A circular disc is rotating about its own axisat constant angular acce

1.	A circular disc is rotating about its own axisat constant angular acceleration. If itsangular velocity increases from 210 rpm to420 rpm during 21 rotations then the angularacceleration of disc is
Answer» Here, the initial angular velocity (ωi) = 210rpm = 210/60 i.e, 7/2rps. So, for converting 7/2rpsto angular velocity, we need to make itθ.We know,θ = 2π x No. of rotations per second. So, 2 x 22/7 x 7/2 =22 And the final angular velocity, (ωf) = 420rpm = 420/60 i.e, 7rps. Similarly, this also has to be converted toθ.2 x 22/7 x 7 =44 Also, we have no. of rotations = 21; which also has to be converted to θi.e, 2 x 22/7 x 21 =132 So we have,(ωi) = 22,(ωf) = 44 andθ = 132. we know,α =(ωf)^2 -(ωi)^2 / 2θ So, we get, (44)^2 - (22)^2 / 132 x2 =5.5 rad/s2

A circular disc is rotating about its own axisat constant angular acceleration. If itsangular velocity increases from 210 rpm to420 rpm during 21 rotations then the angularacceleration of disc is

Answer»

Here, the initial angular velocity (ωi) = 210rpm = 210/60 i.e, 7/2rps.

So, for converting 7/2rpsto angular velocity, we need to make itθ.We know,θ = 2π x No. of rotations per second.

So, 2 x 22/7 x 7/2 =22

And the final angular velocity, (ωf) = 420rpm = 420/60 i.e, 7rps.

Similarly, this also has to be converted toθ.2 x 22/7 x 7 =44

Also, we have no. of rotations = 21; which also has to be converted to θi.e, 2 x 22/7 x 21 =132

So we have,(ωi) = 22,(ωf) = 44 andθ = 132.

we know,α =(ωf)^2 -(ωi)^2 / 2θ

So, we get, (44)^2 - (22)^2 / 132 x2 =5.5 rad/s2

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