Question

2. A body is rotating with angular velocity, w=(3i-4j-k). What is the linear velocity of a point having position vector, r=(5-6 j +6 k)? ​

Answer 1

Answer:

The relation between linear velocity and angular velocity is v = ω ×

35j^−4−6k 16v =−18 i^ −13 j^ +2 k^

Explanation:

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Answer 2

Answer:

The linear velocity of a point having position vector is$-30 \hat{i}-23 \hat{j}+2 \hat{k}$

Explanation:

The angular velocity is the angular displacement in relation to time.

The linear velocity of a rotating body is $V=\omega \times r_{1}$

where r is the radius vector and v is the angular velocity.

Given : $w=(3i-4j-k)$

position vector,$r=(5-6 j +6 k)$

$\begin{aligned}&\text { } \vec{\omega}=3 \hat{i}-4 \hat{j}\hat{-k} \\&\vec{r}=5 \hat{i}-6 \hat{j}+6 \hat{k}\end{aligned}$

Now , $\vec{v}=\vec{\omega} \times \vec{r}$

$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\3 & -4 & -1 \\5 & -6 & 6\end{array}\right|$

$\hat{i\:}\left(-24-\left(6\right)\right)+\hat{j\:}\left(-5-18\right)+\hat{k\:}\left(-18-\left(-20\right)\right)$

$-30 \hat{i}-23 \hat{j}+2 \hat{k}$

The linear velocity of a point having position vector is$-30 \hat{i}-23 \hat{j}+2 \hat{k}$