Question

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A solid dise and a ring, both the radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to10π rad/s. Which of the two will start to roll earlier? The coefficient of kinetic friction is μk = 0.2.
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Answer 1

★ QUESTION :

A solid dise and a ring, both the radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10π rad/s. Which of the two will start to roll earlier? The coefficient of kinetic friction is μk = 0.2

GIVEN :

➠ Radius of a solid dice and a ring = 10 cm

➠ Initial angular speed = 10π rad/s

➠ The coefficient of kinetic friction is (μk) = 0.2

TO FIND :

➠ The two will start to roll earlier = ?

STEP - BY - STEP EXPLAINATION :

➠ V = 0

➠ μk × mg = ma

➠ (and here as we know that Acceleration) = μkg

➠ V = U + at and V = 0 + μkgt

➠ μkmg × R = - Ia

➠ a (angular speed) = -μkmgR / 1

➠ w = w0 + at ; w = w0 - μkmgRt / 1

➠ v (Velocity) = Rw

➠ V = Rw

➠ μkgt = Rw0 - μkmgR^2t / I

➠ μkgt = Rw0 – 2μkgt

= 2μkgt = Rw0

➠ t = w0R / 2μkg

Now, here we will prove for Disc :---

➠ I = 1/2 mr^2

➠ μgktd = Rw0 × μkgmr^2t / 1/2mr^2

➠ Rw0 – 2μkgt

➠ 3μkgtd = Rw0

➠ td = Rw0 / 3μkg

= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53sec.

Now, here we will prove for Ring :---

➠ I = mr^2

➠ μgt = Rw0 - μkgmr^2t / mr^2

➠ Rw0 - μkgmtr

➠ 2μkgt = Rw0

➠ tr = Rw0 / 2μkg

= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80sec.

Therefore, we saw here that Disc will start to roll earlier rather than ring.

Answer 2

Given

A solid disc and a ring has radius 10 cm .Both are given initial angular velocity equal to 10π rad/s.

Solution

• Let the radius be R and angular speed be $\omega$ . Since,both the objects have same data,it would not matter much.

• Also,assume the rotation to be clockwise.

• The object has no translational velocity.

• For pure rolling,V = R$\omega$.

• Consider figure - I ,the bottom point of the objects do not have any velocity in right direction.Accordingly,friction should act in right direction.

• But the torque would act in the opposite direction to the angular velocity. Use R × F to find out the direction.

• Therefore, translational velocity would increase and angular velocity would decrease.

• A stage will be reached when V = R$\omega$.

Let us find out that Which object will roll earlier.For this we would need some formulas that are listed here :-

• F = ma

• Torque = R × F = I$\alpha$ where I is moment of inertia and $\alpha$ is angular acceleration.

• Moment of inertia of disc = MR²/2

• Moment of inertia ring = MR²

Disc

Let the friction be f.Let us assume that disc starts rolling after time t.

Angular acceleration of disc = $\alpha \:=\:\dfrac{fR}{I}$

$\implies \alpha\:=\: \dfrac{2fR}{MR^2}$

$\implies \alpha\:=\: \dfrac{2f}{MR}$

Angular velocity of disc after time t = $w' \:=\: \omega \:-\: \alpha t$

translational acceleration (a) = $\dfrac{f}{m}$

Hence, translational velocity a time t = V= at.

For, pure rolling Rw' = v

$\implies \: \: R( \omega \:-\: \alpha t)\:=\: at$

putting the value of $\alpha$ and a .

$\implies \: \:R( \omega \:-\: \dfrac{2ft}{MR} )\:=\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:-\: \dfrac{2ft}{M} \:=\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:=\: \dfrac{2ft}{M} \:+\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:=\: \dfrac{3ft}{M} \:$

$\implies \: t\:=\: \dfrac{\omega RM}{3f}$

Ring

Let the friction be f.Let us assume that ring starts rolling after time t.

Angular acceleration of disc = $\alpha \:=\:\dfrac{fR}{I}$

$\implies \alpha\:=\: \dfrac{fR}{MR^2}$

$\implies \alpha\:=\: \dfrac{f}{MR}$

Angular velocity of disc after time t = $w' \:=\: \omega \:-\: \alpha t$

translational acceleration (a) = $\dfrac{f}{m}$

Hence, translational velocity a time t = V= at.

For, pure rolling Rw' = v

$\implies \: \:R( \omega \:-\: \alpha t)\:=\: at$

putting the value of $\alpha$ and a .

$\implies \: \:R( \omega \:-\: \dfrac{ft}{MR} )\:=\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:-\: \dfrac{ft}{M} \:=\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:=\: \dfrac{ft}{M} \:+\: \dfrac{ft}{m}$

$\implies \: \:R \omega \:=\: \dfrac{2ft}{M} \:$

$\implies \: t\:=\: \dfrac{\omega RM}{2f}$

Hence,we observe that $t_{disc}\:>\: t_{ring}$ . Therefore,disc will roll earlier.