Spinning up a DVD

We know that, if $\tau$ is the Torque, $I$ is the Moment of Inertia, and $\alpha$ is the angular acceleration, we have $\tau = I \alpha.$ This can easily be derived from Lagrangian Mechanics, just as $F = ma$ is. We know that, for a thin uniform cylinder, the moment of inertia is $I = \dfrac {1}{2} mr^2,$ where $m$ is the mass and $r$ is the radius of rotation. So, we have: $\tau = \dfrac {1}{2} mr^2 \alpha \implies \alpha = \dfrac {2 \cdot \tau}{mr^2}.$ We now substitute:

$\implies \tau = 10^{-3} \, \text{N} \cdot \text{m}$

$\implies m = 0.015 \, \text{kg}$

$\implies r = 0.06 \, \text{m}$

$\implies \alpha = \boxed {37 \, \text{rad/s}^2}$ .

Firstly, lets define some variables.

$\tau=$ the torque exerted on the DVD $=1\times10^{-3}N\times m$

$m=$ the mass of the DVD $=0.015kg$

$r=$ the radius of the DVD $=0.06m$

$I=$ the moment of inertia of the DVD about it's axis of rotation.

$\omega=$ the angular velocity of the DVD about it's axis of rotation.

$L=$ the angular momentum of the DVD about it's axis of rotation $=I\omega$ .

$\alpha=$ the angular acceleration of the DVD about it's axis of rotation $=\frac{d\omega}{dt}$ .

According to wikipedia (http://en.wikipedia.org/wiki/List of moments of inertia), $I=\frac{1}{2}mr^2=\frac{1}{2} \times 0.015 \times 0.06^2=0.000027kg \cdot m^2$

Because $\tau=\frac{dL}{dt}=\frac{d(I\omega)}{dt}$ , and that $I$ is constant as time progresses,

$\tau=I\frac{d\omega}{dt}=I\alpha$

This means that $\alpha$ , the value to be determined is:

$\alpha=\frac{\tau}{I}=\frac{1\times 10^{-3}}{0.000027}=\frac{1000}{27}=37.0rad\cdot s^{-2}$ to three significant figures.

We have $\tau = I\alpha$ , where $\tau$ is the torque, $I$ is the moment of inertia and $\alpha$ is the angular acceleration. On the other hand, $I = \frac{1}{2}mr^2$ (since the DVD can be modeled as a thin, uniform cylinder). Therefore, we have $\alpha = \frac{\tau}{I} = \frac{2\tau}{mr^2} = \frac{2 \cdot 10^{-3}}{0.015 \cdot 0.06^2} = \frac{1000}{27} \approx 37.037.$

Torque=MomentumOfInertia*AngularAcceleration, then Let the momentum of inertia be "m", torque "t" and angular acceleration " $\theta$ " => $m=1/2*0.015*(0.06)^2$ => $\theta=m/t$ => $\theta=37.037$

We know that radius of gyration,* I = 1/2 Mr^2* where M = 0.015 kg r= 0.06 m So , I = 2.7E-5 We know that Angular accelaration = torque/ I =10^-3/2.7^-5 =37.037 rad / s^2

Using $\tau = I \alpha$ and $I_{disk} = \dfrac{1}{2} m r^2$ , we obtain the equation $\dfrac{1}{2} (0.015)(0.06)^2 \alpha = 10^{-3}$ Solving for $\alpha$ , we obtain $37.037\dfrac{rad}{s^2}$ , our final answer.

Torque=moment of inertia x angular acceleration

torque=0.001

moment of inertia of a uniform cylinder=1/2(mass)(radius)^2 =0.000027

Therefore, Angular acceleration=0.001/0.000027 =37.037

$\tau = I\alpha$ $\alpha = \frac {\tau} {I}$ $I = \frac {1} {2} m r^{2}$ $\alpha = \frac {2 \tau} {m r^{2}}$ $\alpha = \frac {2 \times 10^{-3} N \cdot m} {0.015 kg (0.06 m)^{2}}$ $\alpha = 37.04 rad \cdot s^{-2}$

Given the torque you have τ = I a.Due to the fact that you have a thin uniform cylinder I = 1/2 * m * R^2,where R = 0.06m and m = 0.015 kg.So we replace these values in the torque's formula and thus we have: τ = 1/2 * 0.015kg *((0.06)^2)m^2 * a (rad/sec^2) => τ = 0.5 0.015 0.0036 a => 0.01 = 27 10^(-6) *a => 10^3/27 = a => a = 0.037 1000 =>a~= 37 rad/sec^2

