Magnitude and direction of velocity

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SUMMARY

The discussion focuses on calculating the magnitude and direction of velocity for a roller at point B and the center of link AB, given a constant downward velocity of 3 m/s at point A. The angle between link AB and the vertical guide is 75 degrees. Key formulas include the Pythagorean theorem for determining distances and the cosine function to relate the angle to the lengths of the link. The differentiation of these formulas using the chain rule provides the necessary rates of change for velocity and angular velocity.

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The end A of the link in a guide has a constant downward velocity of 3 m/s as shown in the diagram. For the instant where the link AB makes an angle of 75 degrees with the vertical guide calculate the magnitude and direction of the velocity of the roller at B, the magnitude and direction of velocity of the centre of the link AB and the angular velocity of the link AB. Is the velocity at B also constant?
 

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That looks like a typical Calculus "related rates" problem!

You need two formulas: by the Pythagorean theorem, a2+ b2= 2002, where b is the distance from B to the corner and a is the distance from A to the corner. Differentiate that formula with respect to time (using the chain rule, of course) to determine db/dt (you are given da/dt), the "magnitude of the velocity of the roller at B". The direction should be obvious.

You also have [itex]cos(\theta)= \frac{a}{200}[/itex] where [itex]\theta[/itex] is the angle AB makes with the vertical guide (75 degrees at the instant in question). Differentiate that to determine "the angular velocity of the line AB", [itex]\frac{d\theta}{dt}[/itex].

For the midpoint, you can do the same thing but use 100 mm instead of 200.
 

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