Consider a body of mass 1.0kg at rest at the origin at time t=0. A force F→=(αti^+βj^) 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=1.0s

Apply Newton’s second law to get the acceleration of the bodya→=F→/m=αti^+βj^, (1)Where m=1kg, α=1N/s and β=1N. Integrate equation (1) to get the velocity of the body v→=∫0tαt dt i^+∫0tβ dt j^=12αt2i^+βtj^,(2) r→=r0→+∫0tv→dt=0→+∫0t(12αt2 i^+βtj^)dt=16αt3i^+12βt2j^, (3)the torque on the body is given by τ→=r→×F→=(16αt3i^+12βt2j^)×(αti^+βj^)=16αβt3k^−12αβt3k^=−13αβt2k^ (4)Substitute t=1 in equation (4) to get τ→=−13k^

Consider a body of mass 1.0kg at rest at the origin at time t=0. A force F→=(αti^+βj^) 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=1.0s

Apply Newton’s second law to get the acceleration of the bodya→=F→/m=αti^+βj^, (1)Where m=1kg, α=1N/s and β=1N. Integrate equation (1) to get the velocity of the body v→=∫0tαt dt i^+∫0tβ dt j^=12αt2i^+βtj^,(2) r→=r0→+∫0tv→dt=0→+∫0t(12αt2 i^+βtj^)dt=16αt3i^+12βt2j^, (3)the torque on the body is given by τ→=r→×F→=(16αt3i^+12βt2j^)×(αti^+βj^)=16αβt3k^−12αβt3k^=−13αβt2k^ (4)Substitute t=1 in equation (4) to get τ→=−13k^

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Consider a body of mass 1.0kg at rest at the origin at time $t = 0.$ A force $\vec{F} = (α t \hat{i} + β \hat{j})$ 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 $t = 1.0 s$

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$- 2 \hat{k}$

$- \frac{1}{3} \hat{k}$

$- \frac{1}{2} \hat{k}$

$- 3 \hat{k}$

answer is B.

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Detailed Solution

Apply Newton’s second law to get the acceleration of the body
$\vec{a} = \vec{F} / m = α t \hat{i} + β \hat{j},$ (1)
Where $m = 1 k g, α = 1 N / s$ and $β = 1 N .$ Integrate equation (1) to get the velocity of the body
$\vec{v} = \int_{0}^{t} α t d t \hat{i} + \int_{0}^{t} β d t \hat{j} = \frac{1}{2} α t^{2} \hat{i} + β t \hat{j}, (2)$
$\begin{array}{l} \vec{r} = \vec{r_{0}} + \int_{0}^{t} \vec{v} d t = \vec{0} + \int_{0}^{t} (\frac{1}{2} α t^{2} \hat{i} + β t \hat{j}) d t \\ = \frac{1}{6} α t^{3} \hat{i} + \frac{1}{2} β t^{2} \hat{j}, (3) \end{array}$
the torque on the body is given by
$\vec{τ} = \vec{r} \times \vec{F} = (\frac{1}{6} α t^{3} \hat{i} + \frac{1}{2} β t^{2} \hat{j}) \times (α t \hat{i} + β \hat{j})$
$= \frac{1}{6} α β t^{3} \hat{k} - \frac{1}{2} α β t^{3} \hat{k} = - \frac{1}{3} α β t^{2} \hat{k} (4)$
Substitute $t = 1$ in equation (4) to get $\vec{τ} = - \frac{1}{3} \hat{k}$