A uniform rod of mass M = 5.09kg and length L = 1.16m can pivot freely (i.e., we

Question

A uniform rod of mass M = 5.09kg and length L = 1.16m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below.

http://i893.photobucket.com/albums/ac132/yonnyg19/gian0850.gif

A)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).

B)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) I = moment of inertia of rod about pivot point = ML^2/3

torue = MgL/2

I* angular acceleration = torque

angular acceleration = torque/I

=( MgL/2)/( ML^2/3)

= 3g/(2*L)

= 3*9.8/(2*1.16)

=12.67 rad/s^2

b) linear accelearation of tip = L*angular acceleration

=12.67*1.16

= 14.7 m/s^2

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