Obviously I do not know what your axes are.
lengths 2 and 1
masses 2 and 1
if along the long rod:
I = (1/3)(1) * 1^2 = 1/3
if along the short rod
I = (1/12)(2) * 2^2 = 8/12 = 2/3
if perpendicular to the plane of the rods at intersection
I = 3 (1/3) (1)(1^2) = 1 = 3/3
so angular accelerations are in the ratio 1,2,3 with the maximum being along the long rod through its center and the minimum (max I) being the last of the three
A long, thin rod is cut into two pieces, one being twice as long as the other. To the midpoint of piece A (the longer piece), piece B is attached perpendicularly, in order to form the inverted âTâ shown in the drawing. The application of a net external torque causes this object to rotate about axis 1 with an angular acceleration of 1.1 rad/s2. When the same net external torque is used to cause the object to rotate about axis 2, what is the angular acceleration?
1 answer
0 answers