Lecture 52. Dynamics - Variable Acceleration

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1 Dynamics - Vaiable Acceleation Lectue 5 Example. The acceleation due to avity at a point outside the eath is invesely popotional to the squae of the distance x fom the cente, i.e., ẍ = k x. Nelectin ai esistance, show that if a paticle is pojected vetically upwads with speed u fom a point on the Eath s suface, its speed v in any position x is iven by v = u R ( 1 R 1 x ) whee R is the adius of the Eath and the acceleation due to avity at the Eath s suface. Show that the eatest heiht H above the Eath s suface eached by the paticle is H = and find the speed needed to escape the Eath s influence. (R = 6400 km, = 10 m/s =0.01 km/s.) Solution. u R R u Initially, v = u and x = R ẍ =. ẍ = k x = kx d dx ( 1 v )= kx 1 v = kx 1 + c = k x + c When v = u, x = R 1 u = k R + c c = 1 u k R 1 v = k x + 1 u k R v = u + k x k R u k( 1 R 1 x ). But when x = R, ẍ = = k R k = R. v = u R ( 1 R 1 x ) eatast heiht is eached when v =0. u R ( 1 R 1 x )=0 R ( 1 R 1 x )=u 1 R 1 x = u R 1 x = 1 R u R = R u R x = R R u eatest heiht above Eath s suface is R R u R = R R(R u ) R u = R R +Ru R u = u R R u. If it escapes the Eath s influence, H (since it keeps oin). So R u = 0 and u =R u = R (i.e., the escape velocity) (0.01)(6400) = 8 km/s 11.3 km/s.

2 Resisted Motion - fom Cooneos, 198a Case 1. - pojected upwads. A paticle of mass m is pojected vetically upwads with velocity u, in a medium whose esistance to the motion vaies as the velocity of the paticle. Pove that the time to each the hihest point of the path is 1 k ln(1 + ku ), whee k is a constant, and find this eatest heiht. { is the acceleation due to avity.} Solution. Let the velocity v and the displacement x be measued upwad ( ) fom the point of pojection O. Since the esistance R to the motion vaies as the velocity v, then R = Kv whee K is a positive constant. (Diection of R will be downwad since esistance is always in the opposite diection since esistance is always in the opposite diection to the velocity.) Fo convenience late, let K = mk and R = mkv. Hence, the foce actin on the paticle in the downwad ( ) diection, is m + R. Now, by Newton s Second Law, the foce on the paticle in the upwad ( ) diection is mẍ. Thus, mẍ = m mkv, i.e., ẍ = kv is the equation of motion of the paticle. Fistly, ẍ = dv dt, dv dt = kv & dt dv = 1 +kv. t = dv +kv = 1 k ln( + kv)+c, whee C is a constant... (1) By data, when t =0,v= u, 0= 1 k ln( + ku)+c, whence c = 1 k ln( + ku) (1) becomes t = 1 k ln( + ku) 1 k ln( + kv) = 1 +ku k ln +kv... () At the hihest point on the path v =0 t = 1 +ku k ln. That is, the time to each the hihest point is 1 k ln(1 + ku ). Secondly, takin ẍ = v dv dv dx, then the equation of motion becomes v dx i.e., dv dx = (+kv) v and dx dv = v +kv... (3) Hence, x = v dv +kv = 1 k (+kv) v +kv dv = 1 k (1 x = 1 k (v k ln( + kv)) + c 1... (4) Now, when x =0,v = u 0= 1 k [u k ln( + ku)] + c 1 Thus, fom (4), we have x = 1 k [u k ln( + ku)] 1 k [v k +kv ) dv = kv, ln( + kv)] The eatest heiht H is iven, when v =0,by H = 1 k [u k ln( + ku)] 1 k [ k ln ] = u k k ln( +ku )= 1 k [uk ln(1 + ku )].

