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PhyMedAdv 2016 • Paper 1

Kinematics and Dynamics of a Moving Particle in Two Dimensions

Suppose the position vector r⃗\vec{r} of a particle having mass mm is governed by the relation r⃗(t)=αt3i^+βt2j^,\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}, where α=10/3 m s−3\alpha = 10/3 \text{ m s}^{-3}, β=5 m s−2\beta = 5 \text{ m s}^{-2}, and m=0.1 kgm = 0.1 \text{ kg}. At time t=1 st = 1 \text{ s}, which of the following statements are correct regarding the particle?

Options

A

The velocity vector v⃗\vec{v} is expressed as v⃗=(10i^+10j^) m s−1\vec{v} = (10\hat{i} + 10\hat{j}) \text{ m s}^{-1}

Correct
B

The angular momentum L⃗\vec{L} about the origin is L⃗=−(5/3)k^ N m s\vec{L} = -(5/3)\hat{k} \text{ N m s}

Correct
C

The force F⃗\vec{F} acting on the particle is F⃗=(i^+2j^) N\vec{F} = (\hat{i} + 2\hat{j}) \text{ N}

D

The torque τ⃗\vec{\tau} relative to the origin is given by τ⃗=−(20/3)k^ N m\vec{\tau} = -(20/3)\hat{k} \text{ N m}

Correct

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumTorqueNewton's Laws

Step-by-Step Solution

To determine the correct options, we analyze the motion of the particle using the given position vector as a function of time tt: r⃗(t)=αt3i^+βt2j^\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}

Given values:

  • α=103 m s−3\alpha = \frac{10}{3} \text{ m s}^{-3}
  • β=5 m s−2\beta = 5 \text{ m s}^{-2}
  • m=0.1 kgm = 0.1 \text{ kg}
  • t=1 st = 1 \text{ s}

Step 1: Velocity Vector v⃗(t)\vec{v}(t)

The velocity vector is obtained by taking the time derivative of the position vector: v⃗(t)=dr⃗(t)dt=3αt2i^+2βtj^\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = 3\alpha t^2 \hat{i} + 2\beta t \hat{j}

At time t=1 st = 1 \text{ s}: v⃗(1)=3(103)(1)2i^+2(5)(1)j^=(10i^+10j^) m s−1\vec{v}(1) = 3 \left(\frac{10}{3}\right) (1)^2 \hat{i} + 2(5)(1) \hat{j} = (10 \hat{i} + 10 \hat{j}) \text{ m s}^{-1}

Thus, Option (A) is correct.

Step 2: Angular Momentum L⃗\vec{L} About the Origin

The position vector of the particle at t=1 st = 1 \text{ s} is: r⃗(1)=103(1)3i^+5(1)2j^=(103i^+5j^) m\vec{r}(1) = \frac{10}{3}(1)^3 \hat{i} + 5(1)^2 \hat{j} = \left(\frac{10}{3} \hat{i} + 5 \hat{j}\right) \text{ m}

The angular momentum L⃗\vec{L} about the origin is given by: L⃗=r⃗×p⃗=m(r⃗×v⃗)\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})

Calculating the cross product r⃗×v⃗\vec{r} \times \vec{v}: r⃗×v⃗=(103i^+5j^)×(10i^+10j^)\vec{r} \times \vec{v} = \left(\frac{10}{3} \hat{i} + 5 \hat{j}\right) \times (10 \hat{i} + 10 \hat{j}) r⃗×v⃗=(103×10)k^−(5×10)k^=(1003−50)k^=−503k^ m2 s−1\vec{r} \times \vec{v} = \left(\frac{10}{3} \times 10\right) \hat{k} - (5 \times 10) \hat{k} = \left(\frac{100}{3} - 50\right) \hat{k} = -\frac{50}{3} \hat{k} \text{ m}^2 \text{ s}^{-1}

Therefore, the angular momentum at t=1 st = 1 \text{ s} is: L⃗=0.1×(−503k^)=−53k^ N m s\vec{L} = 0.1 \times \left(-\frac{50}{3} \hat{k}\right) = -\frac{5}{3} \hat{k} \text{ N m s}

Thus, Option (B) is correct.

Step 3: Force F⃗\vec{F} Acting on the Particle

The acceleration vector a⃗(t)\vec{a}(t) is the derivative of velocity: a⃗(t)=dv⃗(t)dt=6αti^+2βj^\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = 6\alpha t \hat{i} + 2\beta \hat{j}

At t=1 st = 1 \text{ s}: a⃗(1)=6(103)(1)i^+2(5)j^=(20i^+10j^) m s−2\vec{a}(1) = 6 \left(\frac{10}{3}\right) (1) \hat{i} + 2(5) \hat{j} = (20 \hat{i} + 10 \hat{j}) \text{ m s}^{-2}

By Newton's second law, the net force F⃗\vec{F} 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}

Thus, Option (C) is incorrect (as it states F⃗=i^+2j^ N\vec{F} = \hat{i} + 2\hat{j} \text{ N}).

Step 4: Torque τ⃗\vec{\tau} Relative to the Origin

The torque τ⃗\vec{\tau} relative to the origin is given by: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

At t=1 st = 1 \text{ s}: τ⃗=(103i^+5j^)×(2i^+j^)\vec{\tau} = \left(\frac{10}{3} \hat{i} + 5 \hat{j}\right) \times (2 \hat{i} + \hat{j}) τ⃗=(103×1)k^−(5×2)k^=(103−10)k^=−203k^ N m\vec{\tau} = \left(\frac{10}{3} \times 1\right) \hat{k} - (5 \times 2) \hat{k} = \left(\frac{10}{3} - 10\right) \hat{k} = -\frac{20}{3} \hat{k} \text{ N m}

Thus, Option (D) is correct.

Conclusion

The correct options are A, B, and D.

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