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A circular platform is mounted on the frictionless vertical axle. Its radius R =2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms-1 relative to the ground. Time taken by the man to complete one revolution is

  • π sec

  • 3π/2 sec

  • 2π sec

  • 2π sec


A.

π sec

For conservation of angular momentum

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A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected  to a torque which produces a constant angular acceleration of 2.0 rad/sec2. It's net acceleration in m/s2 at the end of 2.0 s is a approximately.

  • 7.0

  • 6.0

  • 3.0

  • 8.0


D.

8.0

A uniform circular disc of radius 50 cm at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be shown by the figure given below:

Therefore,

Angular acceleration, 
Angular speed, 
Centripetal acceleration, ac
= 42 x 0.5
= 16 x 0.5
= 8 m/s2

Linear acceleration at the end of 2 s is,

at
Therefore, the net acceleration at the end of 2.0 sec is given by,



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A mass m moves in a circle on a smooth horizontal plane with velocity vo at a radius Ro. The mass is attached to a string which passes through a smooth hole in the plane as shown.

The tension in the string is increased gradually and finally m moves in a circle in a circle of radius Ro/2. The final values of the kinetic energy is

  • mv02

  • 1/4mvo2

  • 2mvo2

  • 2mvo2


C.

2mvo2

Conserving angular momentum
Li = Lf
mvoRo = mv' (Ro/2)

v'=2vo

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Three identical spherical shells, each of mass m and radius r are placed as shown in the figure. Consider an axis XX', which is touching to two shells and passing through the diameter of the third shell. Moment of inertia of the system consisting of these three spherical shells about XX' axis is

  • 11/5 mr2

  • 3 mr2

  • 16/5 mr2

  • 16/5 mr2


D.

16/5 mr2

The total moment of inertia of the system is 


I = I1 +I2 +I3

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A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

  • wx over straight d
  • wd over straight x
  • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight x end fraction
  • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight x end fraction

D.

fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight x end fraction

As the weight w balances the normal reactions.


So, 

w= N1 +N2   ----- (i) 
Now balancing torque about the COM,
i.e. anti -clockwise momentum= clockwise momentum

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