A 2.0-m measuring stick of mass 0.135 kg is resting on a table. A mass of 0.500

Question

A 2.0-m measuring stick of mass 0.135 kg is resting on a table. A mass of 0.500 kg is attached to the stick at a distance of 67.0 cm from the center. Both the stick and the table surface are frictionless. The stick rotates with an angular speed of 6.40 rad/s. (a) If the stick is pivoted about an axis perpendicular to the table and passing through its center, what is the angular momentum of the system? (b) If the stick is pivoted about an axis perpendicular to it and at the end that is furthest from the attached mass, and it rotates with the same angular speed as before, what is the angular momentum of the system?

Explanation / Answer

Here ,

a)

angular velocity , w = 6.4 rad/s

angular momentum = moment of inertia * w

angular momentum = (m*l^2/12 + M * d^2) * w

angular momentum = (0.135 * 2^2/12 + 0.5 * 0.67^2) * 6.4

angular momentum = 1.724 kg.m^2/s

the angular momentum is 1.724 kg.m^2/s

b)

Now , for the angular momentum about the furthest

angular momentum = moment of inertia * w

angular momentum = (m*l^2/3 + M * d^2) * w

angular momentum = (0.135 * 2^2/3 + 0.5 * 1.67^2) * 6.4

angular momentum = 10.1 kg.m^2/s

the angular momentum is 10.1 kg.m^2/s

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