Hoework 8 probles: 0.0, 0.7,.55,.
Proble 0.0 A block of ass kg an a block of ass 6 kg are connecte by a assless strint over a pulley in the shape of a soli isk having raius R0.5 an ass M0 kg. These blocks are allowe to ove on a fixe wege of angle θ0 as shown in the figure. The coefficient of kinetic friction is 0.6 for both blocks. Draw free-boy iagras of both blocks an the pulley. Deterine (a) the acceleration of the two blocks an (b) the tension in the string on both sies of the pulley. y f k, N T N -T T x g Mg f k, -T N y θ x g It is convenient to consier each block in a ifferent reference frae. In the reference frae in the figure, a nuber of scalar coponents of the consiere vector quantities assue a zero value. Using Newton s secon law of otion, we can relate the interaction of each object with its otion. Only horizontal coponent (x ) of the acceleration of block s center of ass has a non-zero value a. Therefore T μkn + 0 + 0 a 0 + 0 + N g 0
Siilarly, only the coponent along the rap of the acceleration of the center of ass on the secon block has a nonzero value of a (sae as the first block) T μkn + g sin θ + 0 a 0 + 0 + N g cos θ 0 The z-coponent of angular acceleration of the pulley, relate to the acceleration of the blocks, is relate to the net torque exerte on the pulley. With the reference point at the center of the pulley a MR T R TR R The above set of linear equations has five unknowns (N, T, N, T an a). The rest is algebra. Solving the equations siultaneously for the unknown coponent of the acceleration (of the center of ass of each block) an the two tensions we obtain a ( + cosθ) sin θ μk g M + + 6kg sin 0 0.6 0kg + kg + 6kg ( kg + 6kg cos0 ) 9.8 s 0.09 s T k s s T 6kg ( a + μ g) kg 0.09 + 0.6 9.8 7.67N (( μ cosθ + sin θ) g a) k s ( 0.6 cos0 + sin 0 ) 9.8 0.09 9.N s
Proble 0.7 A string is woun aroun a unifor isk of raius R an ass M. The isk is release fro rest with the string vertical an its top en tie to a fixe support (Figure P.7). As the isk escens, show that (a) the tension in the string is one thir the weight of the isk, (b) the agnitue of the acceleration of the center of ass is g/, an (c) the spee of the center of ass is (gh/) /. Verify your answer to (c) using the energy approach. a) Using Newton's secon law for rotational otion we can relate the net y torque about the instantaneous axis with the angular acceleration of the cyliner. For the irection of the axis of rotation h M (see the figure) () I A α z τ net 0 - MgR I use here the fact that the total gravitational torque is such that the entire z gravitational force was applie at the center of gravity (coinciing with the center of ass). R x Using also the parallel axis theore we can eterine the rotational inertial of the isk about the instantaneous axis an fro () fin the angular acceleration () α z MgR MR + MR Now we can relate the acceleration (of the center of ass) with the above angular acceleration an use Newton's secon law for the translational otion of a syste of particles. For the vertical coponents () T - W Ma y Mα z R Therefore g R
g () T W + Mαz R Mg M R Mg R b) (In fact we neee the acceleration in part (a).) The otion is in a vertical irection therefore the horizontal coponents of the acceleration are zero. Recalling again the relationship between the acceleration an the angular acceleration the agnitue of the acceleration is g (5) a 0 + 0 + ( αr) (The inus sign inicates that consistent with y rawing the acceleration is own.) c) Solution. Since we foun the tension force in part (b), the easiest approach woul be to use the work-energy theore for the translational kinetic energy. The change in the translational kinetic energy is irectly relate to the spee of the center of ass Mv (6) Δ K c T 0 (The initial kinetic energy is zero). Tension an gravity alone prouce external forces which are exerte on the cyliner. The work one on the center of ass ue to these interactions is (7) ΔWc T h + (ig Δri ) Mgh + Mgh Mgh i Fro the work-energy theore (for translational kinetic energy)
