A long uniform rod of length 1.00 m and mass 4.00 kg is pivoted about a horizont

Question

A long uniform rod of length 1.00 m and mass 4.00 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.61 Pivot Figure P10.61 (a) What is the angular speed of the rod at the instant it is horizontal? 5.422 rad/s (b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal? 14.7 rad/s^2 (c) Find the components of the acceleration of the rod's center of mass. ax =-14.7 m/s^2 a,--7.35 m/s*2 (d) Find the components of the reaction force at the pivot.

Explanation / Answer

a) w = sqrt [3 g / L] = sqrt [ 3 * 9.8 / 1] = 5.42 rad/s

b) angular acceleration = 3/2 * g / L = 3/2 * 9.8/1 = 14.7 rad/s2

c) ax = 3 g / 2 = 3 * 9.8 / 2 = -14.7 m/s2

ay = 3 / 4 g = 3/4 * 9.8  = -7.35 m/s2

d) components of reaction force

Rx = 3 M g / 2 = 3 * 9.8 * 4.00 / 2

Rx = -58.8 N

Ry = 1/4 M g = 1/4 * 4 * 9.8

Ry= 9.8 N

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