8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector shall change the component of velocity u|| along the acceleration vector. v 2 r = an Radius of curvature rmin means v is minimum and an is maximum. This is at point P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0. 2 2  R =  = a 2 = 8 meters. 8.6 (C) x2 = 4ay Differentiating w.r.t. y, we get dy x dx = 2a  At (2a, a), dy dx = 1  hence  = 45° the component of weight along tangential direction is mg sin . hence tangential acceleration is g sin  = g 2 8.7 (D) The nature of the motion can be determined only if we know velocity and acceleration as function of time. Here acceleration at an instant is given and not known at other times so D is the correct option 8.8 (C) By energy conservation between A & B  Mg 2R 5 MgR + 0 = 5 + 1 Mv2 2 v = Now, radius of curvature r = 2   ar 2gR / 5  R gcos 37 2 8.9 (D) The friction force on coin just before coin is to slip will be : f = µs mg Normal reaction on the coin ; N = mg The resultant reaction by disk to the coin is = = = mg = 40 × 10–3 × 10 × = 0.5 N 8.10 (D) As 2T sin but dm =  = dm 2 r (for small angle sin 2 m  r ℓ   2  ) As ℓ = 2r  T = m2r/2 Put m = 2 kg  = 10  radian/s and r = 0.25 m  T = 250 N 8.11 (A) when he applies brakes v2 s1  2a if  is the friction coefficient then a = g v 2  s1  2g mv2   when he takes turn mg r v 2 r = g then we can see r > s1 hence driver can hit the wall when he takes turn due to insufficient radius of curvature. 8.12 (A) As tengential acceleration a = dv/dt = dr/dt but  = 4 and dr/dt = 1.5 (reel is turned uniformly at the rate of 2 r.p.s.)  a = 6 , Now by the F.B.D. of the mass. W T – W = g a  T = W (1 + a/g) put a = 6  T = 1.019 W 8.13 (C) For anti-clockwise motion, speed at the highest point should be . Conserving energy at (1) & (2) : 1 mv 2 = mg R  1 m(gR) 2 a 2 2  va2 = gR + gR = 2gR  va = For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B. At the initial position : mv2 T + mgcos600 = c ; R vC being the initial speed in clockwise direction. For vC min : Put T = 0 ;  v =  v /v = = 1  v 2 C : va = 1 : 2 Ans. 8.14 (D) 3R The bob of the pendulum moves in a circle of radius (R + Rsin300) = 2 Force equations :  3R  2 Tsin300 = m    2  Tcos300 = mg 3 2R  tan300 = 2 g =   = Ans. 8.15 (C) vmin = = = 10 m/s 8.16 (A) T cos  + N = mg ...(1) and T sin = m 2 r ...(2) A but T = Kx T = 1.47 × 102 (0.1 sec  – 0.1) (K = 1.47 × 102 N/m) Also r = 0.1 tan  put T, r, m &  in equation (2) we have cos  = 3/5 and T = 9.8 N T cos   0.1m T r M P  Q mg 8.17 (C) T – mg sin  = mv2 R m.(u2  2gℓsin 30º )  3 mg – mg sin30º = 0 ℓ  u0 = 8.18 (B) When the acceleration of bob is horizontal, net vertical force on the bob will be zero. T cos  – mg = 0 The tangential force at that instant is = mg sin  = mg = mg T 8.19 (B) From length constraint on AB a cos 45º = b cos 45º a = b T sin 45º = m(a) mg – T sin 45º = mb T sin 45º a A T sin 45º T T cos 45º mg – ma = ma g T B 45º 45º mg C 2ma = mg a = 2 T cos 45º b mg  mg T = 8.20 (C) V = 4 2  (20)2 = 10 × 100 × tan   tan  = 10 = 5   = tan–1 (2/5) 8.21 (B) In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position. The condition for bead to complete the vertical circle is, its speed at top position vtop  0 From conservation of energy 1 2 2 m top + mg (2R) = 1 mv2 or v = 2 8.22 (A) | V | = → | v | v = v 3 v 2 v 2 ai v 2R  aav = t = t = R  a = ; i R aav = R  3 v 2 = 3 8.23 (A) inertial force M(3g/4) 530 A max Fnet mg B Tmin 530 Fnet is shown in the figure. So, tension will be max. at point A and will be min. at point B. 8.24 (B) For the ring to move in a circle at constant speed the net force on it should be zero. Here spring force will provide the necessary centripetal force.  