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Figure shows a uniform smooth solid cyli...

Figure shows a uniform smooth solid cylinder `A` of radius `4 m` rolling without slipping on the `8 kg` plank which, in turn, is supported by a fixed smooth surface. Block `B` is known to accelerate down with `6 m//s^(2)` and block `C` moves down with acceleration `2 m//s^(2)`.

What is the angular acceleration of the cylinder?

A

`4/5rads^(-2)`

B

`6/5 rads^(-2)`

C

`2rads^(-2)`

D

`1 rads^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
D

Acceleration of the bottom of the cylinder = acceleration of the plank `=2m//s^(2)` towards the right
Acceleration of the top of te cylinder = acceleration of `B=6m//s^(-2)` towards the left
Let linear acceleratiion of the cylinder be a towards the left and angular acceleration of it `alpha` in an anticlockwise sense. writing constraint, we get
`a+Ralpha=6` and `Ralpha-a=2`
`impliesalpha=1rads^(-2),a=2m//s^(2)`
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