6here have we to find alpha at the end of the rod??
A rod and a block are of same mass. Initially rod is in horizontal position. What will be acceleration of tip of the rod when system is released from the position.
Elaborate your answer.
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6 Answers
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for the hanging mass,
balancing forces,
we get!! mg-T=ma
so for the string, T=mg-ma
for rod, balancing forces,
mg=T+R
for rod balancing torque abt hinge
T.L-mg.L2=I.α
or, (mg-ma)L-mg.L2=13mL2.aL
or, mg-ma-12mg=13ma
or, 12mg=43ma
giving, a=38g