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System of Particles and Rotational Motion

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω=10 π rad s–1 

Coefficient of kinetic friction, μk = 0.2 

Initial velocity of both the objects, u = 0 

The frictional force causes motion between two objects. 

As per Newton’s second law of motion,

Frictional force, 
f =ma 

                  μkmg= ma 

where, 

 a = Acceleration produced in the objects 

= Mass 

Therefore,

    
a = μkg                                           … (i)

As per the first equation of motion,

Final velocity of the objects can be obtained as,

v = u + at 

   = 0 + μkgt 

   = μkgt                                             … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed. 

Torque, τ= – I α 

where, 

α = Angular acceleration 

μkmgr = –I α  



 
α = fraction numerator negative straight mu subscript straight k space end subscript mgr over denominator straight I end fraction                                   ...(iii) 

Using the first equation of rotational motion to obtain the final angular speed, 

ω = ω0 + α

   = ω0 + (-μkmgr / I)t                          ...(iv) 

Rolling starts when linear velocity, 

            v = rω 

Therefore, 

  v = r (ω0 μkmgrt / I)                         ...(v) 

Equating equations (ii) and (v), we get

μkgt = r (fraction numerator straight omega subscript straight o space minus space straight mu subscript straight k mgrt over denominator straight I end fraction

      = rω0 - fraction numerator straight mu subscript straight k mgr squared straight t over denominator straight I end fraction                          ...(vi)

For the ring, 

       I = mr

∴ μkgt = rω0 -fraction numerator straight mu subscript straight k space mgr squared straight t over denominator mr squared end fraction
        = rω0 - μkgt 

2μkgt = rω

∴    t = fraction numerator rω subscript straight o over denominator 2 straight mu subscript straight k straight g end fraction 

       = fraction numerator 0.1 space straight x space 10 space straight x space 3.14 over denominator 2 space straight x space 0.2 space straight x space 9.8 end fraction

       =0.80 s                                       ...(vii)

For the disc:

          I = open parentheses 1 half close parenthesesmr2

∴    μkgt = rω0 - μkmgr2t / 1 halfmr

            = rω0 - 2μkgt 

    3μkgt = rω

Therefore,

 
t = fraction numerator rω subscript straight o over denominator 3 space straight mu subscript straight k straight g end fraction

    = fraction numerator 0.1 space straight x space 10 space straight x space 3.14 space over denominator 3 space straight x space 0.2 space straight x space 9.8 end fraction space equals space 0.53 space sec      ...(viii) 

Since td > tr, the disc will start rolling before the ring. 
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Is it necessary for centre of mass to lie within the body?

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Is it necessary that there should be matter at the centre of mass of system?

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What is the need of centre of mass?

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Define centre of mass.

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