Question

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

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

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

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

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

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

Answer 1

Hint: 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; \[\vec v = \dfrac{{d\vec r}}{{dt}}\]

\[ \Rightarrow \vec v = \dfrac{{d[\alpha {t^3}\hat i + \beta {t^2}\hat j]}}{{dt}} = 3\alpha {t^2}\hat i + 2\beta t\hat j\]

Putting values of \[\alpha \& \beta \]

We have velocity at \[t = 1s\]

\[\vec v = 10\hat i + 10\hat j\]

Hence option (A) is correct.

Now linear momentum 

$\vec p = m\vec v$

$\Rightarrow \vec p = 0.1[10\hat i + 10\hat j] = \hat i + \hat j$

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

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

i.e; \[\vec L = \vec r \times \vec p = [\dfrac{{10}}{3}\hat i + 5\hat j] \times [\hat i + \hat j] \]

\[ \Rightarrow \vec L = 0 + \dfrac{{10}}{3}\hat k + [ - 5\hat k] + 0 = - \dfrac{5}{3}\hat k] \]

Hence, option (B) is correct.

Now acceleration is rate of change of velocity

i.e; \[vec a = \dfrac{{d\vec v}}{{dt}}\]

$\Rightarrow \vec a = \dfrac{{d[3\alpha {t^2}\hat i + 2\beta t\hat j]}}{{dt}} = 6\alpha t\hat i + 2\beta \hat j$

putting value of \[\alpha \& \beta \] values of acceleration at \[t = 1s\] is

\[\vec a = 20\hat i + 10\hat j\]

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

Hence option (C) is incorrect.

Torque is cross product of position vector and force

So, \[\vec \tau = \vec r \times \vec F\]

$ \Rightarrow \vec \tau = \left[ {\dfrac{{10}}{3}\hat i + 5\hat j} \right] \times \left[ {2\hat i + \hat j} \right]$

\[ \Rightarrow \vec \tau = 0 + \dfrac{{10}}{3}\hat k + [ - 10\hat k] + 0 = - \dfrac{{20}}{3}\hat k\]

Hence option (D) is also correct.

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.