kx = mx2   = = = 10 rad/sec. Ans. 8.25 (B) dT = dm(ℓ – x)2 dT = m .dx (ℓ  x)2 ℓ T ℓ / 2 m 2  dT =  ℓ (ℓ – x) dx 0 0 m2  x2 ℓ / 2 m2 ℓ2 ℓ2  = ℓx   ℓ  0 = ℓ  2  8   Tension at mid point is : 3 3mℓ2 3mℓ2 T = 8 mℓ 2  stress = 8A  strain = 8AY 8.26 (A) At A ; NA – mg = NA = mg + and At B ; NB = mg – and At C ; NC = mg + mV2 RA mV 2 RA mV 2 RB mV 2 RC As by energy conservation ; RA < RC  N is greatest among all. 8.27 (A, C) As N sin  = mg N cos  = m2 r g tan  = 2r  T2  tan   when  increases T also increases Also T2  r tan  but r = h tan   T2  h tan2  for constant  T2  h Thus when h increases T also increases 8.28 (A, B, C, D) Let N be the normal reaction (Reading of the weighing machine) mv2 at A  N – mg = r Put v  N – mg = mg  N = 2mg = 2W mv2 Also, at E, NE + mg = r = mg  N = 0 hence N > NE by 2W Now at G, NG = mg = W = NC Also NE  0 NA and NA  2 NC 8.29 (A, B, C) Between A and B mgL cos  =  v 2 = 2gL cos A 1 mv2 2 B L Now ar = v 2 B = 2g cos C L and a = g sin  a = = g mv 2 Now, at B TB – mg cos = B L Put VB  T = 3 mg cos When total acceleration vector directed horizontally at tan (90 – ) = ar gsin = 2gcos = 1 tan 2 On solving  = cos–1 1/ 8.30 (A, D) 5 For case : 1 = 6 rad/sec. A/T = 5 6 rad/sec. v B/G = R = 3.14 3  = 3 rad/sec. T/G  = – 6 rad/sec (in opposite direction) 5     4 2 A/G = A/T + T/G = 6   =  6  6  3 rad/s.  =  –  = 2    = rad/sec. A/B A B 3 3 3  and A/B = 30° = 6 rad/sec. Using ;  rel = i (rel) t + 1 2 rel t2    t  0  t = 0.5 sec. Ans. 6 3 8.31 (A) For conical pendulum of length ℓ, mass m moving along horizontal circle as shown T cos = mg (1) T sin = m2ℓ sin (2) g From equation 1 and equation 2, ℓ cos = 2 ℓ cos is the vertical distance of sphere below O point of suspension. Hence if  of both pendulums are same, they shall move in same horizontal plane. Hence statement-2 is correct explanation of statement-1. 8.32 (D) The normal reaction is not least at topmost point, hence statement 1 is false. 8.33 (A) (Moderate) Let the minimum and maximum tensions be Tmax and Tmin and the minimum and maximum speed be u and v. mu2  Tmax = R mv2 Tmin = R  u2 + mg – mg v 2   T = m    + 2 mg.  From conservation of energy u2  v2 = 4g  is indepenent of u. R R and T = 6 mg.  Statement-2 is correct explanation of statement-1. 8.34 (B) vB = If vC = 2vB and vC = Then 2gL = 4 (2gL sin) or sin = 1 or  = sin-1 1 4 4 8.35 (B) Tangential acceleration is a = g cos, which decreases with time. Hence the plot of at versus time may be as shown in graph. Area under graph in time interval t1 = vB – 0 = vB Area under graph in time interval t2 = vC – vB =vB t Hence area under graph in time t1 and t2 is same. A t B C t2  t < t 8.36 (B) → → v B  v C  = v2  v2 – 2v v sin = v 2 = vB B C B C B v = 2v sin  = 2 2gℓ sin  sin   1 1/ 3  sin3 = 1  sin =   4  1 1/ 3  = sin–1  4   4    8.37 (B) Putting h = 0 and the values we have T = 164 N 8.38 (B) Putting h = 2R we get T = 144 – 5gR = 44 N. 8.39 (B) At  = 60°, h = R – R cos 60° = R 2 Putting h = R 2 in v2 = u2 – 2gh We get the result. 