A particle of mass 0.350 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.350 kg is attached to the 100-cm mark of a meterstick of mass 0.200 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 2.00 rad/s. (a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark. The moment of inertia of several objects is the sum of the moments of inertia of the individual objects. kg. m^2/s (b) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark.

Explanation / Answer

a)moment of Inertia of the system, I = I_stick + I_particle

= m*L^2/12 + m*(L/2)^2

= 0.2*1^2/12 + 0.35*0.5^2

= 0.104 kg.m^2

Angular momentum, L = I*w

= 0.104*2

= 0.208 kg.m^2/s

b)

moment of Inertia of the system, I = I_stick + I_particle

= m*L^2/3 + m*L^2

= 0.2*1^2/3 + 0.35*1^2

= 0.417 kg.m^2

Angular momentum, L = I*w

= 0.417*2

= 0.834 kg.m^2/s

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