Rotational and translational motion

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A uniform rod AB of length 7m is experiencing combined motion, with point A having a velocity of 11 m/s and point C (the center of mass) moving at 3 m/s, both perpendicular to the rod and in opposite directions. To solve for the acceleration of point A and the angular velocity of the rod, it's suggested to draw a free body diagram and consider a new frame of reference, particularly the center of mass. Since the velocities of both point C and the angular velocity of the rod remain constant, it implies that both the acceleration of point C and the angular acceleration of the rod are zero. This simplification allows for a clearer analysis of the motion dynamics involved.
Kaushik
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Homework Statement
A Uniform rod AB of length 7m is undergoing combined motion such that, at some instant, velocities at top most point A is perpendicular to the rod and magnitude is 11 m/s. The mid point/ centre of mass ,say C, has a velocity of 3 m/s and is also perpendicular to the rod. If both the velocities are in opp. direction. Determine the acceleration of point A and angular velocity of the rod. ( Velocity of the centre C and angular velocity of the rod remains constant.
Relevant Equations
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A Uniform rod AB of length 7m is undergoing combined motion such that, at some instant, velocities at top most point A is perpendicular to the rod and magnitude is 11 m/s. The mid point/ centre of mass ,say C, has a velocity of 3 m/s and is also perpendicular to the rod. If both the velocities are in opposite direction. Determine the acceleration of point A and angular velocity of the rod. ( Velocity of the centre C and angular velocity of the rod remains constant.

Could you please help me out in starting with the problem? HINTS on how to proceed would be appreciated. Thanks!
 
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Kaushik said:
Homework Statement: A Uniform rod AB of length 7m is undergoing combined motion such that, at some instant, velocities at top most point A is perpendicular to the rod and magnitude is 11 m/s. The mid point/ centre of mass ,say C, has a velocity of 3 m/s and is also perpendicular to the rod. If both the velocities are in opp. direction. Determine the acceleration of point A and angular velocity of the rod. ( Velocity of the centre C and angular velocity of the rod remains constant.
Homework Equations: .

A Uniform rod AB of length 7m is undergoing combined motion such that, at some instant, velocities at top most point A is perpendicular to the rod and magnitude is 11 m/s. The mid point/ centre of mass ,say C, has a velocity of 3 m/s and is also perpendicular to the rod. If both the velocities are in opposite direction. Determine the acceleration of point A and angular velocity of the rod. ( Velocity of the centre C and angular velocity of the rod remains constant.

Could you please help me out in starting with the problem? HINTS on how to proceed would be appreciated. Thanks!
The first order of business is to draw a free body diagram.

After that, you might want to consider using changing to a new frame of reference.

That last part of the problem statement, "Velocity of the centre C and angular velocity of the rod remains constant," is important. What does that tell you about the acceleration of point C? And similarly, what does that tell you about the angular acceleration, \alpha, of the rod?
 
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collinsmark said:
What does that tell you about the acceleration of point C? And similarly, what does that tell you about the angular acceleration, ##α##, of the rod?
Acceleration of C and angular acceleration of the rod = 0?
 
Kaushik said:
Acceleration of C and angular acceleration of the rod = 0?
Correct. So, how can you simplify this problem (perhaps considering a new frame of reference)?
 
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collinsmark said:
perhaps considering a new frame of reference
Centre of mass is the best frame of reference in this case. Isn't?
 
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Kaushik said:
Centre of mass is the best frame of reference in this case. Isn't?
It does have advantages here. :wink:
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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