The linear velocity of a rotating body is given by v - r- where - is the angular velocity and r is the radius vector- The angular velocity of a body is -2 -2 k- and the radius vector r -4 -3 k- then vA- -29 unitsB- -31 unitsC- -41 unitsD- -37 units

The correct option is A √(29) unitsWe know that, v=ω×r=(î 2ĵ+2k̂) × (4ĵ 3k̂) v= | î amp; ĵ amp; k̂ nbsp; 1 amp; 2 amp; 2 nbsp; 0 amp; 4 ...

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Byju's Answer

Standard XII

Physics

Relation between Linear and Angular Speed

The linear ve...

Question

The linear velocity of a rotating body is given by $\to v = \to ω \times \to r$ , where $\to ω$ is the angular velocity and $\to r$ is the radius vector. The angular velocity of a body is $\to ω =^i - 2^j + 2^k$ and the radius vector $\to r = 4^j - 3^k$ , then $∣ ∣ \to v ∣ ∣$ is

\sqrt{29}

units

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\sqrt{31}

units

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\sqrt{37}

units

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\sqrt{41}

units

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Solution

The correct option is A $\sqrt{29}$ units
We know that, $\to v = \to ω \times \to r = (^i - 2^j + 2^k) \times (4^j - 3^k)$

$\to v = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 2 0 & 4 & - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\to v = - 2^i + 3^j + 4^k$
Thus, the magnitude of $\to v$ is, $∣ ∣ \to v ∣ ∣ = \sqrt{(- 2)^{2} + 3^{2} + 4^{2}} = \sqrt{29}$ units

Hence option A is the correct answer

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Similar questions

Q. The linear velocity of a rotating body is given by

\to v = \to ω \times \to r

, where

\to ω

is the angular velocity and

\to r

is the radius vector. The angular velocity of a body is

\to ω =^i - 2^j + 2^k

and the radius vector

\to r = 4^j - 3^k

, then

| \begin{matrix} \to v \end{matrix} |

is:

Q. The linear velocity of a rotating body is given by

\to v = \to ω \times \to r

, where

\to ω

is the angular velocity and

\to r

is the radius vector. The angular velocity of a body is

\to ω =^i - 2^j + 2^k

and the radius vector

\to r = 4^j - 3^k

, then

∣ ∣ \to v ∣ ∣

Q. For a body, angular velocity

(\to ω) =^i - 2^j + 3^k

and radius vector

(\to r) =^i +^j +^k

, then its velocity is

Q. The angular velocity of a rotating body is

\to ω = 4^i +^j - 2^k

. The linear velocity of the body whose position vector is 2

^i + 3^j - 3^k

is:

Q. Angular velocity of a body moving on a circular track is given by

\to ω =^i +^j +^k

and the radius vector

\to r = 2^i - 2^j - 2^k

at the instant of time

‘ t^{'}

. Then the magnitude of linear velocity of the body at that instant is.

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The linear velocity of a rotating body is given by →v=→ω×→r, where →ω is the angular velocity and →r is the radius vector. The angular velocity of a body is →ω=^i−2^j+2^k and the radius vector →r=4^j−3^k, then ∣∣→v∣∣ is

The correct option is A √29 unitsWe know that, →v=→ω×→r=(^i−2^j+2^k)×(4^j−3^k) →v=∣∣ ∣ ∣∣^i^j^k1−2204−3∣∣ ∣ ∣∣ →v=−2^i+3^j+4^k Thus, the magnitude of →v is,∣∣→v∣∣=√(−2)2+32+42=√29 units Hence option A is the correct answer

The linear velocity of a rotating body is given by $\to v = \to ω \times \to r$ , where $\to ω$ is the angular velocity and $\to r$ is the radius vector. The angular velocity of a body is $\to ω =^i - 2^j + 2^k$ and the radius vector $\to r = 4^j - 3^k$ , then $∣ ∣ \to v ∣ ∣$ is