Question

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

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.​

Answer 1

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

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

Answer 2

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

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

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

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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,

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

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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!

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