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JEE Advanced 2021 · System of Particles and Rotational Motion

A particle of mass M=0.2 kgM = 0.2\,\text{kg} is initially at rest in the xyxy-plane at a point (x=−l,y=−h)(x = -l, y = -h), where l=10 ml = 10\,\text{m} and h=1 mh = 1\,\text{m}. The particle is accelerated at time t=0t = 0 with a constant acceleration a=10 m/s2a = 10\,\text{m/s}^2 along the positive xx-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by L⃗\vec{L} and τ⃗\vec{\tau}, respectively. i^\hat{i}, j^\hat{j} and k^\hat{k} are unit vectors along the positive xx, yy and zz-directions, respectively. If k^=i^×j^\hat{k} = \hat{i} \times \hat{j} then which of the following statement(s) is(are) correct?

Answers: A. The particle arrives at the point (x=l,y=−h)(x = l, y = -h) at time t=2 st = 2\,\text{s}. · B. τ⃗=2 k^\vec{\tau} = 2\,\hat{k} when the particle passes through the point (x=l,y=−h)(x = l, y = -h) · C. L⃗=4 k^\vec{L} = 4\,\hat{k} when the particle passes through the point (x=l,y=−h)(x = l, y = -h)

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Written solution

Step 1: Write down the equations of motion.

The particle starts from rest at (−10,−1) m(-10, -1)\,\text{m} and moves with a constant acceleration a=10 m/s2a = 10\,\text{m/s}^2 in the positive xx-direction. The position vector r⃗(t)\vec{r}(t) and velocity vector v⃗(t)\vec{v}(t) at any time t≥0t \ge 0 are given by:

x(t)=−10+12(10)t2=−10+5t2x(t) = -10 + \frac{1}{2}(10)t^2 = -10 + 5t^2 y(t)=−1y(t) = -1 r⃗(t)=(−10+5t2)i^−j^\vec{r}(t) = (-10 + 5t^2)\hat{i} - \hat{j} v⃗(t)=10ti^\vec{v}(t) = 10t\hat{i}

Step 2: Find the time taken to reach the point (l,−h)=(10,−1) m(l, -h) = (10, -1)\,\text{m}.

Set the xx-coordinate to 1010:

−10+5t2=10-10 + 5t^2 = 10 5t2=205t^2 = 20 t=2 st = 2\,\text{s}

Thus, the particle reaches (l,−h)(l, -h) at t=2 st = 2\,\text{s}.

Step 3: Calculate the torque τ⃗\vec{\tau} about the origin.

The constant force acting on the particle is:

F⃗=Ma⃗=0.2×10i^=2i^\vec{F} = M\vec{a} = 0.2 \times 10\hat{i} = 2\hat{i}

The torque about the origin is:

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F} τ⃗=(xi^−hj^)×2i^\vec{\tau} = (x\hat{i} - h\hat{j}) \times 2\hat{i} τ⃗=−2h(j^×i^)\vec{\tau} = -2h(\hat{j} \times \hat{i})

Using j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k} and h=1 mh = 1\,\text{m}:

τ⃗=2k^\vec{\tau} = 2\hat{k}

The torque is constant and independent of position and time. Thus, the torque is 2k^2\hat{k} at both (l,−h)(l, -h) and (0,−h)(0, -h).

Step 4: Calculate the angular momentum L⃗\vec{L} about the origin.

The angular momentum is given by:

L⃗=r⃗×Mv⃗\vec{L} = \vec{r} \times M\vec{v} L⃗=(xi^−hj^)×(0.2×10ti^)\vec{L} = (x\hat{i} - h\hat{j}) \times (0.2 \times 10t\hat{i}) L⃗=(xi^−hj^)×2ti^\vec{L} = (x\hat{i} - h\hat{j}) \times 2t\hat{i} L⃗=−2ht(j^×i^)\vec{L} = -2ht(\hat{j} \times \hat{i})

Using h=1 mh = 1\,\text{m}:

L⃗=2tk^\vec{L} = 2t\hat{k}

At t=2 st = 2\,\text{s} (when the particle is at (l,−h)(l, -h)):

L⃗=4k^\vec{L} = 4\hat{k}

Answer: The correct statements are:

  1. The particle arrives at the point (l,−h)(l, -h) at time t=2 st = 2\,\text{s}.
  2. Torque τ⃗\vec{\tau} about the origin when the particle is at (l,−h)(l, -h) is 2k^ N⋅m2\hat{k}\,\text{N}\cdot\text{m}.
  3. Angular momentum L⃗\vec{L} about the origin when the particle is at (l,−h)(l, -h) is 4k^ kg⋅m2/s4\hat{k}\,\text{kg}\cdot\text{m}^2/\text{s}.

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