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Jul 19, 2008

Problem No. 6/131

Some hints

The KE's for gear B and arm OA are given by:

T_B = (1/2)*m_B*v_A^2 + (1/2)*I_A*omega_B^2
T_OA = (1/2)*I_O*omega_OA^2

with T_total = T_B + T_OA.

Same thing with potential energy. Find the potential energy for B and OA individually and add together.

The work done on the total system is:

U = +M*(pi/2)

You need KINEMATICS to relate omega_B to v_A and omega_OA to v_A. Consider the IC's for gear B and for arm OA for these relationships.

2 comments:

bkorty said...

Is it correct to consider the mass moment of inertia of the arm OA to be m*k_o^2 and treat it more like a disk? They even give a radius of gyration, k_o, in the problem statement, which makes me think so...

CMK said...

Yes, you need to use I_O = m_OA*k_O^2 where k_O is the radius of gyration of OA about O. This is the definition of radius of gyration and is not so much treating the arm like a disk.