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Question

Angular velocity of a body moving on a circular track is given by ω=^i+^j+^k and the radius vector r=2^i2^j2^k at the instant of time t. Then the magnitude of linear velocity of the body at that instant is.

A
2.01 m/s
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B
4.23 m/s
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C
6.21 m/s
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D
5.65 m/s
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Solution

The correct option is D 5.65 m/s
Given ω=^i+^j+^k and
r=2^i2^j2^k
v=ω×r
=^i^j^k111222
=4^j4^k=42+42=5.65 m/s

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