We know that $\tau = I\alpha$ Since we are modeling the DVD as a thin, uniform cylinder, we know the moment of inertia and get $\tau = \frac{mr^2}{2} \alpha$ $10^{-3} = \frac{(0.015)(0.06)^2}{2} \alpha$

Solving for $\alpha$ , we get $\alpha \approx \boxed{37.037}$

Given the torque you have τ = I a.Due to the fact that you have a thin uniform cylinder I = 1/2 * m * R^2,where R = 0.06m and m = 0.015 kg.So we replace these values in the torque's formula and thus we have: τ = 1/2 * 0.015kg *((0.06)^2)m^2 * a (rad/sec^2) => τ = 0.5 0.015 0.0036 a => 0.01 = 27 10^(-6) *a => 10^3/27 = a => a = 0.037 1000 =>a~= 37 rad/sec^2"

Angular acceleration = torque/mass moment of inertia

Mass moment inertia of a cylinder I = 1/2 MR^2

Hence I = 2.7 x 10^-5

τ/I = 10^-3 / (2.7 x 10^-5) = 37.037

T = I@

T = ( 0.5mr^2 ) x @

1E-3 = 0.5 x 0.015 x 0.06 x 0.06 x @

@ = 37.0370

Here, I=mr^2/2 & Torque = Iw. So, w=37.037

MR^2%2*a=torque so a=37.037

We know τ=I∝ ∝=τ/I Where I=1/2 Mr^2 ∴ ∝=τ/(1/2 Mr^2 )=2τ/(Mr^2 ) Given M=0.015kg , r=0.06m and τ=〖10〗^(-3) N ∴ ∝=37.03

First, calculate the moment of inertia for a cylindrical body by formula; I=(mr^2)/2, using m=0.015, r=0.06 I=2.7E-5 Now, we know that torque=Ixa (where a is angular acceleration) we are given that torque is 1E-3, and we have calculated the moment of inertia, so by rearranging the equation and solving for a we have a=37.037.

The moment of inertia of a thin, uniform cylinder of mass $m$ and radius $r$ is $I = \frac{1}{2}mr^2.$ Let $\tau$ be the torque and let $\alpha$ denote the angular acceleration. Then since $\tau = I\alpha$ , we have $\alpha = \frac{10^{-3}}{(.5)(.015)(.06)^2} = 37.037$ .

For a solid cylinder, the moment of inertia, $I=\frac{mr^{2}}{2}$ Where $m$ is the mass of the rigid body And from rotational mechanics we have the torque $\tau=I\alpha$ {also called torque equation} Where $\tau$ is the torque and $\alpha$ is the angular aceelaration. So for this case, $\tau=\frac{mr^{2} \alpha}{2}$ . $\Rightarrow \alpha=\frac{2\tau}{mr^{2}}$ Plugging in the numbers, we have $\alpha=\frac{1000}{27}=\boxed{37.037}$

Since the DVD may be modeled as a thin, uniform cylinder, its moment of inertia is $mr^{2}/2 = (0.015)(0.06)^{2} / 2 = 2.7*10^{-5} kg*m^{2}$ .

Now $\tau = I\alpha$ , so $\alpha = \tau / I = 10^{-3} / (2.7*10^{-5}) = 37.037 rad/s^{2}$ .

Torque $\tau = I \times \alpha$ , $I$ being moment of inertia of DVD (thin cylinder) $I = \frac{1}{2} m r^2$ , & $\alpha$ being angular acceleration. Substitute values to get answer.

As, Torque(T)=Moment of Inertia(I).Angular Acceleration(A) ........(1)

Since, 'I' of a disc=(1/2)Mass of disc.(square of radius)

As, mass of disc=.015 kg

and radius of disc=0.06 m

Then, moment of inertia of DVD (i.e. a disc)=0.000027 kg.m^{2} And torque is given = 10^{-3} N.m

Hence, putting these values in equation (1) we get:

A=37.037 rad/{s}^2

Torque = Inertia x angular acceleration

10^-3 N m = 1/2 0.015Kg (0.06m)^2 x angular acceleration

angular acceleration = 37.037 rad/s^2

The angular accelaration is $\alpha$ , $\frac{\tau}{I}$ , the $I$ is the moment inertia of the model. In this problem, the model is DVD, that can be modeled as a thin, uniform cylinder. You can look at this , $I$ = $\frac{1}{2}mR^{2}$ , $m$ is the mass of model and $R$ is the radius.

So, you will get the angular acceleartion by finding first the moment inertia of model.

$\alpha = \frac {\tau}{I} = \frac {\tau}{\frac {m r^2}{2}}$