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Rotational Motion question

2022 · 28 Jun · Shift 1 · Q64
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  5. /2022 · 28 Jun · Shift 1 · Q64

Rotational Motion question

2022 · 28 Jun · Shift 1 · Q64

JEE MainPhysicsRotational MotionNumerical+4 / −1
The position vector of 1 kg object is r→=(3i^−j^)m\overrightarrow r = \left( {3\widehat i - \widehat j} \right)mr=(3i−j​)m and its velocity v→=(3j^+k^)ms−1\overrightarrow v = \left( {3\widehat j + \widehat k} \right)m{s^{ - 1}}v=(3j​+k)ms−1. The magnitude of its angular momentum is x\sqrt xx​ Nm where x is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 91

  1. Given data
  • Mass: m=1 kgm = 1\,\text{kg}m=1kg
  • Position vector: r⃗=(3i^−j^) m=(3,−1,0)\vec r = (3\hat i - \hat j)\,\text{m} = (3,-1,0)r=(3i^−j^​)m=(3,−1,0)
  • Velocity vector: v⃗=(3j^+k^) m s−1=(0,3,1)\vec v = (3\hat j + \hat k)\,\text{m s}^{-1} = (0,3,1)v=(3j^​+k^)m s−1=(0,3,1)
  1. Formula for angular momentum

Angular momentum of a particle about the origin is L⃗=r⃗×p⃗=m(r⃗×v⃗)\vec L = \vec r \times \vec p = m(\vec r \times \vec v)L=r×p​=m(r×v) Since m=1 kgm=1\,\text{kg}m=1kg, L⃗=r⃗×v⃗\vec L = \vec r \times \vec vL=r×v

  1. Compute the cross product
r⃗×v⃗=∣i^j^k^3−10031∣\vec r \times \vec v = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 0\\ 0 & 3 & 1 \end{vmatrix}r×v=​i^30​j^​−13​k^01​​

Expanding,

r⃗×v⃗=i^[(−1)(1)−0(3)]−j^[(3)(1)−0(0)]+k^[(3)(3)−(−1)(0)]\vec r \times \vec v = \hat i[(-1)(1)-0(3)] - \hat j[(3)(1)-0(0)] + \hat k[(3)(3)-(-1)(0)]r×v=i^[(−1)(1)−0(3)]−j^​[(3)(1)−0(0)]+k^[(3)(3)−(−1)(0)] =−i^−3j^+9k^= -\hat i - 3\hat j + 9\hat k=−i^−3j^​+9k^

So,

L⃗=−i^−3j^+9k^\vec L = -\hat i - 3\hat j + 9\hat kL=−i^−3j^​+9k^
  1. Magnitude of angular momentum
∣L⃗∣=(−1)2+(−3)2+92|\vec L| = \sqrt{(-1)^2+(-3)^2+9^2}∣L∣=(−1)2+(−3)2+92​ =1+9+81=91= \sqrt{1+9+81} = \sqrt{91}=1+9+81​=91​

Given that magnitude is x\sqrt{x}x​, we get

x=91⇒x=91\sqrt{x} = \sqrt{91} \Rightarrow x=91x​=91​⇒x=91
  1. Comparison with stored answer

Stored correct answer: 919191

Our derived answer is also 919191.

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