System of Particles and Rotational Motion

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Question 1

A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C

  • 2

  • 3

  • 4

Solution

C.

4

The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.

The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the midpoints of the horizontal and vertical portions of the T.

Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by

r = (m1r1 + m2r2)/(m1 + m2) = (m×2L + 2m×L)/(m + 2m) = 4L/3

[ Note that we have used the equation, r = (m1r1 + m2r2)/(m1 + m2) for the position vector r of the centre of mass in terms of the position vectors r1 and r2 of the point masses m1 and m2. We could use the simple equation involving the distances from C since the points are collinear].

Question 2

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension

  • angular momentum is conserved

  • angular momentum changes in magnitude but not in the direction

  • angular momentum changes in direction but not in magnitude

  • angular momentum changes both in direction and magnitude

Solution

C.

angular momentum changes in direction but not in magnitude

Question 3

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc.The value of α is

  • 1/3

  • 1/2

  • 1/6

  • 1/4

Solution

A.

1/3

In this question distance of centre of mass of new disc is αR not α/R.

negative fraction numerator 3 straight M over denominator 4 end fraction space straight alpha space straight R space plus space straight M over 4 space straight R space equals 0
rightwards double arrow space straight alpha space equals space 1 third
Question 4

A cylindrical tube, open at both ends, has a fundamental frequency, f, in the air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

  • m1r1:m2r2

  • m1 :m2

  • r1 :r2

  • 1:1

Solution

C.

r1 :r2

As their period of revolution is same, so its angular speed is also same. Centripetal acceleration is circular path,
a= ω2r
Thus, 
straight a subscript 1 over straight a subscript 2 space equals space fraction numerator straight omega squared straight r subscript 1 over denominator straight omega squared straight r subscript 2 end fraction space equals space straight r subscript 1 over straight r subscript 2

Question 5

A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by (n is an integer)

  • fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared straight n squared straight h squared over denominator 2 straight m subscript 1 superscript 2 straight m subscript 2 superscript 2 straight r squared end fraction
  • fraction numerator straight n squared straight h squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction
  • fraction numerator 2 straight n squared straight h squared over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight r squared end fraction
  • fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Solution

D.

fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight h squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction

Rotational kinetic energy of the two body system rotating about their centre of mass is
RKE space equals space 1 half μω squared straight r squared
where comma space straight mu space equals space fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction equals space reduced space mass
and space angular space momentum comma space straight L space equals space μωr squared space equals space fraction numerator nh over denominator 2 straight pi end fraction
straight omega space equals fraction numerator nh over denominator 2 πμr squared end fraction
therefore space RKE space equals space 1 half μω squared straight r squared space equals space 1 half space straight mu. open parentheses fraction numerator nh over denominator 2 πμr squared end fraction close parentheses squared straight r squared
space equals fraction numerator straight n squared straight h squared over denominator 8 straight pi squared μr squared end fraction space equals space fraction numerator straight n squared straight ħ squared over denominator 2 μr squared end fraction space open parentheses where comma straight ħ space equals space fraction numerator straight h over denominator 2 straight pi end fraction close parentheses
fraction numerator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight n squared straight ħ squared over denominator 2 straight m subscript 1 straight m subscript 2 straight r squared end fraction