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Angular velocity when an object slides on a tilted rotating surface

ÀÛ¼ºÀÚ Uploader : billy ÀÛ¼ºÀÏ Upload Date: 2019-10-25º¯°æÀÏ Update Date: 2020-01-16Á¶È¸¼ö View : 272

When an object starts to slide at a distance r (m) away from the center of the rotating plate which is tilted by ¥è (degrees) to the horizontal direction, as shown in the figure, the angular velocity ¥ø (rad/s)  of the object is obtained as follows.

The coefficient of friction between the object and the rotating plate is ¥ì.

The rotational acceleration generated as the object rotates is :

a = ¥ø^2 * r

This is divided into the acceleration a * cos¥è outward along the inclined plane and the acceleration a * sin¥è in the upward direction perpendicular to the plane.

On the other hand, the magnitude of acceleration generated by the acceleration of gravity is divided into g * sin ¥è in the outward direction along the inclined plane and g * cos ¥è in the downward direction perpendicular to the plane.

Accordingly, the force Fs to slide in the outward direction and the frictional force Ff can be obtained by multiplying the acceleration occurring along the inclined plane and the acceleration in the direction perpendicular to the plane by the mass m of the object.

Fs = m(g*sin¥è + a*cos¥è)

Ff = ¥ì*m(g*cos¥è - a*sin¥è)

When the object starts to slide, Fs = Ff, thus,

m(g*sin¥è + a*cos¥è) = ¥ì*m(g*cos¥è - a*sin¥è)

a*(cos¥è + ¥ì*sin¥è) = g*(¥ì*cos¥è - g*sin¥è)

¥ø^2 * r = g*(¥ì*cos¥è - g*sin¥è) / (cos¥è + ¥ì*sin¥è)

¥ø = (g*(¥ì*cos¥è - g*sin¥è) / (r*(cos¥è + ¥ì*sin¥è)))^(1/2)

*** Âü°í¹®Çå[References] ***

¥ø = (g*(¥ì*cos(¥è) - sin(¥è)) / (r*(cos(¥è) + ¥ì*sin(¥è))))^(1/2)
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