A wheel `A` is connected to a second wheel `B` by means of inextensible string, passing over a pulley `C`, which rotates about a fixed horizontal axle `O`, as shown in the figure. The system is released from rest. The wheel `A` rolls down the inclined plane `OK` thus pulling up wheel `B` which rolls along the inclined plane `ON`. Determine the velocity (in m/s) of the axle of wheel `A`, when it has travelled a distance `s = 3.5 m` down the to be slope. Both wheels and the pulley are assumed homogeneous disks of identical weight and radius. Neglect the weight of the string. The string does not over `C`. [Take `alpha =53^@` and `beta=37^@`]

A wheel `A` is connected to a second wheel `B` by means of inextensibl

1.	A wheel `A` is connected to a second wheel `B` by means of inextensible string, passing over a pulley `C`, which rotates about a fixed horizontal axle `O`, as shown in the figure. The system is released from rest. The wheel `A` rolls down the inclined plane `OK` thus pulling up wheel `B` which rolls along the inclined plane `ON`. Determine the velocity (in m/s) of the axle of wheel `A`, when it has travelled a distance `s = 3.5 m` down the to be slope. Both wheels and the pulley are assumed homogeneous disks of identical weight and radius. Neglect the weight of the string. The string does not over `C`. [Take `alpha =53^@` and `beta=37^@`]
Answer» Correct Answer - 2 Applying conservation of energy we get `mgs(sinalpha-sinbeta)=1/2Iomega^(2)+2(1/2mv^(2)+1/2Iomega^(2))` where `omega=v/r` and `I=mr^(2)//2` putting values and solving we get `v=2m//s`

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