Question

6. (9pt) A disk and a ring are mounted on low-friction bearings on the same axle and can

be brought together so that they couple and rotate as one unit. The mass is uniformly

distributed throughout the objects. The disk and the ring both have a radius of 10cm and a

mass of 4kg. The objects are set spinning about their central axis, the disk at 400rev/min

in clockwise direction and the ring at 800rev/min in the opposite direction. They then

couple together and rotate with an angular velocity of 400rev/min in counter-clockwise

direction. Determine whether the angular momentum of the system is conserved, and

what average torque might have acted on the system if it was applied steadily over a

period of 1 minute. The moments of inertia of a solid disk and a ring with respect to their

central axis are ID =

MR2

2

, IR = MR2 .

Answer 1

$\mathrm{Taking\ counter-clockwise \ to \ be \ +\hat k\ and\ clockwise\ as\ -\hat k }$

$\mathrm{I_{disk} = \frac{mR^2}{2} = \frac{4 \times (0.1)^2}{2} = 0.02 \ kg.m^2 }$

$\mathrm{I_{ring} = {mR^2} = {4 \times (0.1)^2} = 0.04 \ kg.m^2 }$

$\mathrm{\omega_{disk} = - 400 \hat k \ rev /m = -\frac{400\times 2\pi}{60} \hat k\ rad/s = -41.89\hat k \ rad/s }$

$\mathrm{\omega_{ring} = + 800 \hat k \ rev /m = \frac{800\times 2\pi}{60} \hat k\ rad/s =+83.78\hat k \ rad/s }$

$\mathrm{\omega_{final} = + 400 \hat k \ rev /m = \frac{400\times 2\pi}{60} \hat k\ rad/s =+41.89\hat k \ rad/s }$

$\mathrm{L_{initial} = I_{disk}\omega_{disk} + I_{ring}\omega_{ring} }$

$\mathrm{L_{initial} = (0.02 \times -41.89 \hat k) + (0.04 \times 83.78 \hat k) =\mathbf{ 2.5134\hat k} \ kg.m^2/s }$

$\mathrm{L_{final} = I_{disk}\omega_{final} + I_{ring}\omega_{final} }$

$\mathrm{L_{final} = (0.02 \times 41.89 \hat k) + (0.04 \times 41.89 \hat k) =\mathbf{ 2.5134\hat k} \ kg.m^2/s }$

$\mathrm{We\ can\ see\ that\ L_{initial} = L_{final}\ ,\ hence\ angular\ momentum\ remains\ conserved }$

$\mathrm{T_{avg} = \frac{\Delta L}{\Delta t} = \frac{0}{60} = \mathbf{0\ N.m}}$