HW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy.

Transcription

1 HW Solutions # MIT - Prof. Kowlski Friction, circulr dynmics, nd Work-Kinetic Energy. 1) 5.80 If the block were to remin t rest reltive to the truck, the friction force would need to cuse n ccelertion of 2.20 m/s 2, ;However, the mximum ccelertion possible to to sttic friction is (0.19)(9.80)=1.86 m/s 2, so the block will ccelerte reltive to the truck. Accelertion with respect to ground is (from F x =+µ s mg = m box ): = µ k g = (0.15)(9.80) = 1.47m/s 2. (1) The ccelertion with respect to truck is: = box truck = = 0.73 (2) The box will move 1.80 m reltive to truck with ccelertion. 2 x 2( 1.80) t = = =2.22s (3) box truck 0.73 In this time, the truck moves the distnce D: 1 1 D = truck t 2 = (2.20)(2.22) 2 =5.43m. (4) 2 2 1

2 2) 5.86 Plese refer to figure (1). Let s denote the common mgnitude of ccelertion s. If block B is to remin t rest on A, the sum of forces cting on B should produce the ccelertion. Setting up the F B = m : B: FBy : N AB m B g = 0 N AB = m B g (5) F Bx : f s = m B µ s N AB (6) NOTE: Here is cse in which friction cuses positive ccelertion. A: F Ay : N A N AB m A g = 0 N A =(m A + m B )g (7) F Ax : T µ k N A f s = m A T = µ k g(m A +m B )+(m A +m B ) (8) T =(m A + m B )( + µ k g) (9) NOTE: The friction force between block A nd B is LESS THAN or EQUAL to µ s N AB. We don t know it yet nd must ssign it n unknown vrible f s. C: FCy = T m C g = m C Cy = m C (10) Replcing T from bove into eqution (10) nd solving for : m C µ k (m A + m B ) = g (11) m A + m B + m C 2

3 To simplify (6) in terms of, we combine (5) nd (6): m B µ s m B g (12) µ s g (13) Using (11): m C µ k (m A + m B ) g µ s g (14) m A + m B + m C Solving the inequlity for m C gives: 3) 5.90 (m A + m B )(µ s + µ k ) m C. (15) 1 µ s Since the lower block hs less coefficient of friction it will ccelerte more rpidly downwrd in the bsence of the string. However the role of string is to mke them hve the sme ccelertion by hving the tension T. S:Smller L:Lrger Writing equtions in the coordinte system which x is prllel to plne nd pointing up nd y perpendiculr to plne nd pointing wy from it. Smller mss: F Sy =0 N S m S g cos θ = 0 (16) F Sx = m S m S g sin θ + N S µ Sk + T = m S (17) Lrger mss: F Ly =0 N L m L g cos θ = 0 (18) F Lx = m L m L g sin θ + N L µ Lk T = m L (19) 3

4 )Adding these two equtions nd solving for : substitution gives (m S + m L )g[cos θ(µ Sk + µ Lk ) sin θ] = (20) m S + m L = g[cos θ(µ Sk + µ Lk ) sin θ] (21) = 2.21m/s 2. (22) b)substituting the bove ccelertion in either of (17) or (19) gives T=2.27 N. c)the upper block will slide down with more ccelertion until they collide! You could solve this section with the bove formlism nd you will get negtive T which mens tht the string must be stiff rod,which supports compressive forces, in order to prevent the collision. 4) Plese refer to figure (2) Mke sure tht you hve understood Exmple 5.22 p.184. ) F y =0 cos βt U cos βt L mg = 0 (23) The net force inwrd is mg T L = T U. (24) cos β F rd =sin βt U +sin βt L =sin β (T U + T L )= m r. (25) b) Solving v 2 4π 2 R F rd = m rd = m = m (26) R τ 2 4

5 (where τ is the period of oscilltions) for period τ : mr τ =2π (27) F rd Using hypotenuse theorem the rdius of oscilltion is R = (1.25) 2 1 =0.75m (28) 3 nd cos β = 4 ; sin β = 5. 5 If you plug in the bove numbers into equtions we derived you ll get: T L =31.0N. (29) so the system mkes 1/ rev/min. τ =1.334s (30) c) When the lower string becomes slck, the system is the sme s the conicl pendulum considered in Exmple 5.22 with cos β = 0.8, the period is (1.25)(0.80) τ slck =2π =2.007s. (31) 9.80 which equivlent to 30 rev/min. d)for oscilltion with less revolutions the Tension in lower string is still Zero nd the problem is gin the conicl pendulum problem; the block will drop to smller ngle. 5

6 5) Work on Sliding Box ) Using work-energy theorem W tot = K 2 K 1 = K (32) K 2 = 0 (33) 1 2 K 1 = mv 0 (34) 2 Here grvity nd norml forcee is perpendiculr to direction of motion nd from the definition W = F. s (35) m g nd N don t do ny work. So W tot is just W f which is the work on the box by friction. 1 W f = W tot =0 1 mv 0 2 = mv 0 2. (36) 2 2 b) Using the definition W = F. s = Fs cos φ (37) becuse friction is lwys ginst the reltive motion in this cse φ = 180 : W tot = W f = F f (x 1 x 0 )( 1) = F f (x 1 x 0 ). (38) Using the result of prt we hve 1 2 F f (x 1 x 0 )= mv 2 0 (39) F f = 1 mv 0 2 (x 1 x 0 ) 2 (40) c)we hve constnt ccelertion sitution in this problem so v i + v f v 0 +0 v 0 v ve = = = (41)

7 (x 1 x 0 )= v ve t stop (42) t stop = 2(x 1 x 0 ) v 0 (43) d)using (32) Here W tot = W person + W f using (35) nd noting tht s = x 2 x 1 we hve W f = F f (x 2 x 1 ). Using the result of prt b we get W f = 1 mv 0 2 x 2 x 1 (44) 2 x 1 Collecting ll the informtion we got we hve 1 2 x 2 x W tot = W person + W f = W person mv 0 =+ mv 2 x (45) 1 2 x 2 x W person = mv 0 + mv 1 (46) 2 x 1 2 7

8 6) U-Control Model Airplne Plese refer to figure (3) We choose the x xis long the ccelertion so we hve 0 ccelertion long the y xis! If you choose the x xis long the wire you hve hve T long nd F. However you should project nd elimintion of two equtions becomes difficult. F = m (47) F y = m y =0 F cos θ T sin θ mg = 0 (48) mv mv mv F x = m x = = F sin θ + T cos θ = (49) R L cos θ L cos θ Eliminting F from (9) nd (10) we get: mv 2 T + mg sin θ = 0 (50) L The physicl condition T 0 will set restriction on θ for fixed v. So v 2 sin θ (51) Lg v 2 θ crit =rcsin (52) Lg Clerly for v 2 = Lg even 90 o will be fine. So there exists finite v sf e : v sf e = Lg (53) 8

9 Figure 1:

10 Figure 2:

11 Figure 3: U-Control Model Airplne 11