An object with a weight of 50.0N is attached to the free end of alight string wr

Question

An object with a weight of 50.0N is attached to the free end of alight string wrapped around a reel of radius 0.250 m and mass 3.00kg. The reel is a solid disk, free to rotate in a vertical planeabout the horizontal axis passing through its center. The suspendedobject is released 6.00 m above the floor.
a) Determine the tension in the string, the acceleration ofthe object, and the speed with which the object hits thefloor.
b) Verify your last answer by using the principle ofconservation of energy to find the speed with which the object hitsthe floor.
a) Determine the tension in the string, the acceleration ofthe object, and the speed with which the object hits thefloor.
b) Verify your last answer by using the principle ofconservation of energy to find the speed with which the object hitsthe floor.

Explanation / Answer

a)the tension in the string T * r = w * S or T = (w * S/r) w = 50.0 N,S = 6.00 m and r = 0.250 m the acceleration of the object a = (v^2/r) v^2 = 2gS g = 9.8 m/s^2 the angular momentum of the object is L = I * w I = (m * r^2/2) w = (v/r) = ((2gS)^(1/2)/r) or M * v * r = (m * r^2/2) * ((2gS)^(1/2)/r) or v = (1/M * r) * (m * r^2/2) * ((2gS)^(1/2)/r) or v = (m/2M) * (2gS)^(1/2) here v is the speed with which the object hits the floor b)the principle of conservation of energy (1/2)m1 * u1^2 + (1/2)m2 * u2^2 = (1/2)m1 * v1^2 + (1/2)m2 *v2^2 u1 = u2 = 0 or (1/2)m1 * v1^2 + (1/2)m2 * v2^2 = 0 or (1/2)m2 * v2^2 = -(1/2)m1 * v1^2 or v2 = -(m1/m2)^(1/2) * v1 = -(m1/m2)^(1/2) *(2gS)^(1/2)

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