IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Here is a slightly more difficult problem in rotational mechanics. Try it out guys! 5 salutes again for the right solution.

A sphere of radius R is projected with a reverse spin w_o down a rough inclined plane with a speed v_o for which coefficient of friction is

>tan@ where @ is the angle of inclination. Find the minimum value of w_o for which it will turn back.

rooney

Is it W₀=( 5V₀

Cos

) / 2MR (

- tan

) ?

karthik2007

Hmm... thats not the ans... but its quite similar to the ans. Nice try rooney

rooney

Damn...I mistyped the answer. This is the answer which i really got (sorry for mistake)

W₀=( 5V₀

) / 2MR (

- tan

)

rooney

Sorry man. made a very small mistake.

W₀=( 5V₀

) / 2R (

- tan

)

waterdemon

Hey Kaarthik just one doubt that I will like to share with U
is regarding the question:

You have given that "the min value of

for which the Sphere
of Radius R will spin back".

My Doubt is:

I have solved the question and also got the answer.
But There is an equation:

mgSin@ - f = ma_x

As per my doubt,

The sphere cannot roll back upwards because of back spin
as the value of :

mgSin@

f

and so there is only possibiltiy of the sphere to stop in
between its downward journey and after that small instant
it will start its downward rolling again.

(Its not a question or doubt actually I am correcting it)

Hope you got it.

Cheers!!!!!!!!!@@!!!!!!!!!

karthik2007

No, waterdemon. Read the question carefully... its given that Mu>tan@

which simply means that f>mgsin@

waterdemon

Oops did not see it. !!!!!!!!!!!
Cheers!!!!!!!!!@@!!!!!!!!!!

Ask & Discuss Questions with Community & Experts