A solid cylinder of mass 'm' rolls without slippling down an inclined

1.	A solid cylinder of mass 'm' rolls without slippling down an inclined plane making an angle θ with the horizontal. The frictional force between the cylinder and the incline is what ?
Answer» ♋♋YOUR ANSWERS♍♍ ✍ $fk = \frac{mg \sin( \alpha ) }{3}$ consider α = θ ════════════════════ EXPLANATION — Consider α = θ We KNOW from NLM That in CASE of linear acceleration in rolling, Force can be written as $mg \sin( \alpha ) - fk = ma$ --------(1) Here, ↪ fk is kinetic friction ↪α is angle of inclination with horizontal Since this frictional force is also responsible for rotation so we can balance the torque τ = I @ [Here @ is ANGULAR acceleration ] ⇒rfk = I @ ⇒@ = (rfk) /I ⇒a / r = (rfk) /I ⇒fk = aI/r² Using equation (1) $mg \sin( \alpha ) - \frac{aI}{ {r}^{2} } = ma \\ \\ moment \: of \: inertia \: of \\ cylinder\: (I) = \frac{1}{2} m {r}^{2}$ $putting \: above \: we \: will \: get \: this \\ a = \frac{2}{3} g \sin( \alpha )$ Now this value again PUT in eqn (1) $mg \sin( \alpha ) - fk = m( \frac{2}{3} g \sin( \alpha ) )$ On solving $fk = \frac{mg \sin( \alpha ) }{3}$ HENCE, the frictional force between the cylinders and the incline is (mgsinα) /3 ╒══════════════════════╕ ♌ ‌ ‌‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ Thanks‌ ,hope you like it !!‌ ‌ ‌‌ ‌♌ ╘══════════════════════╛

A solid cylinder of mass 'm' rolls without slippling down an inclined plane making an angle θ with the horizontal. The frictional force between the cylinder and the incline is what ?

Answer»

♋♋YOUR ANSWERS♍♍

✍ $fk = \frac{mg \sin( \alpha ) }{3}$

consider α = θ

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EXPLANATION —

Consider α = θ

We KNOW from NLM

That in CASE of linear acceleration in rolling, Force can be written as

$mg \sin( \alpha ) - fk = ma$

--------(1)

Here,

↪ fk is kinetic friction

↪α is angle of inclination with horizontal

Since this frictional force is also responsible for rotation so we can balance the torque

τ = I @ [Here @ is ANGULAR acceleration ]

⇒r*fk = I @

⇒@ = (r*fk) /I

⇒a / r = (r*fk) /I

⇒fk = a*I/r²

Using equation (1)

$mg \sin( \alpha ) - \frac{aI}{ {r}^{2} } = ma \\ \\ moment \: of \: inertia \: of \\ cylinder\: (I) = \frac{1}{2} m {r}^{2}$

$putting \: above \: we \: will \: get \: this \\ a = \frac{2}{3} g \sin( \alpha )$

Now this value again PUT in eqn (1)

$mg \sin( \alpha ) - fk = m( \frac{2}{3} g \sin( \alpha ) )$

On solving

$fk = \frac{mg \sin( \alpha ) }{3}$

HENCE, the frictional force between the cylinders and the incline is (mgsinα) /3

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♌ ‌ ‌‌ ‌ ‌ ‌ ‌ ‌ ‌ ‌ Thanks‌ ,hope you like it !!‌ ‌ ‌‌ ‌♌

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