8.40 (A) – r, t ; (B) – q,s ; (C) – p ; (D) – q,r (A) F = constant and →  →  0 Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease. When F and u are antiparallel then particle will come to rest for an instant and will return back (B) →  →  0 and F = constant initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing. (C) → F  0 means instantaneous velocity is alway perpendicular to force. Hence the speed will remain constant. And also be circular. → = constant. Since the particle moves in one plane, the resulting motion has to (D) →  2 ˆi  3ˆj and →  6 ˆi  9ˆj . Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed. 8.41 (A) q, (B) q, t (C) q, t (D) p, s v = 2t2 Tangential acceleration at = 4t v 2 2t 4 Centripetal acceleration ac = R  R Angular speed  = v = R 4t , tan  = R at = ac 4tR  R 4t4 t3 8.42 (A) – q,s ; (B) – p,t ; (C) p,t ; (D) q,r From graph (a)   = k where k is positive constant d angular acceleration =  d = k × k = k2  angular acceleration is non uniform and directly proportional to .  (A) q, s From graph (b)  2 = k . Differentiating both sides with respect to . d d k 2 d = k or  d = 2 k is slope of curve hence angular acceleration is uniform.  (B) p, t From graph (c)   = kt d angular acceleration = dt = k k is slope of curve hence angular acceleration is uniform  (C) p, t From graph (d)   = kt2 d angular acceleration = dt = 2kt k is slope of curve hence angular acceleration is non uniform and di- d rectly proportional to t. Slope of the curve is constant (can be seen in given graph) but  = dt increasing with time.  (D) q,r = 2kt 8.43 (20) 8.44 (a) (5) v 2 R = a u2 sin2  = g = 20 m. As a rod AB moves, the point ‘P’ will always lie on the circle.  its velocity will be along the circle as shown by ‘v ’ in the figure. If the point P has to lie on the rod ‘AB’ also then it should have component in ‘x’ direction as ‘v’.  v sin  = v  v = v cosec  here cos = x R 1 3R 3 = R . 5 = 5  sin = 4  cosec  = 5 5 4  v = 5 v ...Ans. x = 5 4 (b)  = VP = 5V R 4R ALTERNATIVE SOLUTION : (a) Let ‘P’ have coordinate (x, y) x = R cos , y = R sin . dx vX = dt = – R sin  d = v  dt d d dt = v R sin   v  and v = R cos  Y dt = R cos   Rsin  = – v cot     v = = = v cosec  ...Ans. 8.45 As the car travels at a fixed speed 1 m/s, hence tangential acceleration will be zero. Therefore, there will be no component of friction along tangent. mv2 Case I : If Mg > ; hence friction force on car r of mass m will be outwards from the centre. mv2 T –  mg = rmax Mg – mg = m (1) rmax Case II : If Mg < mv2 r ; hence friction force on car of mass m will be towards centre. mv2 T + mg = rmin m Mg + mg = rmin .... (2) From equations (1) and (2) rmax  rmin M  m = M  m 8.46 By Newton’s law at B mv2 T – mg cos  = ℓ By energy conservation b/w A and B mgℓ (1 – cos) + 1 2 1 mv2 = 2 m (5ℓg) mv2 = m 5ℓg – 2mgℓ (1 – cos) T = mg cos + m 5 g – 2mg (1 – cos) = 3 mg + 3 mg cos mv2 putting value of of c in equation (i) 3mg (1 + cos) = 6 mg cos2 (/2) 8.47 The free body diagram of the block is (a) For block not to slide along wedge, applying Newton's second law along incline we get mg sin  = m  (ℓ cos ) cos    =