Physics, asked by Anonymous, 11 months ago

\huge{\underline{\underline{\mathfrak{Question :}}}}

A cylinder is released from rest from the top of an incline of inclination theta and length L the cylinder rolls without slipping prove that its speed when it reaches to the bottom will be

V = √(4 by 3) gL sin theta.​

Answers

Answered by ItSdHrUvSiNgH
12

Explanation:

 \huge\bf{\mid{\underline{\overline{Question:-} \mid}}}

A cylinder is released from rest from the top of an incline of inclination theta and length L the cylinder rolls without slipping prove that its speed when it reaches to the bottom will be

v =  \sqrt{ \frac{4gl \sin( \theta) }{3} }

 \huge\bf{\mid{\underline{\overline{Answer:-}\mid}}}

 \huge \blue{formula \:  \: used} \\  \\ v = r \times  \omega \implies  \omega =  \frac{v}{r}  \\  \\ I =  \frac{1}{2}  \times m \times  {r}^{2}  \\  \\ P. E = mgh \:  \:  \: \boxed{ here \: h = l \sin( \theta) }

K. E will be maximum at bottom

 \\  \\ P. E = K. E_{max} \\  \\ mg(l \sin( \theta) ) =   \frac{1}{2} \times I \times   { \omega}^{2}  +  \frac{1}{2}  \times m {v}^{2}  \\  \\ mg(l \sin( \theta) ) =  \frac{1}{2} ( \frac{1}{2}  \times m \times \cancel{ {r}^{2} }) \times ( {\frac{v}{\cancel{r}}})^{2}  +  \frac{1}{2}  \times m \times  {v}^{2}  \\  \\  \cancel{m}g(l \sin( \theta) ) =(  \frac{1}{4}  +  \frac{1}{2} ) \cancel{m }{v}^{2}  \\  \\  {v}^{2}   \times   \frac{3}{4}  = gl  \sin( \theta)  \\  \\  {v}^{2}  =  \frac{4(gl \sin( \theta) }{3}  \\  \\ \huge \boxed{ v =  \sqrt{ \frac{4 \times l \times g \times  \sin( \theta) }{3} } }

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Answered by ShivamKashyap08
15

{ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}

A cylinder is released from rest from the top of an incline of inclination theta and length L the cylinder rolls without slipping prove that its speed when it reaches to the bottom will be V = \large{\sqrt{\dfrac{4}{3}gL sin \theta}}

\huge{\bold{\underline{\underline{....Answer....}}}}

\huge{\bold{\underline{Given:-}}}

  • Let the Height of Inclined plane be "h".
  • Angle of inclination is θ
  • Distance travelled by the Cylinder = L

\huge{\bold{\underline{Explanation:-}}}

\rule{300}{1.5}

Applying Law of conservation of Energy,

\large{\boxed{\tt W = K.E_{Transnational} + K.E_{Rolling}}}

\large{\tt \leadsto mgh = \dfrac{1}{2} mv^2_{CM} + \dfrac{1}{2} I \omega^2}

I here is Moment of Inertia,

I = MK² (K = Radius of gyration)

Substituting,

\large{\tt \leadsto mgh = \dfrac{1}{2} mv^2_{CM} + \dfrac{1}{2} m . K^2.\omega^2}

V = Rω

ω = V/R

\large{\tt \leadsto mgh = \dfrac{1}{2}m \bigg[v^2_{CM} + K^2 . \bigg(\dfrac{V_{CM}}{R}\bigg)^2 \bigg]}

\large{\tt \leadsto \cancel{m}gh = \dfrac{1}{2}\cancel{m} \bigg[v^2_{CM} + K^2 . \bigg(\dfrac{V_{CM}}{R}\bigg)^2 \bigg]}

\large{\tt \leadsto gh = \dfrac{1}{2} \bigg[v^2_{CM} + K^2 . \bigg(\dfrac{V_{CM}}{R}\bigg)^2 \bigg]}

Applying Trigonometric ratios,

we get,

sinθ = h/s

h = s × sin θ

Simplifying,

\large{\boxed{\tt V_{rolling} = \sqrt{\dfrac{2g. s \: sin \theta}{1 + \dfrac{K^2}{R^2}}}}}

\rule{300}{1.5}

\rule{300}{1.5}

According to the question

  • S = L
  • It is A Solid cylinder.

Now,

\large{\boxed{\tt I = \dfrac{MR^2}{2}}}

(For Solid cylinder)

\large{\tt \leadsto MK^2 = \dfrac{MR^2}{2}}

\large{\tt \leadsto \cancel{M}K^2 = \dfrac{\cancel{M}R^2}{2}}

\large{\tt \leadsto K^2 = \dfrac{R^2}{2}}

\large{ \leadsto {\underline{\underline{ \tt \dfrac{K^2}{R^2} = \dfrac{1}{2}}}} \: \tt -----(1)}

Substituting the values in the above obtained Formula,

\large{\boxed{\tt V_{rolling} = \sqrt{\dfrac{2g. s \: sin \theta}{1 + \dfrac{K^2}{R^2}}}}}

\large{\tt V_{rolling} = \sqrt{\dfrac{2g. L \: sin \theta}{1 + \dfrac{(1)}{(2)}}}}

\large{\tt V_{rolling} = \sqrt{\dfrac{2g. L \: sin \theta}{ \dfrac{2 + 1}{2}}}}

\large{\tt V_{rolling} = \sqrt{\dfrac{2g. L \: sin \theta}{\dfrac{3}{2}}}}

\huge{\boxed{\boxed{\tt V_{rolling} = \sqrt{\dfrac{4}{3}gL sin \theta} }}}

Hence derived!

\rule{300}{1.5}

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