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A particle starts from rest at top of a solid hemisphere as shown in Fig Q10.The hemisphere has a radius R and is fixed on the horizontal ground.(a)If a solid ball of radius r rolls down the surface of the hemisphere without slipping,find the angle θ,where the ball leaves the surface of the hemisphere. The moment... 顯示更多 A particle starts from rest at top of a solid hemisphere as shown in Fig Q10.The hemisphere has a radius R and is fixed on the horizontal ground. (a)If a solid ball of radius r rolls down the surface of the hemisphere without slipping,find the angle θ,where the ball leaves the surface of the hemisphere. The moment of inertia of a solid ball is I=(2/5)mr^2
最佳解答:
(1) v = ? for c = r + R E1 = Rotating energy = 0.5*I*w^2 = 0.5 * 2/5 * mr^2 * [v/(R+r)]^2 = m/5 * [r/(R+r)]^2 * v^2 E2 = Kinetic energy = 0.5 * m * v^2 E3 = Potential energy = m*g*(r+R)(1-cosθ) E3 = E1 + E2 => v^2 = g(r+R)(1-cosθ)/0.5+0.2(r/r+R)^2 => v^2 = 10g(1-cosθ)c^3/(5c^2+2r^2) ... (a) (2) θ = ? for Ball leaving Fr = Centrifugal force * cosθ = m*v^2*cosθ/c = m*g => v^2 = g*c/cosθ = Eq.(a) => k = (5c^2 + 2r^2)/10c^2 = (1-cosθ)cosθ = cosθ - (cosθ)^2 => (cosθ)^2 - cosθ + k = 0 => cosθ = [1 + √(1-4k)]/2 => θ = acos{[1 + √(1-4k)]/2}
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