Physics 4A Solutions to Chapter 11 Homework

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1 Physics 4A Solutions to Chapter 11 Homework Chapter 11 Questions:, 8, 10 Exercises & Problems: 1, 14, 4, 7, 37, 53, 66, 81, 83 Answers to Questions: Q 11- (a) 5 and 6 (b) 1 and 4 tie, then the rest tie Q 11-8 (a) 4, 6, 7, 1, then, 3, and 5 tie (zero) (b) 1, 4, and 7 Q b, then c and d tie, then a and e tie (zero) Answers to Problems: P 11-1 Using the floor as the reference position for computing potential energy, mechanical energy conservation leads to Substituting I 1 1 Urelease = Ktop + Utop mgh= mvcom + Iω + mg( R). = 5 mr (Table 10-(f)) and ω = v com r (Eq. 11-), we obtain vcom com com mgh = mv + mr + mgr gh = v + gr 5 r 10 where we have canceled out mass m in that last step. (a) To be on the verge of losing contact with the loop (at the top) means the normal force is vanishingly small. In this case, Newton s second law along the vertical direction (+y downward) leads to vcom mg = mar g = R r

2 where we have used Eq for the radial (centripetal) acceleration (of the center of mass, which at this moment is a distance R r from the center of the loop). Plugging the result vcom = gbr rg into the previous expression stemming from energy considerations gives 7 gh = bgb g R rg + gr 10 which leads to h=.7r 0.7r.7 R. With R = 14.0 cm, we have h = (.7)(14.0 cm) = 37.8 cm. (b) The energy considerations shown above (now with h = 6R) can be applied to point Q (which, however, is only at a height of R) yielding the condition 7 gb6rg= v com + gr 10 which gives us vcom = 50gR 7. Recalling previous remarks about the radial acceleration, Newton s second law applied to the horizontal axis at Q leads to which (for R vcom 50gR N = m = m R r 7( R r) >> r ) gives 4 50mg 50( kg)(9.80 m/s ) N = = N. 7 7 (b) The direction is toward the center of the loop. P To find the center of mass speed v on the plateau, we use the projectile motion equations of Chapter 4. With v oy = 0 (and using h for h ) Eq. 4- gives the time-of-flight as t = h/g. Then Eq. 4-1 (squared, and using d for the horizontal displacement) gives v = gd /h. Now, to find the speed v p at point P, we apply energy conservation, that is, mechanical energy on the plateau is equal to the mechanical energy at P. With Eq. 11-5, we obtain 1 mv + 1 I com ω + mgh 1 = 1 mv p + 1 I com ω p. Using item (f) of Table 10-, Eq. 11-, and our expression (above) v = gd /h, we obtain gd /h + 10gh 1 /7 = v p which yields (using the values stated in the problem) v p = 1.34 m/s.

3 P 11-4 If we write r = x i + y j + z k, then (using Eq. 3-30) we find r F is equal to d yf zf + zf xf + xf yf z yii b x zg j d y xik. (a) Here, r = r where ˆ r = 3.0i.0j ˆ+ 4.0k, ˆ and F = F 1. Thus, dropping the prime in the above expression, we set (with SI units understood) x = 3.0, y =.0, z = 4.0, F x = 3.0, F y = 4.0, and F z = 5.0. Then we obtain e j m. τ = r F 1 = 60. i 30. j 60. k N (b) This is like part (a) but with F = F. We plug in F x = 3.0, F y = 4.0, and F z = 5.0 and obtain τ = r F = 6 i+ 30. j 18k N e j m. (c) We can proceed in either of two ways. We can add (vectorially) the answers from parts (a) and (b), or we can first add the two force vectors and then compute τ = r df1+ Fi (these total force components are computed in the next part). The result is τ = r F + F = 3i 4k N m. ( ) ( ˆ ˆ 1 ) (d) Now r = r r o where r o = 3.0i ˆ+.0j ˆ+ 4.0k. ˆ Therefore, in the above expression, we set x = 0, y = 4.0, z = 0, and Fx = = 0 Fy = = 8.0 Fz = = 0. We get τ = r F + F = d 1 i 0. P 11-7 Let r = xˆi+ yˆj+ z ˆk be the position vector of the object, v = v ˆi ˆ x + vyj+ vz ˆk and m its mass. The cross product of r and v is (using Eq. 3-30) its velocity vector, r v = yv zv + zv xv + xv yv ˆk. ( ) ˆ ( ) ˆ z y i x z j ( y x)