Mv (8) c Mgh Solving for the unknown spee we obtain gh (9) v c Note. The proble was not properly forulate. You can only verify the forula given in the proble if h represents the isplaceent of the center of ass. In the inclue figure h shoul be arke as the istance fro the ceiling to the top of the cyliner an not as the istance to its center. Solution. (work-energy theore for total kinetic energy) Using the work-energy theore for the total kinetic energy of the cyliner, we will be able to relate the angular velocity (axial coponent) of the cyliner (all its particles) with the isplaceent of the center of ass. In this version of the theore only gravitational work is perfore on the syste. (0) W 0 + W ( g Δ ) Mgh ext g i r i i (The gravitational work again epens only on the isplaceent of the center of ass.) We can relate the change in the total kinetic energy with the spee of the cyliner v () Δ K tot Mv + IAω Mv + MR Mv R Fro the work-energy theore for the total kinetic energy () Mv Mgh we fin the sae value for the spee
() v gh Solution (fro Newton s secon law) The center of ass is oving with constant acceleration. Consistent with the figure, the istance travele by the center of ass of the cyliner is relate with tie as follows at () h Therefore, the center of ass is isplace by istance h at instant h (5) t a We can fin the vertical coponent of the velocity of the center of ass fro the inverse relationship between acceleration an velocity (6) v at a h a hg
Proble.55 Two astronauts, each having ass, are connecte by a rope of length having negligible ass. They are isolate in space, orbiting their center of ass at spee v. Treating the astronauts as particles, calculate (a) the agnitue of the angular oentu of the two-astronaut syste an (b) the rotational energy of the syste. By pulling on the rope, one of the astronauts shortens the istance between the to /. (c) What are the astronauts new spees? () How uch cheical potential energy in the boy of the astronauts was converte to echanical energy in the syste when he shortene the rope? l L p l r r p a,b) Directly fro the efinition, the agnitue of astronaut s angular oentu is l r v v sin 90 v an his or her kinetic energy v Ki Since the angular oentu of the secon astronaut is in the sae irection, the agnitue of the total angular oentu of the two astronauts is Li L i v The total kinetic energy of the two astronauts is K K + K v tot,i,i,i
c) Since the syste is isolate, the total angular oentu is conserve. However, when the astronauts are closer, their spee will ouble to copensate for the change in the raius of their paths Lf Li vi vf v ) In the absence of external influence, the (internal) work perfore by the astronauts is equal to the change in the total kinetic energy of the syste Δ W K K v (0.) int tot,f tot,i Solution. Calculation of the internal work irectly fro its efinition is possible but a bit ore coplicate. In the process, each astronaut gets closer to the center of ass while the tension force, holing the astronaut in his or her circular otion, ust be equal to the require centripetal force. ΔW int v [ v() r ] r r ( r) r v / r r v r r v 6 v
Proble. A hungry bear weighing 700 N walks out on a bea in an attept to retrieve a basket of foo hanging at the en of the bea. The bea is unifor, weights 00 N an 6 long; the basket weights 80 N. (a) Draw a free-boy iagra to the bea. (b) When the bear is at x, fin the tension in the wire an coponent of the force exerte by the wall on the left en of the bea. (c) If the wire can withstan a axiu tension of 900 N, what is the axiu istance the bear can walk before the wire breaks? y a) The following forces are exerte on the bea x F N W T T x F noral force exerte by the wall N noral force exerte by the bear (equal to the weight of the bear) W gravitational force exerte by the earth T the tension force exerte by the wire T the tension force exerte by the hanging basket (equal to the weight of the basket) b) Since the bea is in (static) equilibriu, the net external torque about any point an the net external force exerte on the bea are zero vectors. If we select the reference point at the wall (the origin of the inicate reference frae), the only non-zero coponents of all the torques are along the z-irection. (The analysis of the x- an y-coponents leas to trivial equations.) For the z-coponents
τ xn ; N,z L τ W,z W ; τ LT ; T,z τ LT sin0 T,z The balance of the torques therefore requires L LT sin0 LT W xn 0 Fro which, the tension in the wire ust be 80N + 00N + 700N T 6 N sin0 c) The sae equation leas to the axiu istance the bear can walk x LT sin0 LT N L W 6 900N sin0 80N 700N 00N 5. Note that the proble coul be solve only fro the balance of torques. Balance of forces allows one to eterine the force F exerte by the wall (not aske).