1. 2. Determine the linear acceleration of the tip of the rod. Assume that the f

Question

1.

2.

Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown.

1. A uniform rod of mass M = 5.00kg and length L = 1.12m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below. The rod is held horizontally and then released. At the moment of release, determine the angular acceleration of the rod. Use units of rad/(s*s). 2. Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown.

Explanation / Answer

a) We know that

torque = I*alpha

moment of inertia of a uniform rod is 1/3ML^2

so torque = MgL/2

alpha= torque/ I = 3g/2L

alpha =3*9.81/ 2*1.12 = 13.125 m/s^2

angular acceleration = 13.125 m/s^2

b) Because the rod starts from rest, there is no radial component of a

linear accelearation  = r*alpha = L (3g/2L)

linear accelearation = 3g/2 = 3*9.81/2

linear accelearation= 14.715 m/s^2

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