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Rotational Motion question

2016 · Shift 1 · Q40
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  5. /2016 · Shift 1 · Q40

Rotational Motion question

2016 · Shift 1 · Q40

JEE AdvancedPhysicsRotational MotionMultiple correct+4 / −2
The position vector r→\overrightarrow rr of a particle of mass m is given by the following equation r→(t)=αt3i^+βt2j^,\overrightarrow r \left( t \right) = \alpha {t^3}\widehat i + \beta {t^2}\widehat j,r(t)=αt3i+βt2j​, where α=103ms−3\alpha = {{10} \over 3}m{s^{ - 3}}α=310​ms−3, β=5 ms−2\beta = 5\,m{s^{ - 2}}β=5ms−2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?
  1. A
    The velocity v→\overrightarrow vv is given by v→=(10i^+10j^)\overrightarrow v = \left( {10\widehat i + 10\widehat j} \right)v=(10i+10j​) ms-1
  2. B
    The angular momentum L→\overrightarrow LL with respect to the origin is given by L→=−(53)k^ N m s\overrightarrow L = - \left( {{5 \over 3}} \right)\widehat k\,N\,m\,sL=−(35​)kNms
  3. C
    The force F→\overrightarrow FF is given by F→=(i^+2j^)N\overrightarrow F = \left( {\widehat i + 2\widehat j} \right)NF=(i+2j​)N
  4. D
    The torque τ→\overrightarrow \tauτ with respect to the origin is given by τ→=−(203)k^ N m\overrightarrow \tau = - \left( {{{20} \over 3}} \right)\widehat k\,N\,mτ=−(320​)kNm
View written solutionFree

Correct answer: A, B, D

  1. Given position vector
r⃗(t)=αt3i^+βt2j^\vec r(t)=\alpha t^3\hat i+\beta t^2\hat jr(t)=αt3i^+βt2j^​

with

α=103 m s−3,β=5 m s−2,m=0.1 kg\alpha=\frac{10}{3}\,\text{m s}^{-3},\qquad \beta=5\,\text{m s}^{-2},\qquad m=0.1\,\text{kg}α=310​m s−3,β=5m s−2,m=0.1kg

We need all quantities at t=1 st=1\,\text{s}t=1s.

  1. Position at t=1t=1t=1 s
r⃗(1)=αi^+βj^=103i^+5j^\vec r(1)=\alpha \hat i+\beta \hat j =\frac{10}{3}\hat i+5\hat jr(1)=αi^+βj^​=310​i^+5j^​

(in metres)

  1. Velocity

Velocity is the time derivative of position:

v⃗=dr⃗dt=3αt2i^+2βtj^\vec v=\frac{d\vec r}{dt}=3\alpha t^2\hat i+2\beta t\hat jv=dtdr​=3αt2i^+2βtj^​

At t=1t=1t=1 s:

v⃗(1)=3αi^+2βj^\vec v(1)=3\alpha \hat i+2\beta \hat jv(1)=3αi^+2βj^​

Now,

3α=3⋅103=10,2β=2⋅5=103\alpha=3\cdot \frac{10}{3}=10, \qquad 2\beta=2\cdot 5=103α=3⋅310​=10,2β=2⋅5=10

Hence,

v⃗=(10i^+10j^) m s−1\vec v=(10\hat i+10\hat j)\,\text{m s}^{-1}v=(10i^+10j^​)m s−1

So, Option A is correct.

