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A uniform thin rod of length L and mass M is free to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position.(a) What is the speed of its center of gravity when the rod reaches its lowest position?(b) What is the tangential speed of the lowest point on the rod when it is in the vertical position?

Sagot :

Tutorconsortium421

There are no torques to spin the pivot when the center of gravity is below it. This indicates that the pivot 2's center of gravity is located somewhere on the red line and somewhere on the blue line.

How do you calculate a rod's angular speed?

The formula =mvd(16M+12m) allows us to determine the angular velocity of the rod after it has been struck by the bullet. Calculated value: = 29.1 s-1

When do objects begin to rotate?

Nothing more than a steady radial orientation toward the center constitutes a revolution. The common point is located on the axis of that motion. The axis of that motion contains the shared point. The axis is perpendicular to the plane of motion at a 90 degree angle.

To know more about center of gravity visit:-

https://brainly.com/question/20662235

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