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Rotational Motion question

2015 · Q143
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Rotational Motion question

2015 · Q143

NEETPhysicsRotational MotionMCQ+4 / −1
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 a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is
  1. A
    W(d−x)x{{W\left( {d - x} \right)} \over x}xW(d−x)​
  2. B
    W(d−x)d{{W\left( {d - x} \right)} \over d}dW(d−x)​
  3. C
    Wxd{{Wx} \over d}dWx​
  4. D
    Wdx{{Wd} \over x}xWd​
View written solutionFree

Correct answer: B

AIPMT 2015 Cancelled Paper Physics - Rotational Motion Question 75 English Explanation

Given situation is shown in figure.

N1 = Normal reaction on A

N2 = Normal reaction on B

W = Weight of the rod

In vertical equilibrium,
N1 + N2 = W …(i)

Torque balance about centre of mass of the rod,

N1x = N2(d – x)

Putting value of N2 from equation (i)

N1x = (W – N1)(d – x)

⇒\Rightarrow⇒ N1x = Wd – Wx – N1d + N1x

⇒\Rightarrow⇒ N1d = W(d – x)

∴\therefore∴ N1=W(d−x)d{N_1} = {{W\left( {d - x} \right)} \over d}N1​=dW(d−x)​

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