Help with physics question about a rod
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Help with physics question about a rod

[From: ] [author: ] [Date: 11-05-12] [Hit: ]
3.If it is rotated about the z axis (which is through the origin and perpendicular to the plane of the figure)?-You can assume the rods are massless and that the masses are particles. You will have three different moments of inertia, as only the masses that are not on the axis of rotation contribute to angular momentum. The moment of inertia,......
The L-shaped object in the figure consists of three masses connected by light rods.
http://session.masteringphysics.com/prob…

1.What torque must be applied to this object to give it an angular acceleration of 1.20 rad/s^2 if it is rotated about the x axis?

2.If it is rotated about the y axis?

3.If it is rotated about the z axis (which is through the origin and perpendicular to the plane of the figure)?

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You can assume the rods are massless and that the masses are particles. You will have three different moments of inertia, as only the masses that are not on the axis of rotation contribute to angular momentum. The moment of inertia, I, is the sum of all mr^2 of all rotating masses, where m is their mass and r is their distance from the axis of rotation (if r is zero, then they add nothing to it).

Ix, the moment of inertia of the whole system rotating in the x-axis, only concerns the 9 kg mass:

Ix = (9 kg)(1 m)^2 = 9 kg*m^2

And Iy only concerns the 2.5 kg mass:

Iy = (2.5 kg)(2 m)^2 = 10 kg*m^2

Then obviously Iz is the sum of Ix and Iy, 19 kg*m^2.

You will use those different I's and the desired angular acceleration, a, to find the torque needed, t, using the rotational version of Newton's Second Law:

t = I*a

Easy enough:

tx = (9 kg*m^2)(1.2 s^-2) = 10.8 N*m

ty = (10 kg*m^2)(1.2 s^-2) = 12.0 N*m

tz = (19 kg*m^2)(1.2 s^-2) = 22.8 N*m
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