1.     Aparticle is moving on a circular path.
2
                                                                                              sin 94
      The angle turned by radius vector is                                  (b)                               cos
                                                                                          R              R
       e -(1 -e ), where is in radian and
      tis in second. The angular acceleration                               (c)               COs 9;          sin 9 j
                                                                                          R               R
      at t= 0 is
            1
      (a) rad s2                                   1                        (d)           i+;
            3
                                         (b)            rad s-2
                                                                                   R               R
                                                                       3.   A particle moves on a circular path of
      (c) 10 rad s-2                     (d) -- rad s-2                     radius 0.2 m with an angular acceleration
                                                   6
                                                                            0.25 rad s-2. The particle starts from
 2. For aparticle in auniform circular motion,                               rest, find the time at which the
      the acceleration a at a point P(R, 0) on the                          magnitude of radial acceleration and
      circle of radius R is (here, 0 is measured
      from the -axis)
                                                                            the magnitude of tangential acceleration
                                                                            are same.
      (a)                                                                    (a) 1 s                            (b) 2 s
                    cos 0i+              sin j
                R                R                                          (c) 3 s                            (d) 4 s
                                                              ANSWERS
                                                    de            /3
 |1. (b) :8= (1-et3), so, ) = dt                              3
                                                                             a=
                                                                                  R
                                                                                          towards centre of circle
                    do                                                                                 [centripetal acceleration)
                         -   e   /   3
      or
                    dt
                             1             -2
                                                                                  a=      (-cosei-sine,)
                                                                                           R
      At t= 0, a=-rad s
                             9
|2. (c): For a particle in a uniform circular                                                  R   -cose i-sn®j
                                                                                                           R
      motion                                                           3.    (b) : a; =a,
                                                                            Or ra = r0² or a = 0² or 0.25 = 2
                                               PR, 0)                             () = V25x10                 -0.5 rad s1
                                                                                  () =at or 0.5 = 0.25t
                                                                                      5x10        10
                                                                                  t=
                                                                                              25x10-2-==2s
                                                                                                       5
1.    In the uniform circular motion,                 5. Three identicai particle as shown in igure
      (a) acceleration andvelocity both remain             tied with string and speed of outermost
          constant                                         particle is v then ratio T,: T,: T, will be
      (b) acceleration and speed both remain               (where I, is tension in outside string)
          constant
      (c) acceleration and velocity both keep                             m
          on changing
      (d) acceleration and speed both change
                                                           (a) 3: 5:7             (b) 3:5:6
2.    If aparticle moves in a circle, describing           (c) 3:4:5              (d) 7:5:3
      equal angles in equal intervals, the velocity
      vector                                          6.   A car passes alorng a curved path of a
      (a) remains constant                                 road according to figure moving with
      (b) changes in magnitude                             speed of 72 km h, radius of curve is 10
      (c) changes in direction                              m. If mass of the car is 500 kg, reaction
      (d) changes both in magnitude and                    force on car at lowest point of curved
           direction                                       part will be
 3.   When a body moves with a constant
      speed along acircle                                                     P
       (a) no work is done on it                           (a) 25 kN              (b) 50 kN
       (b) no acceleration is produced in the              (c) 75 kN              (d) None of these
           body
      (c) no force acts on the body
                                                      7.   A motorcycle is going on an over bridge of
                                                           radius R. The driver maintainsa constant
      (d) its velocity remains constant
                                                           speed. As the motorcycleis ascending on
 4.   A mass is supported on a frictionless                the over bridge, the normal force on it
       horizontal surface. It is attached to               (a)   increases
      string and rotates about aixed centre                (b) decreases
      at an angular velocity o. lf the length              (c) remains same
      of the string and angular velocity are                (d) tluctuates
        doubled, the tension in the string which      8.   A car is travelling with linear velocity r
      was initially T, is now                              on acircular road of radius r. If the speed
      (a) To                (b) T/2                        is increasing at the rate of am s, then
      (c) 4To               (d) 8T,                        the resultant acceleration will be
                                                      (a) 3.7 m s²                   (b)   2.7 ms2
     (a)                     (b)                      (c) 1.8 m s2        (d) 2.0 ms
                                                  10. A particle describes a horizontal circle
     (c)                     (d)                      in a conical funnel whose inner surface
                                                      is smooth with speed of 0.5 m s. What
9.   A car is moving with speed of 30 m s             is the height of the plane of circle from
     on a circular path of radius 500 m. If it        vertex of the funnel?
     increases its speed at the rate of 2 ms-?        (a) 0.25 cm                    (b) 2 cm
     the resultant acceleration will be
                                                      (c) 4 cm                       (d) 2.5 cm
                                              ANSWERS
1. (c)                  2.     (c)
3. (a)
4. (d) :T, =mar
   and T = m(2o)(2r) = 8T,                           a=
5. (b): T =mo(3)
     I,- T, - mo{2/)                              9. (b) :v=30 ms-l, r =500 m,
           T,- ma(51)                                 du
                                                           =2 m s2
     T, - T, - mal                                    dt
     ’     T,= mo(6/)                                                30
     I:T, :T, =3:5:6                                       |2+|                 - 2.7 ms-2
                                                                     500
                     72x 1000            1
                                                                                my
6. fa) : a, =                        X
                                                  10. (d) :N sin 0 =
                       3600              10
                  = 40 m s2                          and N cos           = ng
     F, = ma, = (500 x 40) = 20000 N= 20 kN
      F, mg= (500 x 10) =5000 N =5 kN                      tan       =
     Reaction force,F= 20 kN +5 &N= 25 kN                                rg                          mg
                                                            h 2
                                                     or
7. (b):                  mg (at highest point)
              dv =A m s-2                                  h=                    =2.5 cm
8. (b) : a, =                                                    8         10
              dt