Question

A body performs uniform circular motion with radius \(\vec{r}=4\hat{i}-2\hat{j}+2\hat{k}\) and angular velocity \(\vec{\omega }=2\hat{i}-3\hat{j}+\hat{k}\). The magnitude of linear velocity of the body is

• A. \(5\sqrt{2}\)
• B. \(4\sqrt{5}\)
• C. \(8\sqrt{2}\)
• D. \(2\sqrt{5}\)

Hint:

Use the vector relation v equals omega cross r and find the magnitude of the result.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In circular motion the linear velocity is the cross product \(\vec{v}=\vec{\omega}\times\vec{r}\).

Step 2: Expand the cross product
\(\vec{\omega}\times\vec{r}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-3&1\\4&-2&2\end{vmatrix}\)
\(=\hat{i}(-6+2)-\hat{j}(4-4)+\hat{k}(-4+12)=-4\hat{i}+8\hat{k}\)

Step 3: Magnitude
\(|\vec{v}|=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\)

Final Answer:
The magnitude of the linear velocity is \(4\sqrt{5}\). \[ \boxed{\text{(B) }4\sqrt{5}} \]