Question

A uniform disk of mass M_{disk} = 4 kg and radius R =
0.24 mhas a small block of mass m_{block} = 2.2 kg on its
rim. It rotates about an axis a distance d = 0.16 m from its center
intersecting the disk along the radius on which the block is
situated.

What is the moment of inertia of the block about the rotation axis?

What is the moment of inertia of the disk about the rotation axis?

When the system is rotating about the axis with an angular velocity of 4.7 rad/s, what is its energy?

If while the system is rotating with angular velocity 4.7 rad/s
it has an angular acceleration of 8.9 rad/s^{2}, what is
the magnitude of the acceleration of the block?

Answer #1

a) Moment of inertia of the block about the rotation axis,

I (block) = md^2

= 2.2*(0.24 - 0.16) ^2

= 0.014 kg m^2

b) Moment of inertia of the disk about the rotation axis,

I(disk) = 0.5MR^2 + MD^2

= 0.5*4*0.23^2 + 4*0.16^2

= 0.21 kg m^2

c) KE = (1/2)Iw^2

= (1/2)*(0.21 + 0.014)*4.7^2

= 2.47 J

d) Acceleration of the block, a = sqrt(a_{r}^2 +
a_{t}^2 )

Where, a_{r} = w^2 r = 4.7^2*(0.24 - 0.16) = 1.77

a_{t} = 8.9*(0.24 - 0.16) = 0.71

a = sqrt(1.77^2 + 0.71^2)

= 1.91 m/s^2

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