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If you push on the center there is less distance to the hinge. The greater the distance from the hinge, the greater the torque to open it for the same force ( force x distance = torque)
I'm thinking center to equator
The greater the distance from the axis of rotation (i.e., pivot point) to the center of gravity of an object, the greater the torque required to rotate the object.
If you exert a force of 100 newtons , and your hands are 10cm (0.1m) from the pivot(the center of the handlebars, the formula is force(100)xPerpendicular distance (0.1) the force would be 10 newtons , but if you increased the distance your hands are from the center (e.g. to 50cm away) the force would be 50 newtons , So, the greater the distance from the center , the greater the turning force (moment).
we can say that tangential speed of the object is linearly proportional to the distance from the center. Increase in the distance results in the increase in the amount of speed. As we move to the center speed decreases, and at the center speed becomes zero.
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