A particle of mass 0.400 kg is attached to the 100 cm mark of a meter stick of m

Question

A particle of mass 0.400 kg is attached to the 100 cm mark of a meter stick of mass 0.175 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 5.00 rad/s. (a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 70.0 cm mark 1.05 Your response differs from the correct answer by more than 100%, kg-m2/s (b) What is the angular momentum when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark? X kg-m2/s Need Help? Talk to a Tutor

Explanation / Answer

(A) Icm = m L^2 / 12 = 0.175 / 12

I1 = Icm + m d^2 = 0.175/12 + 0.175 (0.70-0.50)^2

I1 = 0.02158 kg m^2 of stick only

I = I1 + I2= 0.02158 + (0.4 x 0.30^2) = 0.05758 kg m^2

L = I w

L = 0.288 kg m^2 / s

(B) I = (0.175 / 3) + (0.4 x 1^2) = 0.4583

L = I w = 0.4583 x 5

L= 2.29 kg m^2 / s

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