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    System of Particles and Rotational Motion
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    A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

    Let W and W′ be the respective weights of the metre stick and the coin.
    The mass of the metre stick is concentrated at its mid-point.

     i.e., at the 50 cm mark. 

    Mass of the meter stick = m’

    Mass of each coin, m = 5 g 

    When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P.

    The centre of mass is located at a distance of 45 cm from point P. 

    The net torque will be conserved for rotational equilibrium about point R. 

    10 × g(45 - 12) - m'g(50 - 45) = 0 

    ∴                                  m' = 66 g 

    Hence, the mass of the metre stick is 66 g.
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    A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
    (a) Will it reach the bottom with the same speed in each case?
    (b) Will it take longer to roll down one plane than the other?
    (c) If so, which one and why?
    (a) Mass of the sphere = m

    Height of the plane = h

    Velocity of the sphere at the bottom of the plane = v

    At the top of the plane, the total energy of the sphere = Potential energy = mgh

    At the bottom of the plane, the sphere has both translational and rotational kinetic energies. 

    Hence,

    Total energy =     mv2 +  I ω2

    Using the law of conservation of energy, 

    mv2 +  I ω2 = mgh 
    For a solid sphere, the moment of inertia about its centre, I = mr2 
    Hence, equation (i) becomes

    mr2 + [mr2] ω2  =  mgh 

                 v2 + r2ω2  =  gh 
    But we have the relation, v = rω 

    ∴      v2 + v2  =  gh 

                      v2 () = gh 

                                 v =gh 

    Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g).

    Both these values are constants.

    Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

    (b) Consider two inclined planes with inclinations θ1 and θ2, related as, 

                            θ1 < θ2 

    The acceleration produced in the sphere when it rolls down the plane inclined at θ1 is, 

                          a1 = g sin θ1 

    The various forces acting on the sphere are shown in the following figure.



     
    R1 is the normal reaction to the sphere.

    Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ2 is, 

                           a2 = g sin θ2 

    The various forces acting on the sphere are shown in the following figure.


     
    R2 is the normal reaction to the sphere. 

    θ2 > θ1; sin θ2 > sin θ1                          ... (i) 

    ∴ a2 > a1                                                … (ii) 

    Initial velocity, u = 0 

    Final velocity, v = Constant 

    Using the first equation of motion, we can obtain the time of roll as

    v = u + at 

    ∴ t ∝ (1/α) 

    For inclination θ1 : 

                          t1     ∝ (1/α1) 

    For inclination θ2 : 

                           t2     ∝ (1/α2) 

    From above equations, we get, 

                               t2 < t1 

    Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination. 
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    Answer
    A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
    Given,

    Radius of the hoop,     r = 2 m

    Mass of the hoop, m = 100 kg

    Velocity of the hoop, v = 20 cm/s = 0.2 m/s

    Total energy of the hoop = Translational K.E. + Rotational K.E.

                                ET = mv2 + I ω2
    Moment of inertia of the hoop about its centre, I = mr2 

                               ET = mv2 + (mr2)ω2 
    Using the relation, v = rω 

    ∴                          ET = mv2 + mr2ω2 

                                   = mv2 +mv2 

                                   = mv2 

    The work required to be done for stopping the hoop is equal to the total energy of the hoop. 

    ∴ Required work to be done, W = mv2 

                                            = 100 × (0.2)2 

                                            = 4 J.
    108 Views
    Answer
    The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
     
    Mass of an oxygen molecule, m = 5.30 × 10–26 kg

    Moment of inertia, I = 1.94 × 10–46 kg m2 

    Velocity of the oxygen molecule, v = 500 m/s 

    The separation between the two atoms of the oxygen molecule = 2r  

    Mass of each oxygen atom = 
    Hence, moment of inertia I, is calculated as

    ()r2 + ()r2 = mr2 

                         r =  
     =  0.60 × 10-10 m 

    It is given that, 

           KErot = KEtrans 

     I ω2 =  × × mv2 

         mr2ω2 =mv2 

               ω =  

                  =

                  = 6.80 × 1012 rad/s.
    184 Views
    Answer
    From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
    Mass per unit area of the original disc = σ 

    Radius of the original disc = R 

    Mass of the original disc, M = πR2σ 

    The disc with the cut portion is shown in the following figure,

                 
    Radius of the smaller disc = R/2 

    Mass of the smaller disc, M’ = π (R/2)2σ

                                              =     π R2σ / 4 

                                               =  M / 4 

    Let O and O′ be the respective centres of the original disc and the disc cut off from the original.

    As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.

    It is given that, 

    OO′= 

    After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses.

    The two masses are, 

    M (concentrated at O), and –M′ concentrated at O′.

    (The negative sign indicates that this portion has been removed from the original disc.)

    Let x be the distance through which the centre of mass of the remaining portion shifts from point O. 

    The relation between the centres of masses of two masses is given as, 

    x =

    For the given system,

    x =
       = 

    (The negative sign indicates that the centre of mass gets shifted toward the left of point O.) 


    The centre of gravity of the resulting flat body is R/6; from the original centre of the body and opposite to the centre of the cut portion.

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