Question

A disc is rotating with an angular velocity ω⁰. A constant retarding torque is apiled on it to stop the disc. The angular velocity becomes (ω⁰/2) after n rotations. How many more rotations will it make before coming to rest?​

Answer 1

Answer:

Use the equations of motion.

W22=w12+2αθ

Where w1 and w2 are initial and final angular velocities, α is angular acceleration and θ is the angular displacement.

n revolutions correspond to 2πn radians

w²/4=w²+4nπα

α=−3w²/16nπ

before coming to rest; let us assune it makes m rotations

using the same equation of motion. We have w2 as zero and w1 as w/2. θ is 2πm radians and we have α above.

0=w²/4+4mπ(−3w²/16nπ)

m=n/3

Answer 2

$\large{\underline{\underbrace{\sf{Question}}}}$

A disc is rotating with an angular velocity ω⁰. A constant retarding torque is apiled on it to stop the disc. The angular velocity becomes (ω⁰/2) after n rotations. How many more rotations will it make before coming to rest?

$\large{\underline{\underbrace{\sf{Given \: that}}}}$

✠ A disc is rotating with an angular velocity ω⁰

✠ The constant retarding torque is aplied on it to stop the disc.

✠ The angular velocity becomes (ω⁰/2) after n rotations.

$\large{\underline{\underbrace{\sf{To \: find}}}}$

✠ Number of rotations that make that disc stoped or in a rest.

$\large{\underline{\underbrace{\sf{Solution}}}}$

✠ Number of rotations that make that disc stoped or in a rest = $n_{1}$ = $\frac{n}{3}$

$\large{\underline{\underbrace{\sf{These \: term \: means}}}}$

✠ $\dot\theta_{1}$ means Initial angular velocity rad/sec

✠ $\dot\theta_{f}$ means Final angular velocity rad/sec

✠ $\alpha$ means Angular acceleration rad/sec²

✠ Δθ means Angular displacement rad

$\large{\underline{\underbrace{\sf{Understanding \: question}}}}$

This question says that there is a disc rotating with an angular velocity ω⁰ afterthat it says that the constant retarding torque is aplied on it to stop the disc ! The angular velocity becomes (ω⁰/2) after n rotations. How many more rotations will it make before coming to rest? In short we have to find the number of rotations that make that disc stoped or in a rest.

$\large{\underline{\underbrace{\sf{Procedure}}}}$

To solve this problem we have to use an equation of motion that is ${\dot\theta_{f}^{2}}$ = ${\dot\theta_{1}^{2}+2\alpha\Delta\theta}$ Afterthat we have to put the values according to this equation. Afterwards we get the 1st era of of motion afterthat same doing for 2nd era we have to use the same equation. And we get our final result that is $n_{1}$ = $\frac{n}{3}$

$\large{\underline{\underbrace{\sf{Answer}}}}$

To solve this question firstly we have to use the given Equation and we already know that this is the equation of motion.We will use following equation of motion

${\dot\theta_{f}^{2}}$ = ${\dot\theta_{1}^{2}+2\alpha\Delta\theta}$

Where, these means

• $\dot\theta_{1}$ = Initial angular velocity rad/sec

• $\dot\theta_{f}$ = Final angular velocity rad/sec.

• $\alpha$ =Angular acceleration rad/sec²Δθ =Angular displacement rad

Now,

Torque remains constant since, the angular retardation.

In the first era of motion

• The initial angular velocity is ω
• Final angular velocity ω/2
• Angular displacement 2πn rad

Using above given equation of motion, we have to put the values and we get the following results

-ω/2² = ω2 + $2\alpha$ (2πn)

This gives $2\alpha$ = −3ω² / 16πn

Second era of motion

• Final velocity is zero
• Initial velocity ω/2
• Say undergoes $n_{1}$ number of rotations before it comes to rest.

Once again using equation of motion

0=ω/2² + 2(−3ω216πn(2πn1)) $0 = \frac{ω}{2}^{2} + 2( - \frac{3ω ^{2} }{16\pi \:n} (2\pi \: n_{1}))$

This give, $n_{1}$ = $\frac{n}{3}$