Question

Multiple Correct Answer Type​

Answer 1

Question :

Consider a body of mass 1.0 kg at rest at the origin at t = 0. A force $\sf{\vec F \: = \: (\alpha t \hat i \: + \: \beta \hat J)}$ is applied on the body. Where α = 1.0 N/s and β = 1.0 N. The torque acting on the body about the origin at time t = 0s is $\vec \tau$. Which of the following statements is (are) true ?

(1) $| \tau |$ = ⅓ Nm

(2) The torque $\tau$ is the direction of the unit vector $\sf{+ \hat k}$ .

(3) The velocity of the body at t = 1s is $\sf{\vec c \: = \: \dfrac{1}{2} \big( \hat i \: + \: 2 \hat j\big) }$ .

(4) The Magnitude of the displacement of the body at t = 1s is 1/6 m

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Answer :

(1) and (3) are the correct Statements

$\rule{150}{0.5}$

Explanation :

$\mathfrak{Given} \begin{cases} \sf{Mass \: (m) \: = \: 1 \: kg} \\ \sf{\vec F \: = \: ( \alpha t \hat i \: + \: \beta \hat j)} \\ \sf{\alpha \: = \: 1.0 \: Ns^{-1}} \\ \sf{\beta \: = \: 1.0 \: N} \end{cases}$

As we are given in question that :

$\longrightarrow \sf{\vec F \: = \: (\alpha t \hat i \: + \: \beta \hat j) \: \: \: \: \: ...(1)}$

Also, we know that the force is :

$\longrightarrow \sf{\vec F \: = \: ma \: \: \: \: \: ...(2)}$

Now, equate (1) and (2)

$\longrightarrow \sf{ma \: = \: (\alpha t \hat i \: + \: \beta \hat j)} \\ \\ \longrightarrow \sf{a \: = \: 1 t \hat i \: + \: 1 \hat j} \\ \\ \longrightarrow {\boxed{\sf{a \: = \: (t \hat i \: + \: \hat j) \: ms^{-2}}}}$

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Now,

$\longrightarrow \sf{\dfrac{dv}{dt} \: = \: \alpha t \hat i \: + \: \beta \hat j} \\ \\ \longrightarrow \sf{dv \: = \: dt(t \hat i \: + \: \hat j)}$

After integration we get,

$\longrightarrow {\boxed{\sf{v \: = \: ( \dfrac{t^2}{2} \hat i \: + \: t \hat j) \: ms^{-1}}}}$

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Now,

$\longrightarrow \sf{\dfrac{dr}{dt} \: = \: \dfrac{t^2}{2} \hat i \: + \: t \hat j} \\ \\ \longrightarrow \sf{dr \: = \: dt(\dfrac{t^2}{2} \hat i \: + \: t \hat j)}$

After Integration,

$\longrightarrow \boxed{\sf{r \: = \: (\dfrac{t^3}{6} \hat i \: + \: \dfrac{t^2}{2} \hat j) \: m}}$

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As we know that,

$\boxed{\boxed{\sf{\tau \: = \: r \: \times \: F}}} \\ \\ \longrightarrow \sf{\tau \: = \: \big( \dfrac{t^3}{6} \hat i \: + \: \dfrac{t^2}{2} \hat j \big) \: \times \: \big(\alpha t \hat i \: + \: \beta \hat j \big) } \\ \\ \longrightarrow \sf{\tau \: = \: \big( \dfrac{t^3}{2} \: - \: \dfrac{t^3}{6} \big) ( \hat k)} \\ \\ \longrightarrow {\boxed{\sf{\tau \: = \: \bigg[ \dfrac{-t^3}{3}( \hat k) \bigg] \: Nm }}}$

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When we put value of t = 1 for Torque.

