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Rotational Motion question

2019 · 10 Apr · Shift 2 · Q46
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  5. /2019 · 10 Apr · Shift 2 · Q46

Rotational Motion question

2019 · 10 Apr · Shift 2 · Q46

JEE MainPhysicsRotational MotionMCQ+4 / −1
The time dependence of the position of a particle of mass m = 2 is given by r→(t)=2ti^−3t2j^\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat jr(t)=2ti−3t2j​ . Its angular momentum, with respect to the origin, at time t = 2 is
  1. A
    36 k^\widehat kk
  2. B
    - 48 k^\widehat kk
  3. C
    −34(k^−i^)- 34\left( {\widehat k - \widehat i} \right)−34(k−i)
  4. D
    48(i^+j^)48\left( {\widehat i + \widehat j} \right)48(i+j​)
View written solutionFree

Correct answer: B

  1. Given position vector
r⃗(t)=2t i^−3t2 j^\vec r(t)=2t\,\hat i-3t^2\,\hat jr(t)=2ti^−3t2j^​

and mass

m=2m=2m=2

We need angular momentum about the origin at t=2t=2t=2.

  1. Find velocity

Velocity is the time derivative of position:

v⃗(t)=dr⃗dt=2 i^−6t j^\vec v(t)=\frac{d\vec r}{dt}=2\,\hat i-6t\,\hat jv(t)=dtdr​=2i^−6tj^​

At t=2t=2t=2,

r⃗(2)=2(2)i^−3(2)2j^=4i^−12j^\vec r(2)=2(2)\hat i-3(2)^2\hat j=4\hat i-12\hat jr(2)=2(2)i^−3(2)2j^​=4i^−12j^​ v⃗(2)=2i^−6(2)j^=2i^−12j^\vec v(2)=2\hat i-6(2)\hat j=2\hat i-12\hat jv(2)=2i^−6(2)j^​=2i^−12j^​
  1. Find linear momentum
p⃗=mv⃗=2(2i^−12j^)=4i^−24j^\vec p = m\vec v = 2(2\hat i-12\hat j)=4\hat i-24\hat jp​=mv=2(2i^−12j^​)=4i^−24j^​
  1. Angular momentum formula

Angular momentum about origin is

L⃗=r⃗×p⃗\vec L = \vec r \times \vec pL=r×p​

So,

L⃗=(4i^−12j^)×(4i^−24j^)\vec L = (4\hat i-12\hat j)\times(4\hat i-24\hat j)L=(4i^−12j^​)×(4i^−24j^​)
  1. Compute cross product

Using

(ai^+bj^)×(ci^+dj^)=(ad−bc)k^( a\hat i+b\hat j )\times( c\hat i+d\hat j )=(ad-bc)\hat k(ai^+bj^​)×(ci^+dj^​)=(ad−bc)k^

we get

L⃗=[4(−24)−(−12)(4)]k^\vec L = [4(-24)-(-12)(4)]\hat kL=[4(−24)−(−12)(4)]k^ L⃗=(−96+48)k^\vec L = (-96+48)\hat kL=(−96+48)k^ L⃗=−48k^\vec L = -48\hat kL=−48k^
  1. Match with options

This corresponds to:

Option B: −48k^-48\hat k−48k^

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