A uniform disk of mass 300kg is rotating...

A uniform disk of mass `300kg` is rotating freely about a vertical axis through its centre with constant angular velocity `omega .` A boy of mass `30kg` starts from the centre and moves along a radius to the edge of the disk. The angular velocity of the disk now is

`(omega_(0))/(6)`

`(omega_(0))/(5)`

`(4omega_(0))/(5)`

`(5 omega_(0))/(6)`

Text Solution

Verified by Experts

The correct Answer is:

As `Sigma tau =0` , angular momentum remains conserved `:`
`:. L=(0+(300R^(2))/(2))`
`omega_(0)=((300R^(2))/(2)+30R^(2)).omega`
`=150 omega _(0)=180 omega`
`rArr omega = 5/6 omega _(0) Ans. `

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A uniform disc of mass 300 kg is rotating freely about a vertical axis through its centre at a constant angular velocity w_(0) . A boy of mass 30 kg starts from the centre and moves along a radius to the edge of the disc. Angular velocity of the disc will now become

`(w_(0))/(6 )`

`(w_(0))/(5 )`

` (4 omega_(0))/(5)`

`(5 omega_(0))/(6)`

A uniform disc of mass M and radius R is rotating about a horizontal axis passing through its centre with angular velocity omega . A piece of mass m breaks from the disc and flies off vertically upwards. The angular speed of the disc will be

`((M-2m)omega)/((M-m))`

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`((M-2m)omega)/((M+m))`

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A ring of radius r is rotating about a vertical axis along its diameter with constant angular velocity omega.A read of mass m remains at rest w.r.t. ring at the position shown in figure. Then w^(2) is:

`(g)/(sqrt(3))`

`(2g)/(sqrt(3r))`

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