The figure shows particles 1 and 2, each of mass m , attached to the ends of a r
ID: 2265046 • Letter: T
Question
The figure shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 0.90 m and L2 = 7.1 m. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) particle 1 and (b) particle 2?
The figure shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 0.90 m and L2 = 7.1 m. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) particle 1 and (b) particle 2?Explanation / Answer
The rod will rotate clockwise because the torque is greater on the right side. Calling this rotation positive, theangular acceleration "aa" is related to the net torque as;
T(net) = I(aa)
Initially the rod is horizontal and the net torque due to the vertical weights is
T(net) = mgL1 - mgL2
The moment of inertia of the system is;
I = mL1^2 + mL2^2
So the angular acceleration, which is the same for both masses is;
(aa) = T(net)/I = g(L1-L2)/(L1^2 + L2^2)
The Linear accelerations are then;
a1 = L1(aa) = gL1(L1-L2)/(L1^2 + L2^2) , down
a2 = L2(aa) = gL2(L1-L2)/(L1^2 + L2^2) , up
use g=980 cm/s^2 to get the a's in cm/s^2. Or use g=9.8 m/s^2 and change lengths to meters to get the a's in m/s^2.
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