3 {Altenatively, the distance tavelled fom v = u to v = 0, i.e., the eatest heiht H, is iven by H = 0 v dv u +kv..., see (3) above.}

4 Lectue 53 Case - dopped down A paticle of mass m falls vetically fom est, in a medium whose esistance is popotional to the velocity. Find the teminal velocity of the paticle and deive expessions in tems of (i) t (ii) v. Solution. ẍ = kv dv dt = kv dv kv = dt 1 k ln( kv) =t + c when t =0,v =0 1 k ln =0+c, c = 1 k ln. t = 1 k ln 1 k ln( kv) = 1 k ln ln kv = kt kv = ekt = e kt kv kv kv = e kt kv = e kt v = k (1 e kt ) & lim v = t k teminal velocity = k (since as t,e kt 0) dx dt = k (1 e kt ) x = k (1 e kt ) x = k (1 e kt ) dt = k (t + 1 k e kt )+c & when x =0,t=0 c = k x = k (t + 1 k e kt ) k = k (kt + e kt 1) ẍ = kv v dv dx = kv v dv kv = dx x = v dv kv = 1 k ( kv)+ k kv dv = ( 1 k + k 1 kv dv) = v k k ln( kv)+c

5 When x =0, v =0 c = k ln x = v k k ln( kv)+ k ln = v k + k ln kv.

6 Lectue 54 Cicula Motion Summay (Physics) Constant anula velocity ω = dθ dt 1. v = ω. T = π ω 3. θ = ωt (anula displacement) 4. ω = v Tanential and Nomal components of Acceleation. Chane in velocity alon tanent at P =(v + δv) cos δθ v but cos δθ 1 v + δv v = δv. Alon nomal at P =(v + δv) sin δθ but sin δθ δθ =(v + δv)δθ = vδθ + δvδθ but δθδv we inoe = vδθ. tanential acceleation fom P to Q = δv δt δv tanential acceleation at P = lim δt 0 δt = dv dt = d(ω) dt = d dt ( dθ dt )=d θ dt = θ -this equals zeo fo constant anula velocity.

7 Method 1-asabove δv Acceleation alon tanent at P = lim δt 0 δt = dv dt = d dω dt (ω) = dt = d dt ( dθ dt )=d θ dt Acceleation alon nomal at P = lim δt 0 vδθ δt Method. Acceleation in nomal/tanential diections. = v dθ dt = ω.ω = ω. = θ Conside OA to be the x axis. x = cos θ, y = sin θ dx dt = dx dθ dθ dt = sin θ. θ dy dt = dy dθ dθ dt = cos θ. θ. Acceleation: d x dt = d dt ( sin θ. θ) = sin θ. d θ dt + θ d dt ( sin θ) = sin θ. θ + θ.( cos θ). dθ dt = sin θ. θ cos θ.( θ) d y dt = cos θ. d θ dt + θ. d dθ dθ ( cos θ). dt = cos θ. θ sin θ.( θ)

8 Lectue 55 Acceleation Components d x dt = sin θ. θ cos θ. θ d y dt = cos θ. θ sin θ. θ acceleation in diection PN =ÿ sin θ +ẍ cos θ = cos θ sin θ. θ sin θ. θ sin θ cos θ. θ cos θ. θ = ( θ) (sin θ + cos θ) = θ = ω diection is towads cente. acceleation in diection PT =ÿ cos θ ẍ sin θ = cos θ. θ sin θ cos θ. θ + sin θ. θ + sin θ cos θ. θ = θ(cos θ + sin θ) = θ (o ω o d θ dt ) Example A stin 50cm lon will beak if a mass exceedin 40k is hun fom it. A mass of k is attached to one end of a stin and it is evolved in a cicle in a hoizontal plane. Find the eatest anula velocity without the stin beakin (avity = ). Foce 40N (i.e., m/sec ) Hoizontal foces T = ma = mω = k.0.5m.ω But T 40 ω 40 (stn is elastic and not massive) ω 40 ad/sec.

9 Conical Pendulum - need to analyse foces vetically and hoizontally. T cos θ = m Hoizontal foces: T sin θ = mω T sin θ mω T cos θ = tan θ = m = ω independant of mass Fom diaam, tan θ = h h = ω h = ω ω = h ω = h But peiod = π ω =π h T sin θ = mω but sin θ = l = l sin θ T sin θ = ml sin θω T = mlω. Example a od of lenth R is attached hoizontally to a otatin vetical suppot. A stin of lenht l with a mass m attached to the end, is attached to the od as shown. The suppot is otated at speed of ω ad/s. Show that ω = tan θ R+l sin θ. Vetically, T cos θ = m, hoizontally, T sin θ = mω,but = R + l sin θ T sin θ mω T cos θ = tan θ = m = ω (R+l sin θ)ω = ω = tan θ R+l sin θ ω = tan θ R+l sin θ

