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GENERAL PHYSICS (PHYS 2205/2206) STANDARD VIEW Your answer is partially correct Try again PRINTER VERSION BACK NEXT The drawi
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Answer #1

Let Moment of inertia of the system when axis is passing through m1 be I1

I1= m2\times 32 + m3\times 52 = 6.1 \times 9 + 7.9 \times 25 = 252.4 kg m2

Let Moment of inertia of the system when axis is passing through m3 be I2

I2 = m1\times 52 + m2\times 42 = 9 \times 25 + 6.1 \times 16 = 322.6 kg m2

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Torque \tau _{1} about axis passing through m1 = r \times F = 3 \times 518 = 1554 N m

r and F are vectors and r \times F is cross product of vectors.

For torque \tau _{1} . Force vector is perpendicular to distance from point of application to axis.

Torque \tau _{2} about axis passing through m3 = r \times F = 0 , because force vector is parallel

to distance from point of application to axis.

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Angular acceleration \alpha is obtained from , \tau = I\times \alpha

for the system-1, angular acceleration \alpha = 1554/ 252.4 = 6.16 rad/s

angular velocity \omega after 5.99s = \alpha t = 6.16*5.99 = 36.9 rad /s

for system-2, \tau _{2} = 0 , hence angular acceleration and angular velocity are zero

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