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  • A circular disc rolles down an inclined plane from a height what's speed of the disc when it reac

A circular disc rolles down an inclined plane from a height \mathrm{h} what's speed of the disc when it reached the bottom of inclined plane.
 

Option: 1

\mathrm{\sqrt{\frac{g h}{3}}}
 


Option: 2

\mathrm{\sqrt{\frac{2 g h}{3}}}
 


Option: 3

\mathrm{\sqrt{\frac{4 g h}{3}}}
 


Option: 4

\mathrm{\sqrt{\frac{2 g h}{30}}}


Answers (1)

best_answer

height given, speed =?

Motion is soling (on rough surface)

still total conservation of energy is applicable.

\mathrm{(T \cdot E \cdot)_{\text {initial }}=(T \cdot E \cdot)_{\text {final }} }

\mathrm{(k \cdot E+P \cdot E)_i=(k \cdot E+P \cdot E)_f}

\mathrm{ {[0+m g(h+R)]=\frac{1}{2} m v_c^2+\frac{1}{2} I_c \omega^2+m g R} }

\mathrm{m g h=\frac{1}{2} m v^2+\frac{1}{2} \cdot \frac{m R^2}{2} \cdot \frac{v^2}{R^2} }

\mathrm{g h=\frac{1}{2} v^2+\frac{1}{4} v^2 }

\mathrm{v^2\left(\frac{1}{2}+\frac{1}{4}\right)=g h }

\mathrm{v^2\left(\frac{2+1}{4}\right)=g h }

\mathrm{v^2=\frac{4 g h}{3} }

\mathrm{v=\sqrt{4 g h / 3} }

Hence option 3 is correct.











 

Posted by

Riya

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