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Question: The position vector \[\vec r\] of a particle of mass m is given by the following equation \[\vec r[t...

The position vector r⃗\vec r of a particle of mass m is given by the following equation r⃗[t]=αt3i^+βt2j^\vec r[t] = \alpha {t^3}\hat i + \beta {t^2}\hat j where, α=10/3ms−3,β=5ms−2\alpha = 10/3m{s^{ - 3}},\beta = 5m{s^{ - 2}} and m=0.1kgm = 0.1kg at t=1st = 1s.

Which one of the following statements is/are true about the particle?

A. The velocity v⃗\vec v is given by v⃗=[10i^+10j^]ms−1\vec v = [10\hat i + 10\hat j]m{s^{ - 1}}

B. The angular momentum L⃗\vec L with respect to the origin is given by L⃗=[−5/3]k^Nms\vec L = [ - 5/3]\hat kNms

C. The force F⃗\vec F is given by F⃗=[i^+2j^]N\vec F = [\hat i + 2\hat j]N

D. The torque τ⃗\vec \tau with respect to origin is given by τ⃗=−[20/3]k^Nm\vec \tau = - [20/3]\hat kNm

Explanation

Solution

Use the fundamental definition of velocity, angular momentum, force and torque to find the required value, to find the force first find the acceleration by double differentiating position vector once you know the acceleration multiply it with mass to get the force.

Complete step by step solution:

We know that rate of change of position vector gives us velocity vector,

i.e; v⃗=dr⃗dt\vec v = \dfrac{{d\vec r}}{{dt}}

⇒v⃗=d[αt3i^+βt2j^]dt=3αt2i^+2βtj^ \Rightarrow \vec v = \dfrac{{d[\alpha {t^3}\hat i + \beta {t^2}\hat j]}}{{dt}} = 3\alpha {t^2}\hat i + 2\beta t\hat j

Putting values of α&β\alpha \& \beta

We have velocity at t=1st = 1s

v⃗=10i^+10j^\vec v = 10\hat i + 10\hat j

Hence option (A) is correct.

Now linear momentum

p⃗=mv⃗\vec p = m\vec v

⇒p⃗=0.1[10i^+10j^]=i^+j^\Rightarrow \vec p = 0.1[10\hat i + 10\hat j] = \hat i + \hat j

At t=1t = 1 value of position vector r⃗[t]=αi^+βj^=103i^+5j^\vec r[t] = \alpha \hat i + \beta \hat j = \dfrac{{10}}{3}\hat i + 5\hat j

We know that angular momentum is a cross product of position vector and linear momentum.

i.e; L⃗=r⃗×p⃗=[103i^+5j^]×[i^+j^]\vec L = \vec r \times \vec p = [\dfrac{{10}}{3}\hat i + 5\hat j] \times [\hat i + \hat j]

⇒L⃗=0+103k^+[−5k^]+0=−53k^]\Rightarrow \vec L = 0 + \dfrac{{10}}{3}\hat k + [ - 5\hat k] + 0 = - \dfrac{5}{3}\hat k]

Hence, option (B) is correct.

Now acceleration is rate of change of velocity

i.e; veca=dv⃗dtvec a = \dfrac{{d\vec v}}{{dt}}

⇒a⃗=d[3αt2i^+2βtj^]dt=6αti^+2βj^\Rightarrow \vec a = \dfrac{{d[3\alpha {t^2}\hat i + 2\beta t\hat j]}}{{dt}} = 6\alpha t\hat i + 2\beta \hat j

putting value of α&β\alpha \& \beta values of acceleration at t=1st = 1s is

a⃗=20i^+10j^\vec a = 20\hat i + 10\hat j

Using F⃗=ma⃗\vec F = m\vec a, we have F⃗=0.1[20i^+10j^]=2i^+j^\vec F = 0.1[20\hat i + 10\hat j] = 2\hat i + \hat j

Hence option (C) is incorrect.

Torque is cross product of position vector and force

So, τ⃗=r⃗×F⃗\vec \tau = \vec r \times \vec F

⇒τ⃗=[103i^+5j^]×[2i^+j^] \Rightarrow \vec \tau = \left[ {\dfrac{{10}}{3}\hat i + 5\hat j} \right] \times \left[ {2\hat i + \hat j} \right]

⇒τ⃗=0+103k^+[−10k^]+0=−203k^ \Rightarrow \vec \tau = 0 + \dfrac{{10}}{3}\hat k + [ - 10\hat k] + 0 = - \dfrac{{20}}{3}\hat k

Hence option (D) is also correct.

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Therefore, Option (A), Option (B) and Option (D) are correct.

Note:

In these type of questions all you need to do is use the very fundamental definition of those physical quantities which are asked, like here first derivative of position vector is velocity and second derivative of position vector is acceleration once you know the position, velocity and acceleration of the particle you can easily find all those quantities which are asked.