Rotation Ques 94
- Consider a body of mass $1.0 kg$ at rest at the origin at time $t=0$. A force $F=(\alpha t \hat{i}+\beta \hat{j})$ is applied on the body, where $\alpha=1.0 Ns^{-1}$ and $\beta=1.0 N$. The torque acting on the body about the origin at time $t=1.0 s$ is $\tau$ Which of the following statements is (are) true?
(2108 Adv.) (a) $|\tau|=\frac{1}{3} N-m$
(b) The torque $\tau$ is in the direction of the unit vector $+\hat{k}$
(c) The velocity of the body at $t=1 s$ is $v=\frac{1}{2}(\hat{i}+2 \hat{j}) ms^{-1}$
(d) The magnitude of displacement of the body at $t=1 s$ is $\frac{1}{6} m$
Show Answer
Answer:
Correct Answer: 94.(a,c)
Solution:
Formula:
- $\quad F=(\alpha t) \hat{i}+\beta \hat{j}$
$$ \text { [at } t=0, v=0, r=0] $$
$$ \alpha=1, \beta=1 \Rightarrow F=t \hat{i}+\hat{j} $$
$m \frac{d v}{d t}=t \hat{i}+\hat{j}$
On integrating,
$$ \begin{aligned} & m v=\frac{t^{2}}{2} \hat{i}+\hat{t} \quad \quad \quad[m=1 kg] \\ & \frac{d r}{d t}=\frac{t^{2}}{2} \hat{i}+\hat{t}=v \Rightarrow v=\frac{1}{2}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \text { at } t=1 s \end{aligned} $$
Again, on integrating,
$$ r=\frac{t^{3}}{6} \hat{i}+\frac{t^{2}}{2} \hat{j} \quad[r=0 \text { at } t=0] $$
At $t=1 s, \tau=(r \times F)=\frac{1}{6} \hat{i}+\frac{1}{2} \hat{j} \times(\hat{i}+\hat{j})=-\frac{1}{3} \hat{\mathbf{k}}$
BETA
