The GPE at the starting point is mgL/2
The GPE at the end is zero.
The gain of KE equals then mgL/2
KE of this rotating rod is 1/2 I w^2
mgL/2=1/2 *1/3 mL^2*w^2
solve for w
b) angular acceleration = 1/I *torque
where torque= mg*L/2
angular acceleration= 3/mL^2 *mgL/2
= 3/2 Lg
check my thinking and math.
A thin rod (length = 1.03 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
2 answers
Small correction for the final result in b)
angular acceleration
= torque / I
= (mgL/2) / (1/3 ML^2)
= 3g / 2L
angular acceleration
= torque / I
= (mgL/2) / (1/3 ML^2)
= 3g / 2L
2 answers