4 Since only the x and z components of the position and velocity vectors are nonzero (i.e., y = 0andv y = 0), the above expression becomes r v = b xvz + zvzg j. As for the torque, writing F = Fˆi+ F ˆj+ Fk, ˆ then we find r F to be x y z (a) With r = (.0 m)i ˆ (.0 m)kˆ and angular momentum of the object is τ = r F = yf zf + zf xf + xf yf ˆk. ( ) ˆ ( ) ˆ z y i x z j ( y x) v = ( 5.0 m/s)i ˆ+ (5.0 m/s)kˆ, in unit-vector notation, the = m xv + zv j = 0.5 kg.0 m 5.0 m s +.0 m 5.0 m s ˆj = 0. ( ) ( ) ˆ ( ) ( )( ) ( )( ) z x (b) With x =.0 m, z =.0 m, F y = 4.0 N, and all other components zero, the expression above yields τ = r F = (8.0 N m)i ˆ+ (8.0 N m)k ˆ. Note: The fact that = 0 implies that r and v are parallel to each other ( r v = 0 ). Using τ = r F = rfsinφ, we find the angle between r and F to be τ 8 N m sinφ = = = 1 φ = 90 rf ( m)(4.0 N) That is, r and F are perpendicular to each other. P (a) A particle contributes mr to the rotational inertia. Here r is the distance from the origin O to the particle. The total rotational inertia is ( ) ( ) ( ) I = m 3d + m d + m d = 14md = 14(.3 10 kg)(0.1 m) = kg m. (b) The angular momentum of the middle particle is given by L m = I m ω, where I m = 4md is its rotational inertia. Thus Lm = md ω = = 3 4 4(.3 10 kg)(0.1 m) (0.85 rad/s) kg m /s. (c) The total angular momentum is Iω= ω= = 3 14md 14(.3 10 kg)(0.1 m) (0.85 rad/s) kg m /s.

5 P The axis of rotation is in the middle of the rod, with r = 0.5 m from either end. By Eq , the initial angular momentum of the system (which is just that of the bullet, before impact) is rmv sinθ where m = kg and θ = 60. Relative to the axis, this is counterclockwise and thus (by the common convention) positive. After the collision, the moment of inertia of the system is I = I rod + mr where I rod = ML /1 by Table 10-(e), with M = 4.0 kg and L = 0.5 m. Angular momentum conservation leads to 1 rmv θ = ML + mr 1 Thus, with ω = 10 rad/s, we obtain ω sin. 1 ( ( 4.0 kg)( 0.5 m) + ( kg)( 0.5 m) )( 10rad/s) 1 ( 0.5 m)( kg) sin 60 v = = m/s. P We make the unconventional choice of clockwise sense as positive, so that the angular velocities (and angles) in this problem are positive. Mechanical energy conservation applied to the particle (before impact) leads to 1 mgh = mv v = gh for its speed right before undergoing the completely inelastic collision with the rod. The collision is described by angular momentum conservation: c mvd = I + md hω where I rod is found using Table 10-(e) and the parallel axis theorem: I Md M d 1 F rod = + H G I 1 K J = Md. 1 3 Thus, we obtain the angular velocity of the system immediately after the collision: rod md gh ω = ( Md /3) + md

6 which means the system has kinetic energy ( ) I + md ω which will turn into potential rod /, energy in the final position, where the block has reached a height H (relative to the lowest point) and the center of mass of the stick has increased its height by H/. From trigonometric considerations, we note that H = d(1 cosθ), so we have ( gh) 1 H 1 md M ( Irod + md ) ω = mgh + Mg = m gd + ( Md /3) + md from which we obtain 1 mh 1 θ = cos 1 = cos 1 ( 1 cosθ ) ( m+ M /) ( m+ M /3) ( 1 + M /m) ( 1 + M /3m) (0 cm/ 40 cm) = = (1+ 1)(1+ /3) = cos 1 cos (0.85) h/ d P As the wheel-axle system rolls down the inclined plane by a distance d, the change in potential energy is Δ U = mgdsinθ. By energy conservation, the total kinetic energy gained is 1 1 Δ U =Δ K =Δ Ktrans +ΔKrot mgdsinθ = mv + Iω. Since the axle rolls without slipping, the angular speed is given by ω = v/ r, where r is the radius of the axle. The above equation then becomes θ 1 mr mr = ω + =Δ rot + mgd sin I 1 K 1 I I (a) With m = 10.0 kg, d =.00 m, r = 0.00 m, and energy may be obtained as I = kg m, the rotational kinetic mgd sin θ (10.0 kg)(9.80 m/s )(.00 m)sin 30.0 Δ Krot = = = 58.8 J. mr (10.0 kg)(0.00 m) I kg m (b) The translational kinetic energy is Δ Ktrans =ΔK Δ Krot = 98 J 58.8 J = 39. J.

7 Note: One may show that mr / I = / 3, which implies that ΔK / Δ K = /3. Equivalently, we may write ΔKtrans / ΔK = /5 and Δ K / K 3/5 rot Δ =. So as the wheel rolls down, 40% of the kinetic energy is translational while the other 60% is rotational. trans rot P We note that its mass is M = 36/9.8 = 3.67 kg and its rotational inertia is Icom = MR (Table 5 10-(f)). (a) Using Eq. 11-, Eq becomes vcom comω com com com K = I + Mv = MR + Mv = Mv 5 R 10 which yields K = 61.7 J for v com = 4.9 m/s. (b) This kinetic energy turns into potential energy Mgh at some height h = d sin θ where the sphere comes to rest. Therefore, we find the distance traveled up the θ = 30 incline from energy conservation: 7 7vcom Mvcom = Mgdsin θ d = = 3.43m g sinθ (c) As shown in the previous part, M cancels in the calculation for d. Since the answer is independent of mass, then it is also independent of the sphere s weight.

Physics 4A Solutions to Chapter 10 Homework

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