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A problem of uniform circular motion in physics As shown in figure a, the acrobat rides a motorcycle to make a uniform circular motion on the inclined circular wall (i.e. running away at high speed). When the speed is V, there is no friction force to provide centripetal force. The inclined circular wall is called o angle with the horizontal plane, and the bottom radius is R. what is the height h of the acrobat from the ground at this time? To find the process, I said I couldn't draw. Come on, please tell me "when the speed is V, there is no friction to provide centripetal force." what does that mean?
When the centripetal force is provided by the friction force, the vector sum of the supporting force [perpendicular to the wall] and gravity of the circular wall is in the horizontal direction, which is mgcotq. The centripetal force is provided by this force. From mgtanq = MVV / R, grtanq = vv
Where R is the radius of circular motion [the circle is on the horizontal plane h above the ground]. R is known from the geometric relationship: r = R / 2 + hcotq, so G [R / 2 + hcotq] tanq = VV, and h can be obtained by solving this equation
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