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16 Oct 2009 21:33:40 IST

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Rotation

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A solid body starts rotating about a stationary axis with an angular acceleration of (2 * 10^-2) t rad/s.

here, t is in seconds. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle of 60^o with its velocity vector?

(A) 6 sec

(B) 7 sec

(D) 5 sec

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Comments (6)

Bhimsen Padalkar

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16 Oct 2009 23:47:33 IST

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w = 0.02 t

Ang. accn @ = dw / dt = 0.02

Therefore tangential accn a_T = r x @ = 0.02 r

Also Radial accn a_r = r w² = 0.04 x 10^-2 t² r

Now given that a_r / a_T = tan 60 = sqrt (3) = 1.732

It gives 0.02 t² = 1.732

So, t = 9 sec

Ajay Maity

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25 Oct 2009 13:18:22 IST

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Sorry to say bt dis is not d right answer, and i didn't even get d steps wat u've done.......cn u pls try it again, if u can??

M. Fuaad Khan

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25 Oct 2009 17:16:34 IST

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By applying eq of motion in rotational analogues

theta = 1/2 alpha t²

pi/3 = 1/2 (2x10^-2)t³

upon solving t=5 second

VARUN RAJ

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26 Oct 2009 07:18:55 IST

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actuallly thers no mistake in the first solution

Ajay Maity

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26 Oct 2009 19:20:42 IST

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ok!! i agree it my mistake of posting it wrongly.....Actually, it is @ = 0.02t and not w = 0.02t.......i have written the units of angular acceleration wrongly......pls apologize....and solve it again....

VARUN RAJ

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26 Oct 2009 20:58:53 IST

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see this is how it should be

dw/dt=0.002t

or w=0.01t^2

or radial acc is w^2r =10^-4t^4

and tangential acc................................

is dw/dt is 0.02t

now proceeding as above

tan 60=radial acc/tang acc

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