10 Lectue 56 Tension in Stin. T sin θ = mω, T cos θ = m, T = mlω T sin θ = m ω 4, T cos θ = m T sin θ + T cos θ = T (sin θ + cos θ)=t = m ω 4 + m = m ( ω 4 + ) T = m ω 4 + Example. [ ] Show θ = cos 1 (M+m) Mlω Resolvin vetical foces at B: T cos θ = m T = m cos θ Resolvin foces at C: Hoizontal: T 1 sin θ + T sin θ = Mω (but = l sin θ) Vetical: M + T cos θ = T 1 cos θ M + m = T 1 cos θ T 1 = M+m cos θ T 1 sin θ + T sin θ = Mlsin θω T 1 + T = Mlω (M+m) cos θ + m cos θ = Mlω (M+m) cos θ = Mlω cos θ = (M+m) Mlω = ( ) M+m cos θ

11 ( ) θ = cos 1 (M+m) Mlω

12 Motion Aound a Cuved Tack Lectue 57 Example - fom Cooneos Supplement Set 4E Q1ii A moto ca of mass t is oundin a cuve of adius 840m on a level tack at 90kmh 1. What is the foce of fiction between the wheels and the ound? Solution Note fictional foce supplies centipetal foce. 90, km/h= 3600 = 5m/s. Fictional foces = mv = kms = 1488N.

13 Motion aound a banked tack Lectue 58 If it is oin slow, it will slide down. If it is oin fast, it will slide up. Thee will be no fictional foces if it does not slide up o down. This is the ideal bankin of the tack. So ideal bankin of a tack implies F =0. We wish to detemine the anle of bankin to avoid side-slip. N cos θ = m. hoizontal foces: N sin θ = mv Eliminate N: mv N sin θ ( N cos θ = tan θ = ) m = v i.e., θ is independant of mass and is dependant on and velocity. Example. A cuved ailway tack of adius 690m is desined fo tains tavellin at an aveae speed of 48 km/h. Find the ideal bankin anle and how much should the oute tack be aised if the ail aue=1.44m (distance between tacks). Solution. 48km/h= 48(1000) 3600 = tan θ = v = ( ) θ =1 30

14 h 1.44 sin 1 30 = h =1.44 sin 1 30 = m= 3.77cm

15 Lectue 59 Cicula Motion aound a cuved tack (cont d) Example 1. A ca of mass m ounds a cuve of adius banked at an anle θ to the hoizontal with speed v. iff is the sideways fictional foce between the tyes and the oad and N is the nomal eaction of the oad on the tye, show that F = m cos θ( v tan θ), N= m cos θ( v tan θ +1). Solution: hoizontally: mv = N sin θ + F cos θ vetically: N cos θ = m + F sin θ eliminate N: N sin θ = mv F cos θ N cos θ = m + F sin θ N sin θ N cos θ = mv F cos θ m+f sin θ mv sin θ F cos θ cos θ = tan θ = m+f sin θ m sin θ + F sin θ = mv F sin θ + F cos θ = mv F = m cos θ( v cos θ F cos θ cos θ m sin θ tan θ). Eliminate F : ( mv = N sin θ + F cos θ) sin θ N cos θ = m + F sin θ) cos θ mv sin θ = N sin θ + F cos θ sin θ. Subtact. N cos θ = m cos θ + F sin θ cos θ mv sin θ N cos θ = N sin θ m cos θ N sin θ + N cos θ = mv sin θ + m cos θ N = m cos θ( v tan θ +1) Example. A cat tavels at vm/s alon a cuved tack of adius Rm. Find the inclination of the tack to the hoizontal if thee is to be no tendancy fo the ca to slip sideways. If the speed of the ca is V m/s pove that the sideways fictional foce on the wheels of the ca of mass m is m(v v ). v4 +R

16 Solution. hoizontally: mv R = N sin θ. vetically: m = N cos θ eliminate N: N sin θ N cos θ = mv R m. So tan θ = v R θ = tan 1 v R. hoizontally: mv R = N sin θ + F cos θ vetically: m = N cos θ + F sin θ F = m cos θ( V R tan θ) = mr ( V v 4 +R R v R )= m(v v ) v 4 +R

Physics 101 Lecture 6 Circular Motion

Physics 101 Lecture 6 Circular Motion Physics 101 Lectue 6 Cicula Motion Assist. Pof. D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Equilibium, Example 1 q What is the smallest value of the foce F such that the.0-kg block will not slide