We get $\sf{\tau \: = \: \dfrac{1}{3} \: Nm}$

And also when we put t = 1 in velocity

We get $\sf{v \: = \: \dfrac{1}{2} \big( \hat i \: + \: 2 \hat j \big) \: ms^{-1}}$

Answer 2

The force acting on the particle is,

$\longrightarrow \vec F=\alpha t\,\hat i+\beta\,\^j$

Since $\alpha=1.0$ and $\beta=1.0,$

$\longrightarrow \vec F=t\,\hat i+\^j$

But $\vec F=m\vec a=\vec a,$ since $m=1.0.$ Then,

$\longrightarrow \vec a=t\,\hat i+\^j$

$\longrightarrow \dfrac{d\vec v}{dt}=t\,\hat i+\^j$

$\longrightarrow d\vec v=(t\,\hat i+\^j)\,dt$

$\displaystyle\longrightarrow \vec v=\int(t\,\hat i+\^j)\,dt$

$\displaystyle\longrightarrow \vec v=\dfrac{t^2}{2}\,\hat i+t\,\^j\quad\quad\dots(1)$

Or,

$\displaystyle\longrightarrow \vec v=\dfrac{1}{2}\left(t^2\,\hat i+2t\,\^j\right)$

At $t=1\ s,$

$\displaystyle\longrightarrow \vec v=\dfrac{1}{2}\left(1^2\,\hat i+2\times1\,\^j\right)$

$\displaystyle\longrightarrow \vec v=\dfrac{1}{2}\left(\hat i+2\,\^j\right)$

From (1),

$\displaystyle\longrightarrow \dfrac{d\vec r}{dt}=\dfrac{t^2}{2}\,\hat i+t\,\^j$

$\displaystyle\longrightarrow d\vec r=\left(\dfrac{t^2}{2}\,\hat i+t\,\^j\right)\,dt$

$\displaystyle\longrightarrow \vec r=\int\left(\dfrac{t^2}{2}\,\hat i+t\,\^j\right)\,dt$

$\displaystyle\longrightarrow \vec r=\dfrac{1}{2}\cdot\dfrac{t^3}{3}\,\hat i+\dfrac{t^2}{2}\,\^j\right)$

$\displaystyle\longrightarrow \vec r=\dfrac{t^3}{6}\,\hat i+\dfrac{t^2}{2}\,\^j\right)$

This is the position vector as well as the displacement vector of the body, since the body was initially at rest at the origin.

$\displaystyle\longrightarrow \vec s=\dfrac{t^3}{6}\,\hat i+\dfrac{t^2}{2}\,\^j\right)$

At $t=1\ s,$

$\displaystyle\longrightarrow \vec s=\dfrac{1}{6}\,\hat i+\dfrac{1}{2}\,\^j\right)$

$\displaystyle\longrightarrow|\vec s|=\sqrt{\left(\dfrac{1}{6}\right)^2+\left(\dfrac{1}{2}\right)^2}$

$\displaystyle\longrightarrow|\vec s|=\sqrt{\dfrac{5}{18}}\ m$

So the torque acting on the particle is,

$\displaystyle\longrightarrow \vec\tau=\vec r\times\vec F$

$\displaystyle\longrightarrow\vec\tau=\left(\dfrac{t^3}{6}\,\hat i+\dfrac{t^2}{2}\,\^j\right)\times\big(t\,\hat i+\^j\big)$

$\displaystyle\longrightarrow \vec\tau=\left|\begin{array}{ccc}\hat i&\^j&\^k\\&&\\\dfrac{t^3}{6}&\dfrac{t^2}{2}&0\\&&\\t&1&0\end{array}\right|$

$\displaystyle\longrightarrow \vec\tau=\left(\dfrac{t^3}{6}-\dfrac{t^3}{2}\right)\,\^k$

$\displaystyle\longrightarrow \vec\tau=-\dfrac{t^3}{3}\,\^k$

But $\vec\tau$ is actually the torque acting on the body at $t=1.0\ s.$ Then,

$\displaystyle\longrightarrow \vec\tau=-\dfrac{1^3}{3}\,\^k$

$\displaystyle\longrightarrow \vec\tau=-\dfrac{1}{3}\,\^k$

$\displaystyle\longrightarrow\left|\vec\tau\right|=\dfrac{1}{3}\ N\,m$

Finally, we see that,

(1)   $\displaystyle\left|\vec\tau\right|=\dfrac{1}{3}\ N\,m$

(2)   The torque $\vec\tau$ is in the direction of the unit vector $-\^k.$

(3)   The velocity of the body at $t=1\ s$ is $\displaystyle\vec v=\dfrac{1}{2}\left(\hat i+2\,\^j\right)\ m\,s^{-1}.$

(4)   The magnitude of displacement of the body at $t=1\ s$ is $\sqrt{\dfrac{5}{18}}\ m.$

Therefore, options (1) and (3) are correct.