Question

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 1

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✍ $fk = \frac{mg \sin( \alpha ) }{3}$

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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Answer 2

Answer:

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

3

mgsin(α)

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 = mamgsin(α)−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)

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

mgsin(α)−

r

2

aI

=ma

momentofinertiaof

cylinder(I)=

2

1

mr

2

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

puttingabovewewillgetthis

a=

3

2

gsin(α)

Now this value again put in eqn (1)

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

3

2

gsin(α))

On solving

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

3

mgsin(α)

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

Hope it is useful for you

thank you