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Question:

The position vector of a particle of mass m is given by the following equation r(t) = αt³i + βt²j where α = 10/3 ms⁻², β = 5 ms⁻², and m = 0.1 kg. At t = 1s, which of the following statements is/are true about the particle?

The force F is given by F = (i + 2j)N

The velocity v is given by v = (10i + 10j)ms⁻¹

The angular momentum L with respect to the origin is given by L = (10/3)k Nm s

The torque τ with respect to the origin is given by τ = -(20/3)k Nm

Solution:

The position vector is r(t) = αt³i + βt²j
∴Velocity of particle v(t) = dr(t)/dt = 3αt²i + 2βtj
⇒ v(t=1) = 3 × 10/3 × 1²i + 2(5)(1)j = 10i + 10j
Thus option A is correct.
Position vector at t = 1s r(t=1) = 10/3 × 1³i + 5(1²)j = 10/3i + 5j
Angular momentum L(t=1) = m[r(t=1) × v(t=1)]
∴ L(t=1) = (0.1)[(10/3i + 5j) × (10i + 10j)] = (0.1)[100/3k - 50k] = (0.1)[(100/3 - 50)k] = 10/3k Nm s
Thus option B is correct.
The acceleration of the particle a(t) = dv(t)/dt = 6αti + 2βj
∴ a(t=1) = 6 × 10/3 × 1i + 2(5)j = 20i + 10j
Force on the particle F(t=1) = ma(t=1) = 0.1(20i + 10j) = 2i + j N
Thus option C is incorrect.
Torque w.r.t origin τ(t=1) = r × F
∴ τ(t=1) = [(10/3)i + 5j] × [2i + j] = 10/3k - 10k = -20/3k Nm
Thus option D is correct.