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More from System of Particles and Rotational Motion
PhyMedAdv 2021 • Paper 1

Kinematics and Torque Angular Momentum of Particle

Consider a particle having mass M=0.2 kgM = 0.2\text{ kg} that is originally at rest in the xyxy-plane at the point (x=−l,y=−h)(x = -l, y = -h), where l=10 ml = 10\text{ m} and h=1 mh = 1\text{ m}. Beginning at t=0t = 0, it accelerates with a constant rate a=10 m/s2a = 10\text{ m/s}^2 in the positive xx-direction. Let its angular momentum and torque around the origin be L⃗\vec{L} and τ⃗\vec{\tau} respectively (in SI units). Let i^,j^,k^\hat{i}, \hat{j}, \hat{k} be the unit vectors along x,y,zx, y, z axes with k^=i^×j^\hat{k} = \hat{i} \times \hat{j}. Select the correct statement(s):

Options

A

The particle arrives at the point (x=l,y=−h)(x = l, y = -h) at time t=2 st = 2\text{ s}

Correct
B

τ⃗=2k^\vec{\tau} = 2\hat{k} when the particle passes through the point (x=l,y=−h)(x = l, y = -h)

Correct
C

L⃗=4k^\vec{L} = 4\hat{k} when the particle passes through the point (x=l,y=−h)(x = l, y = -h)

Correct
D

τ⃗=k^\vec{\tau} = \hat{k} when the particle passes through the point (x=0,y=−h)(x = 0, y = -h)

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumTorque

Step-by-Step Solution

To determine the correct statement(s), we analyze the motion of the particle step-by-step.

1. Kinematics of the Particle

The particle has a mass M=0.2 kgM = 0.2\text{ kg}, starts from rest at t=0t = 0 at the initial position (x0,y0)=(−l,−h)=(−10 m,−1 m)(x_0, y_0) = (-l, -h) = (-10\text{ m}, -1\text{ m}), and accelerates with a constant acceleration a⃗=10i^ m/s2\vec{a} = 10\hat{i}\text{ m/s}^2.

The position vector of the particle as a function of time tt is: r⃗(t)=(x0+12at2)i^+y0j^=(−10+5t2)i^−j^ (in meters)\vec{r}(t) = \left(x_0 + \frac{1}{2}at^2\right)\hat{i} + y_0\hat{j} = \left(-10 + 5t^2\right)\hat{i} - \hat{j}\text{ (in meters)}

The velocity vector as a function of time tt is: v⃗(t)=dr⃗dt=10ti^ (in m/s)\vec{v}(t) = \frac{d\vec{r}}{dt} = 10t\hat{i}\text{ (in m/s)}

2. Time to Reach the Point (x=l,y=−h)=(10 m,−1 m)(x = l, y = -h) = (10\text{ m}, -1\text{ m})

Setting the xx-coordinate to x=10 mx = 10\text{ m}: −10+5t2=10  ⟹  5t2=20  ⟹  t2=4  ⟹  t=2 s-10 + 5t^2 = 10 \implies 5t^2 = 20 \implies t^2 = 4 \implies t = 2\text{ s}

Thus, Option (A) is correct.

3. Torque τ⃗\vec{\tau} About the Origin

The net force acting on the particle is: F⃗=Ma⃗=0.2×10i^=2i^ N\vec{F} = M\vec{a} = 0.2 \times 10\hat{i} = 2\hat{i}\text{ N}

The torque τ⃗\vec{\tau} about the origin at any position r⃗(t)\vec{r}(t) is given by: τ⃗=r⃗(t)×F⃗\vec{\tau} = \vec{r}(t) \times \vec{F} τ⃗=[(−10+5t2)i^−j^]×(2i^)\vec{\tau} = \left[\left(-10 + 5t^2\right)\hat{i} - \hat{j}\right] \times \left(2\hat{i}\right) τ⃗=−(j^×2i^)=2k^ N⋅m\vec{\tau} = -(\hat{j} \times 2\hat{i}) = 2\hat{k}\text{ N}\cdot\text{m}

Since τ⃗=2k^\vec{\tau} = 2\hat{k} is constant and independent of time and position:

  • When the particle passes through (x=l,y=−h)(x = l, y = -h), τ⃗=2k^\vec{\tau} = 2\hat{k}. Thus, Option (B) is correct.
  • When the particle passes through (x=0,y=−h)(x = 0, y = -h), τ⃗=2k^≠k^\vec{\tau} = 2\hat{k} \neq \hat{k}. Thus, Option (D) is incorrect.

4. Angular Momentum L⃗\vec{L} About the Origin at (x=l,y=−h)(x = l, y = -h)

At t=2 st = 2\text{ s}, the position vector is: r⃗(2)=10i^−j^ m\vec{r}(2) = 10\hat{i} - \hat{j}\text{ m}

The velocity vector is: v⃗(2)=10(2)i^=20i^ m/s\vec{v}(2) = 10(2)\hat{i} = 20\hat{i}\text{ m/s}

The linear momentum vector is: p⃗=Mv⃗(2)=0.2×20i^=4i^ kg⋅m/s\vec{p} = M\vec{v}(2) = 0.2 \times 20\hat{i} = 4\hat{i}\text{ kg}\cdot\text{m/s}

The angular momentum L⃗\vec{L} about the origin is: L⃗=r⃗(2)×p⃗=(10i^−j^)×(4i^)=−(j^×4i^)=4k^ kg⋅m2/s\vec{L} = \vec{r}(2) \times \vec{p} = (10\hat{i} - \hat{j}) \times (4\hat{i}) = -(\hat{j} \times 4\hat{i}) = 4\hat{k}\text{ kg}\cdot\text{m}^2/\text{s}

Thus, Option (C) is correct.

Conclusion

The correct options are (A), (B), and (C).

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