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bharath devasani Grade: 12

Two heavy metallic plates are joined together at 90o to each other. A laminar sheet of mass 30 Kg is hinged at the line AB joining the two heavy metallic plated. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its centre of mass is 1.2 Kg m2. Two rubber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5 m from the line AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacle is 6 N-s,


(a) Find the location of the centre of mass of the laminar sheet from AB.
(b) At what angular velocity does the laminar sheet come back after the first impact?
(c) After how many impacts, does the laminar sheet come to rest?

7 years ago

Answers : (1)

askIITianExpert_ Sukumar_ IITKharagpur
askIITians Faculty
34 Points

Let the initial angular velocity of the laminar sheet be w

Let w’ be the angular velocity with which sheet turns back after the impact with the hinge 

Now, Change in angular moment = moment of impulse

=> IAB w - IAB (-w’) = 0.5 * 6     

=> IAB (w+w’) = 3 Nm-s …..............eq 1

Now, as per question,  the hinge doesn’t impart any impulse, so we can say that

     Change in linear momentum of laminar sheets = impulse on the sheet due to each obstacle

 => ml (w+w’) = 6 N-s ……..............eq 2

where l is the distance of the centre of mass of the laminar sheet from AB

   By using the theorem of parallel axes, we can find out the moment of inertia of the laminar sheet about AB 

     IAB = I0 +ml2                 ……........eq 3

Dividing eq 1 by eq 2 , we get

IAB /ml =  1/2

=> (I0 +ml)/ml =1/2 (by using eq 3 )

=> 2ml2 –ml +2Io =0

put the value of m and Io from the question and solve the quadratic equation

Solving above quadratic equation , we get l =  0.4 m or 0.1 m

now, if we put l=0.4 in eq 2, we get negative w’ which is not possible (as the sheet must come back)

so we get l = 0.1m

corresponding w’ = 1 rad/s

ans. a) l=0.1 m

ans. b) w’= 1 rad/s

ans. c) Sheet will never come to rest

7 years ago
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