System of Particles and Rotational Motion

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Question
CBSEENPH11020654

straight A space force space straight F space equals space straight alpha stack space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 straight i with hat on top space minus space 6 space straight j with hat on top space minus 12 space straight k with hat on top. The value of α for which angular momentum about origin is conserved is
  • -1

  • 2

  • zero

  • 1

Solution

A.

-1

When the resultant external torque acting on a system is zero, the total angular momentum of a system is zero, the total angular momentum of a system remains constant.This is the principle of the conservation of angular momentum.
Given,
straight F space equals space straight alpha space straight i with hat on top space plus 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 space straight i with hat on top space minus 6 space straight j with hat on top space minus 12 space straight k with hat on top
As comma space angular space momentum space about space origin space is space consverved
straight i. straight e. comma
Torque space equals space constant
rightwards double arrow space Torque space equals space 0 space rightwards double arrow space straight r space xF space equals space 0

open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 6 end cell cell negative 12 end cell row straight alpha 3 6 end table close vertical bar space equals space 0

rightwards double arrow space left parenthesis negative 36 space plus 36 right parenthesis straight i with hat on top space minus space left parenthesis 12 space plus space 12 straight alpha right parenthesis space straight j with hat on top space plus left parenthesis 6 space plus 6 straight alpha right parenthesis space straight k with hat on top space equals 0

rightwards double arrow space 0 space straight i with hat on top space minus space 12 space straight i with hat on top space left parenthesis space 1 plus straight alpha right parenthesis straight i with hat on top space plus space 6 space left parenthesis 1 space plus space straight alpha right parenthesis straight k with hat on top space equals 0

6 space left parenthesis 1 space plus straight alpha right parenthesis space equals 0 space rightwards double arrow space straight alpha space equals negative 1

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Question
CBSEENPH11020737

A bar magnet having a magnetic moment of 2 x 104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10-4 T exists in the space. The work  is done in taking the magnet slowly from a direction parallel to the field to a direction 60o from the field is

  • 0.6 J

  • 12 J

  • 6 J

  • 2 J

Solution

C.

6 J

The work done in rotating a magnetic dipole against the torque acting on it, when placed in a magnetic field is stored inside it in the form of potential energy. 
When magnetic dipole is rotated from initial position θ = θ1 to final position θ = θ2, then work done = MB (cos θ1 = cos θ2)
space equals space MB space open parentheses 1 minus 1 half close parentheses
space equals space fraction numerator 2 space straight x space 10 to the power of 4 space straight x space 6 space straight x space 10 to the power of negative 4 end exponent over denominator 2 end fraction space equals space 6 space straight J

Question
CBSEENPH11020641

A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is 

  • 1.5 Mv

  • 2 Mv

  • zero

  • Mv

Solution

B.

2 Mv

Impulse |J| = |Δp|
= Mv - (- Mv)
 = 2 Mv

Question
CBSEENPH11020680

A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed.straight omega subscript straight i Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with constant angular speed.straight omega subscript straight fThe energy lost by the initially rotating disc due to friction is 

  • 1 half fraction numerator straight I subscript straight b superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction space straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight t superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b minus straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Solution

D.

1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Loss of energy, 
increment straight E space equals space 1 half space straight I subscript straight t space straight omega subscript straight i superscript 2 minus 1 half fraction numerator straight I subscript straight t superscript 2 straight omega subscript straight i superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction

equals space 1 half fraction numerator straight I subscript straight b straight I subscript straight t straight omega subscript straight i superscript 2 over denominator 2 left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction

Question
CBSEENPH11020551

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

  • π/2 sec

Solution

A.

π sec

For conservation of angular momentum
Iω space equals space mvr
200 space straight x space straight omega space equals space 50 space straight x space 2 space straight x space 1
space straight omega space equals space 1 half space rad divided by straight s
straight v equals space rω space equals space 1 space straight m divided by straight s
straight T equals space fraction numerator 2 πr over denominator 1 minus left parenthesis negative 1 right parenthesis end fraction space equals space fraction numerator 2 πr over denominator 2 end fraction space equals space πr space sec

3