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System of Particles and Rotational Motion

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. 

(a) How far will the cylinder go up the plane? 

(b) How long will it take to return to the bottom?

Initial velocity of the solid cylinder, v = 5 m/s 

Angle of inclination, θ = 30° 

Let the cylinder go up the plane upto a height h

Using the relation,

1 half space mv squared space plus space 1 half space straight I space straight omega squared space equals space mgh
1 half space mv squared space plus space 1 half open parentheses 1 half space mr squared close parentheses straight omega squared space equals space mgh space

3 over 4 mv squared space equals space mgh space

straight h space equals 3 fraction numerator straight v squared over denominator 4 straight g end fraction space equals space 3 space straight x space fraction numerator 52 over denominator 4 space straight x space 9.8 end fraction

space space space equals space 1.913 space straight m space

If s is the distance up the inclined plane, then as

      sin θ =open parentheses straight h over straight s close parentheses 

          s =open parentheses fraction numerator straight h over denominator sin space straight theta end fraction close parentheses  

            =
open parentheses fraction numerator 1.913 over denominator sin space 30 to the power of straight o end fraction close parentheses space equals space 3.826 space straight m

Therefore, time taken to return to the bottom of the inclined plane,  


  straight t space equals space square root of fraction numerator 2 space straight s space open parentheses 1 space plus space bevelled straight k squared over straight r squared close parentheses over denominator straight g space sin space straight theta end fraction end root space

space space equals space fraction numerator 2 space straight x space 3.826 space open parentheses 1 space plus space bevelled 1 half close parentheses over denominator 9.8 space sin space 30 to the power of straight o end fraction space

space space equals space 1.53 space sec

1.53 sec is required for the balloon to return to the bottom. 





 
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