Question

Consider a body of mass 1.0kg at rest at the origin at time t=0. A force $\overrightarrow F = \alpha t\widehat i + \beta \widehat j$ is applied on the body, where α=1.0N/s and β=1.0N. The torque acting on the body about the origin at time t=1.0s is τ. Which of the following statements is (are) true?
A. $\left| {\vec \tau } \right| = \dfrac{1}{3}N.m$
B. The torque $\vec \tau $is in the direction of unit vector $ + \widehat k$
C. Velocity of the body at t= 1sec is $\overrightarrow v = \dfrac{1}{2}(\widehat i + 2\widehat j)$m/s
D. The magnitude of displacement of the body at t= 1s is $\dfrac{1}{6}m$

Answer 1

Hint: From the vector form of 2nd law of Newton,$\overrightarrow F = m\overrightarrow a = m\dfrac{{d\overrightarrow v }}{{dt}}$ and we know,$\overrightarrow v = \dfrac{{d\overrightarrow r }}{{dt}}$
From the definition of torque,$\overrightarrow \tau = (\overrightarrow r \times \overrightarrow F )$
Where, $\overrightarrow \tau = $torque acting on the body
$\overrightarrow F = $Force acting on the body
$\overrightarrow r = $Displacement of the body
$\overrightarrow v = $Velocity of the body
$m = $Mass of the body
Using these two equations we will come to the solution of the above problem.

Complete step by step answer:
Mass of the body, $m = 1kg$
At $t = 0s,\overrightarrow v = 0,\overrightarrow r = 0$
Also,$\overrightarrow F = \alpha t\widehat i + \beta \widehat j$, $\alpha = 1N/s,\beta = 1N$
So, $\overrightarrow F = t\widehat i + \widehat j$
From Newton’s second law,$\overrightarrow F = m\overrightarrow a = m\dfrac{{d\overrightarrow v }}{{dt}}$
Now, $m\dfrac{{d\overrightarrow v }}{{dt}} = t\widehat i + \widehat j$\[\]
Or, $md\overrightarrow v = (t\widehat i + \widehat j)dt$
Integrating both sides,$m\int\limits_{v = 0}^{v = \overrightarrow v } {d\overrightarrow v = \int\limits_{t = 0}^{t = t} {(t\widehat i + \widehat j)dt} } $
Or, $\overrightarrow v = \dfrac{{{t^2}}}{2}\widehat i + t\widehat j[\because m = 1kg]$
Or,$\dfrac{{d\overrightarrow r }}{{dt}} = \dfrac{{{t^2}}}{2}\widehat i + t\widehat j$
Or,$d\overrightarrow r = (\dfrac{{{t^2}}}{2}\widehat i + t\widehat j)dt$
Integrating both sides, $\int\limits_{r = 0}^{r = \overrightarrow r } {d\overrightarrow r = \int\limits_{t = 0}^{t = t} {\left( {\dfrac{{{t^2}}}{2}\widehat i + t\widehat j} \right)} dt} $
Or, $\overrightarrow r = \dfrac{{{t^3}}}{6}\widehat i + \dfrac{{{t^2}}}{2}\widehat j$
At t= 1sec,
$\overrightarrow r = \dfrac{{{1^3}}}{6}\widehat i + \dfrac{{{1^2}}}{2}\widehat j = \dfrac{1}{6}\widehat i + \dfrac{1}{2}\widehat j$
$\overrightarrow v = \dfrac{{{1^2}}}{2}\widehat i + 1\widehat j = \dfrac{1}{2}\widehat i + 1\widehat j = \dfrac{1}{2}(\widehat i + 2\widehat j)$………………………….(1)
$\overrightarrow F = \widehat i + \widehat j$
$
  \overrightarrow \tau = (\overrightarrow r \times \overrightarrow F ) \\
   = (\dfrac{1}{6}\widehat i + \dfrac{1}{2}\widehat j) \times (\widehat i + \widehat j) \\
   = (\dfrac{1}{6} - \dfrac{1}{2})\widehat k \\
   = - \dfrac{1}{3}\widehat k................................(2) \\
$
$\left| {\overrightarrow r } \right| = \sqrt {{{(\dfrac{1}{6})}^2} + {{(\dfrac{1}{2})}^2}} = \dfrac{{\sqrt {10} }}{6}$……………………………………(3)
Now, from (2) it is clear that at t= 1sec, $|\overrightarrow \tau | = \dfrac{1}{3}$.
So, option (A) is correct.
Direction of $\overrightarrow \tau $is towards unit vector $ - \widehat k$from (2)
So, option (B) is incorrect.
From (1), Velocity of the body at t= 1sec is $\overrightarrow v = \dfrac{1}{2}(\widehat i + 2\widehat j)$m/s
So, option (C) is correct.
From (3), the magnitude of displacement of the body at t= 1s is $\dfrac{{\sqrt {10} }}{6}m$
So, option (D) is incorrect.

So, the correct answers are “Options A and C”.

Note:
It is to be noted that, $\overrightarrow \tau = (\overrightarrow r \times \overrightarrow F )$. $(\overrightarrow r \times \overrightarrow F ) \ne (\overrightarrow F \times \overrightarrow r )$. So, $\overrightarrow \tau \ne (\overrightarrow F \times \overrightarrow r )$.
If the body is not at rest initially and let have a speed of $\overrightarrow {{v_1}} $, then limits of the integration will change, like,$m\int\limits_{v = \overrightarrow {{v_1}} }^{v = \overrightarrow v } {d\overrightarrow v = \int\limits_{t = 0}^{t = t} {(t\widehat i + \widehat j)dt} } $.