Question:
What is the value of linear velocity, if angular velocity is $3 \hat i - 4 \hat j + \hat k$ and distance from the centre is $5 \hat i - 6 \hat j + 6 \hat k$ ?
What is the value of linear velocity, if angular velocity is $3 \hat i - 4 \hat j + \hat k$ and distance from the centre is $5 \hat i - 6 \hat j + 6 \hat k$ ?
Updated On: Jun 18, 2022
- $6 \hat i + 6 \hat j - 8 \hat k$
- $-18 \hat i - 13 \hat j + 2 \hat k$
- $4 \hat i - 13 \hat j - 6 \hat k$
- $6 \hat i - 2 \hat j + 8 \hat k$
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The Correct Option is B
Solution and Explanation
Linear vector $ v = \omega \times r$
Given, $\omega = 3 \hat i - 4 \hat j + \hat k \ \ and \ \ r = 5 \hat i - 6 \hat j + 6 \hat k$
$\therefore v = \begin{vmatrix}
\hat i & \hat j & \hat k \\
3 & -4 & 1 \\
5 & -6 & 6 \\
\end{vmatrix}$
$ =\hat i (-24 + 6) - \hat j (18 - 5) + \hat k (-18 + 20)$
$ = -18 \hat i - 13 \hat j + 2 \hat k$
Given, $\omega = 3 \hat i - 4 \hat j + \hat k \ \ and \ \ r = 5 \hat i - 6 \hat j + 6 \hat k$
$\therefore v = \begin{vmatrix}
\hat i & \hat j & \hat k \\
3 & -4 & 1 \\
5 & -6 & 6 \\
\end{vmatrix}$
$ =\hat i (-24 + 6) - \hat j (18 - 5) + \hat k (-18 + 20)$
$ = -18 \hat i - 13 \hat j + 2 \hat k$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration