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System of Particles and Rotational Motion Test - 40

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System of Particles and Rotational Motion Test - 40
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  • Question 1
    1 / -0
    The moment of inertia of a body rotating about a given axis is 12.0kgm212.0 kg m^2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 55cm and the unit of mass is 10g10g?
    Solution
    I=12Kgm2I=12Kg{ m }^{ 2 }
    New system \Rightarrow unit length =5cm=5cm  and unit mass =10g=10g.
    I=12×1000×(104)gcm2I=12\times 1000\times \left( { 10 }^{ 4 } \right) g{ cm }^{ 2 }
    In II unit length =5cm=5cm.    unit mass =10g=10g.
    I=12×10000×10325×10(10g)(5cm2)I=12\times \dfrac { 10000\times { 10 }^{ 3 } }{ 25\times 10 } \left( 10g \right) \left( 5{ cm }^{ 2 } \right)
    I=48000(10g)(5cm) 2I=48000\left( 10g \right) { \left( 5cm \right)  }^{ 2 }
       =4.8×105=4.8\times { 10 }^{ 5 }
  • Question 2
    1 / -0
    A long slender rod of mass 2 kg2\ kg and length 4 m4\ m is placed on a smooth horizontal table. Two particles of masses 2 kg2\ kg and 1 kg1\ kg strike the rod simultaneously and stick to the rod after collision as shown in Figure.
    Velocity of the centre of mass of the rod after collision is

    Solution
    By conservation of linear momentum,
    2×10+1×10=(2+12)v2\times 10+1\times 10=\left( 2+12 \right) v
    v=305v=\cfrac { 30 }{ 5 }
    v=6v=6㎧
  • Question 3
    1 / -0
    A string is wrapped around a cylinder of mass MM and radius RR. The string is pulled vertically upwards to prevent the centre of mass from falling as the cylinder unwinds the string, The work done on the cylinder for reaching an angular speed ω\omega is:
    Solution
    Work done is the rotational KE acquired be cylinder,
    =12Iω2=\dfrac{1}{2} I\omega ^2
    =12MR22ω2=\dfrac{1}{2}\dfrac{MR^2}{2}\omega ^2
    =MR24ω2.=\dfrac{MR^2}{4}\omega ^2.
    Hence, the answer is MR24ω2.\dfrac{MR^2}{4}\omega ^2.

  • Question 4
    1 / -0
    The radius of gyration of a ring of mass 80 g and diameter 6 cm, about an axis passing through its center of gravity and perpendicular to the plane is:
    Solution

  • Question 5
    1 / -0
    Four particles of equal masses are placed on the vertices of a square and are rotated with a uniform angular velocity about one of the edges (A) as shown in the figure. Which particle will have a larger angular momentum

    Solution
    Angular momentum L = IωI\omega. I is the moment of inertia. For discrete particles, I = mr2mr^2.; r is the distance of the particle from the axis of rotation. 

    The distance of the mass at C is at a larger distance from the origin as compared to other masses. Thus, moment of inertia for mass C is larger and hence angular momentum for C is the largest as compared to other masses
  • Question 6
    1 / -0
    The figure below shows a pattern of two fishes. Write the mapping rule for the rotation of Image A to Image B.
     

    Solution
    The dimensions of the fishes are same in A and B and hence no need to change the dimensions and only rotation by 90 degrees should be carried out
  • Question 7
    1 / -0
    The angular momentum of a particle rotating with an angular velocity ω\omega distant r from a given origin is I. If the origin is shifted by 2r, then the new angular momentum of the particle with same angular velocity will be 
    Solution
    The angular momentum (L) of a particle of mass m distant r from a fixed origin, rotating with an angular velocity ω\omega is given by L=Iω=mr2ωL=I\omega = mr^2 \omega

    If the distance is increased by 2r, then the total distance from the fixed orgin will be 3r

    Thus, angular momentum increases by 9 times 

    Thus, the correct option is b
  • Question 8
    1 / -0
    Three particles of equal masses are placed at the corners of an equilateral triangle as shown in the figure. Now particle A starts with a velocity ν1\nu_1 towards line ABAB, particle BB starts with a velocity ν2\nu_2 towards line BCBC and particle CC starts with velocity ν3\nu_3 towards line CACA. The displacement of CMCM of three particle A, B and C after time tt will be (given if ν1=ν2=ν3\nu_1 = \nu_2 = \nu_3)

    Solution
    REF.Image.
    v1x=v12,v1y =v132 v_{1x} = \dfrac{-v_{1}}{2}, v_{1y}  = \dfrac{-v_{1}\sqrt{3}}{2}

    v2x=v2,v2y=0 v_{2x} = v_{2} , v_{2y} = 0

    v3x=v332,v3y=v332 v_{3x} = \dfrac{-v_{3}\sqrt{3}}{2} , v_{3y} = v_{3}\dfrac{\sqrt{3}}{2}

    using , vcm=i^vx+j^vy \vec{v}_{cm} = \hat{i}v_{x} + \hat{j}v_{y}

    where,
    Vx=mv1/2+v2mmv3/23m=0 V_{x} = \dfrac{-mv_{1/2}+ v_{2}m - mv_{3/2}}{3m} = 0

    vy=mv13/2+mv33/23m=0 v_{y} = \dfrac{-m v_{1}\sqrt{3}/2 + mv_{3}\sqrt{3}/2}{3m} = 0

    \therefore Net velocity = 0 (using vector theory)
    i.e. CM is at rest.

    So, displacement of CM is zero.

    so, (a) is correct.

  • Question 9
    1 / -0
    A vertical rod AC, with C at ground and A being the top most point is moving in such a way, that A moves with a speed of 8 m/s, mid point of the rod B moves with a speed of 4 m/s. along the same direction, with what velocity will the point C move
    Solution
    Since C is at ground and the other points B and A which are equidistant from each other also has velocities that are proportional to their distances from the C. Thus the rod is rotating with an uniform angular velocity about point C. Thus, the speed of point C is zero
  • Question 10
    1 / -0
    In the below example, identify the motion of the objects.

    Solution
    A and B are fixed pulleys and hence can only rotate through the conveyor belt, while the mass C is in translational motion. Thus A and B exhibit pure rotation, while C exhibits pure translation
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