1ct. dt. a) The second hand of the smoothly running watch turns through π radians during 60 s. Thus,