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Jatin Anand
Asked: 2022-11-08T19:46:31+05:30 In:

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A uniform rod of mass `m` and length `l` is applied pivoted at point `O`. The rod is initially in vertical position and touching a block of mass `M` which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point `O` this causes the block to move forward as shown The rod loses contact with the block at `theta=30^(@)` all surfaces are smooth now answer the following questions.
Q. The hinge reaction at `O` on the rod when it loses contact with the block is
A. `(3mg)/(4)(hati+hatj)`
B. `((mg)/(4))hatj`
C. `((mg)/(4))hati`
D. `(mg)/(4)(hati+hatj)`

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A uniform rod of mass `m` and length `l` is applied pivoted at point `O`. The rod is initially in vertical position and touching a block of mass `M` which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point `O` this causes the block to move forward as shown The rod loses contact with the block at `theta=30^(@)` all surfaces are smooth now answer the following questions.
Q. The hinge reaction at `O` on the rod when it loses contact with the block is
A. `(3mg)/(4)(hati+hatj)`
B. `((mg)/(4))hatj`
C. `((mg)/(4))hati`
D. `(mg)/(4)(hati+hatj)`
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1 Answer

  1. `a_(C)=((3sqrt(3))/(8)cos60^(@)-(3)/(8)gcos30^(@))hati`
    `+((3sqrt(3)g)/(8)gcos30^(@)+(3)/(8)gcos60^(@))(-hatj)`
    `=-((3g)/(4))hatj`
    Now, `mg(-hatj)+F_(“hinge”)=ma_(C)`
    substituting the value we get
    `F_(“Hinge”)=((mg)/(4))hatj`

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