  1. Angular momentum about origin
L⃗=r⃗×p⃗=r⃗×mv⃗\vec L=\vec r\times \vec p=\vec r\times m\vec vL=r×p​=r×mv

First,

mv⃗=0.1(10i^+10j^)=(i^+j^)m\vec v=0.1(10\hat i+10\hat j)=(\hat i+\hat j)mv=0.1(10i^+10j^​)=(i^+j^​)

So,

L⃗=(103i^+5j^)×(i^+j^)\vec L=\left(\frac{10}{3}\hat i+5\hat j\right)\times (\hat i+\hat j)L=(310​i^+5j^​)×(i^+j^​)

Using cross product:

i^×i^=0,j^×j^=0,i^×j^=k^,j^×i^=−k^\hat i\times \hat i=0,\quad \hat j\times \hat j=0,\quad \hat i\times \hat j=\hat k,\quad \hat j\times \hat i=-\hat ki^×i^=0,j^​×j^​=0,i^×j^​=k^,j^​×i^=−k^

Therefore,

L⃗=103(i^×j^)+5(j^×i^)\vec L=\frac{10}{3}(\hat i\times \hat j)+5(\hat j\times \hat i)L=310​(i^×j^​)+5(j^​×i^) L⃗=103k^−5k^\vec L=\frac{10}{3}\hat k-5\hat kL=310​k^−5k^ L⃗=−53k^ kg m2s−1\vec L=-\frac{5}{3}\hat k\,\text{kg m}^2\text{s}^{-1}L=−35​k^kg m2s−1

Since kg m2s−1=N m s\text{kg m}^2\text{s}^{-1}=\text{N m s}kg m2s−1=N m s,

L⃗=−53k^ N m s\vec L=-\frac{5}{3}\hat k\,\text{N m s}L=−35​k^N m s

So, Option B is correct.

  1. Acceleration
a⃗=dv⃗dt=6αti^+2βj^\vec a=\frac{d\vec v}{dt}=6\alpha t\hat i+2\beta \hat ja=dtdv​=6αti^+2βj^​

At t=1t=1t=1 s:

a⃗(1)=6αi^+2βj^\vec a(1)=6\alpha \hat i+2\beta \hat ja(1)=6αi^+2βj^​

Now,

6α=6⋅103=20,2β=106\alpha=6\cdot \frac{10}{3}=20, \qquad 2\beta=106α=6⋅310​=20,2β=10

Thus,

a⃗=(20i^+10j^) m s−2\vec a=(20\hat i+10\hat j)\,\text{m s}^{-2}a=(20i^+10j^​)m s−2

Force is

F⃗=ma⃗=0.1(20i^+10j^)=(2i^+j^) N\vec F=m\vec a=0.1(20\hat i+10\hat j)=(2\hat i+\hat j)\,\text{N}F=ma=0.1(20i^+10j^​)=(2i^+j^​)N

Option C states (i^+2j^) N(\hat i+2\hat j)\,\text{N}(i^+2j^​)N, which is incorrect.

So, Option C is false.

  1. Torque about origin

Torque is

τ⃗=r⃗×F⃗\vec \tau=\vec r\times \vec Fτ=r×F

Using

r⃗=103i^+5j^,F⃗=2i^+j^\vec r=\frac{10}{3}\hat i+5\hat j, \qquad \vec F=2\hat i+\hat jr=310​i^+5j^​,F=2i^+j^​

we get

τ⃗=(103i^+5j^)×(2i^+j^)\vec \tau=\left(\frac{10}{3}\hat i+5\hat j\right)\times (2\hat i+\hat j)τ=(310​i^+5j^​)×(2i^+j^​) τ⃗=103(i^×j^)+10(j^×i^)\vec \tau=\frac{10}{3}(\hat i\times \hat j)+10(\hat j\times \hat i)τ=310​(i^×j^​)+10(j^​×i^) τ⃗=103k^−10k^\vec \tau=\frac{10}{3}\hat k-10\hat kτ=310​k^−10k^ τ⃗=−203k^ N m\vec \tau=-\frac{20}{3}\hat k\,\text{N m}τ=−320​k^N m

So, Option D is correct.

  1. Final evaluation of options
  • A: True
  • B: True
  • C: False
  • D: True

Therefore, the correct options are:

A, B, D\boxed{A,\ B,\ D}A, B, D​
  1. Comparison with stored correct answer

Stored correct answer: A, B, D

Derived answer: A, B, D

They match.

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