A body is rotating with angular velocity ω=(3 hati-4 hatj+ hatk). The linear velocity of a point having position vector r=(5 hati-6 hatj+6 hatk) is

Question

A body is rotating with angular velocity ω=(3 hati-4 hatj+ hatk). The linear velocity of a point having position vector r=(5 hati-6 hatj+6 hatk) is (A) 6 hati+2 hatj-3 hatk (B) 18 hati+3 hatj-2 hatk (C) -18 hati-13 hatj+2 hatk (D) 6 hati-2 hatj+8 hatk

Tardigrade Admin · Accepted Answer

Here, vecω =3 hati-4 hatj+ hatk r=5 hati-6 hatj+6 hatk As upsilon= vecω ×r =| hati& hatj& hatk 3&-4&1 5&-6&6| = hati (-24-(-6))+ hatj (5-18)+ hatk (-18-(-20)) =-18 hati-13 hatj+2 hatk

Q. A body is rotating with angular velocity $ω = (3 \hat{i} - 4 \hat{j} + \hat{k})$ . The linear velocity of a point having position vector $r = (5 \hat{i} - 6 \hat{j} + 6 \hat{k})$ is

$6 \hat{i} + 2 \hat{j} - 3 \hat{k}$

$18 \hat{i} + 3 \hat{j} - 2 \hat{k}$

$- 18 \hat{i} - 13 \hat{j} + 2 \hat{k}$

$6 \hat{i} - 2 \hat{j} + 8 \hat{k}$

Here, $ω = 3 \hat{i} - 4 \hat{j} + \hat{k}$
$r = 5 \hat{i} - 6 \hat{j} + 6 \hat{k}$
As $υ = ω \times r$
$= ∣ ∣ \hat{i} 35 \hat{j} - 4 - 6 \hat{k} 16 ∣ ∣$
$= \hat{i} (- 24 - (- 6)) + \hat{j} (5 - 18) + \hat{k} (- 18 - (- 20))$
$= - 18 \hat{i} - 13 \hat{j} + 2 \hat{k}$

Q. A body is rotating with angular velocity ω=(3i^−4j^​+k^). The linear velocity of a point having position vector r=(5i^−6j^​+6k^) is

6i^+2j^​−3k^

18i^+3j^​−2k^

−18i^−13j^​+2k^

6i^−2j^​+8k^