Question 760: Centre of Mass and Rotational Motion

Question

Consider a body of mass $$1.0$$ $$kg$$ at rest at the origin at time $$t=0.$$ A force $$\overrightarrow F  = \left( {\alpha t \widehat i + \beta \widehat j} ight)$$ is applied on the body, where $$\alpha  = 1.0N{s^{ - 1}}$$ and $$\beta  = 1.0\,N.$$ The torque acting on the body about the origin at time $$t=1.0s$$ is $$\overrightarrow 	au  .$$ Which of the following statements is (are) true?

Accepted Answer

Given, force applied on body
$$\overrightarrow F  = (\alpha t\widehat i + \beta \widehat j)$$
Here, $$\alpha$$ = 1.0 N s
$$-$$1
and $$\beta$$ = 1.0 N. Therefore,
$$\overrightarrow F  = t\widehat i + \widehat j$$
By Newton's second law of motion, F = ma. Given m = 1.0 kg. Therefore,
$$\overrightarrow F  = \overrightarrow a  \Rightarrow \overrightarrow a  = (t\widehat i + \widehat j)$$
We know $$\overrightarrow a  = {{d\overrightarrow v } \over {dt}} \Rightarrow \overrightarrow v  = \int {\overrightarrow a dt} $$
$$ \Rightarrow \overrightarrow v  = \int {(t\widehat i + \widehat j)dt = {{{t^2}} \over 2}\widehat i + t\widehat j} $$ ..... (1)
Also, $$\overrightarrow v  = {{\overrightarrow {dr} } \over {dt}} \Rightarrow \overrightarrow r  = \int {\overrightarrow v dt} $$
$$ \Rightarrow \overrightarrow r  = \int {\left( {{{{t^2}} \over 2}\widehat i + t\widehat j} ight)dt = {{{t^3}} \over 6}\widehat i + {{{t^2}} \over 2}\widehat j} $$ ..... (2)
Now, $$\overrightarrow 	au   = \overrightarrow r  	imes \overrightarrow F  = \left| {\matrix{
   {\widehat i} & {\widehat j} & {\widehat k}  \cr 
   {{t^3}/6} & {{t^2}/2} & 0  \cr 
   t & 1 & 0  \cr

} } ight|$$
$$\overrightarrow 	au   = \widehat i(0) - \widehat j(0) + \widehat k\left( {{{{t^3}} \over 6} - {{{t^3}} \over 2}} ight) = {{ - 1} \over 3}{t^3}\widehat k$$ N m
At t = 1.0 s, $$\overrightarrow 	au   = {{ - 1} \over 3}\widehat k$$ N m
Hence, option (A) is correct.
Now, from Eq. (1), $$\overrightarrow v  = {{{t^2}} \over 2}\widehat i + t\widehat j$$
At t = 1.0 s, $$\overrightarrow v  = \left( {{1 \over 2}\widehat i + \widehat j} ight)$$ m s
$$-$$1
= $$ = {1 \over 2}(\widehat i + 2\widehat j)$$ m s
$$-$$1
Hence, option (C) is correct.

Answer

$$\left| {\overrightarrow 	au  } ight| = {1 \over 3}\,Nm$$

Answer

The torque $$\overrightarrow 	au  $$ is in the direction of the unit vector $$ + \,\widehat k$$

Answer

The velocity of the body at $$t = 1s$$ is $$\overrightarrow v  = {1 \over 2}\left( {\widehat i + 2\widehat j} \right)m{s^{ - 1}}$$

Answer

The magnitude of displacement of the body at $$t = 1s$$ is $${1 \over 6}m$$

Centre of Mass and Rotational Motion

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