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Question

# The centre of mass of a uniform rod of length 10 m is moving with translational velocity of 50 m/s on a frictionless horizontal surface as shown in the figure and the rod rotates about its centre of mass with an angular velocity of 5 rad/sec. Find out the velocities of points A and B.

A
VA=75^i
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B
VB=25^i
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C
VA=25^i
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D
VB=75^i
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Solution

## The correct options are C −→VA=25^i D −→VB=75^iGiven, −−→Vcom=50 m/s (^i) Angular velocity about COM ω=5 rad/s (^k) (by right hand thumb rule) Length of the rod L=10 m As we know −→VB=−−→Vcom+→ω×−−→rBO −→VB=50^i+5^k×5(−^j) −→VB=50^i+25^i {∵ ^k×^j=−^i} −→VB=75^i Similarly, −→VA=−−→Vcom+→ω×−−→rAO −→VA=50^i+5(^k)×5(^j) −→VA=50^i+25(−^i) {∵ ^k×^j=−^i} −→VA=25^i

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