Spinning disk

The disk initially has an angular velocity, $\omega_i$ . The initial angular momentum about its central axis is, $I\omega_i$ , where $I=\frac{1}{2}MR^2$ , is the disk’s moment of inertia about its central axis.

When it is placed on the table, it only experiences one force, friction $f$ . The friction $f$ has two effects: first, it causes an linear acceleration, which increases the disk’s linear velocity from $0$ to $v_f$ ; second, it causes a torque about the central axis of the disk, which slows down its angular velocity from $\omega_i$ to $\omega_f$ .

Of course, we can use the standard equation of $F=ma$ to solve the problem. A simpler and more direct way is to use the relation between impulse and linear momentum and the relation between torque and angular momentum.

During the slipping time, the friction causes an impulse, which equals the change in linear momentum:

$\int f \mathrm {d}t=Mv_f-0$

During the same amount of time, the friction causes a torque, the integral of which decreases the angular momentum:

$\int -fR \mathrm {d} t=R\int f \mathrm {d}t=I(\omega_f-\omega_i)$

Given the equations above as well as the non-slipping condition $v= R\omega$ and thus $v_f=R\omega_f$ , we get:

$-Rmv_f=-RM(R\omega_f)=I(\omega_f-\omega_i)=\frac{1}{2}MR^2(\omega_f-\omega_i)$

$\implies -\omega_f=\frac{1}{2}(\omega_f-\omega_i)$

$\implies \frac{\omega_f}{\omega_i}=\frac{1}{3}$

We consider the change in linear and angular momentum from the point the disk is placed on the table to the point the disk begins to roll without slipping. When the disk rolls without slipping, it should have a linear velocity of $v_f = R \cdot ω_f$ . The moment of inertia of the disk is $\frac{1}{2}MR^2$ . We also let the change in linear momentum be $Δp$ .

Change in linear momentum = Final linear momentum - Initial linear momentum $\Rightarrow Δp = Mv_f - 0 = Mv_f = MRω_f$

Change in angular momentum = Final angular momentum - Initial angular momentum $\Rightarrow -R \cdot Δp = \frac{1}{2}MR^2ω_f - \frac{1}{2}MR^2ω_i = \frac{1}{2}MR^2(ω_f-ω_i)$

Substituting the first equation into the second, we have $-MR^2ω_f = \frac{1}{2}MR^2(ω_f-ω_i)$ . So $-ω_f = \frac{1}{2}ω_f-\frac{1}{2}ω_i \Rightarrow \frac{1}{2}ω_i = \frac{3}{2}ω_f$ . Therefore, $\frac{ω_f}{ω_i} = \frac{1}{3} = 0.333$ .

Let the angular velocity of the disk be $\omega_{i}$ when it just touches the table. Let the acceleration at the centre of mass be $a$ , the angular acceleration be $\alpha$ . Let the frictional force acting on the disk when it is slipping be $F$ .

By Newton's Law of Motion, $ma=F$ and $I\alpha=FR$ . Hence, $a=\frac{F}{m}$ and $\alpha=\frac{FR}{I}$

By kinematics, and given initial linear velocity is 0, $v=v_{i}+at=at=\frac{F}{m}t$ and $\omega=\omega_{i}+\alpha t=\omega_{i}+\frac{FR}{I}t$

When its motion changes to rolling without slipping, $-\omega R=v$ has to hold.

Hence, when it is rolling without slipping, $-\frac{F}{m}t=(\omega_{i}+\frac{FR}{I}t)R$ . Since it is a solid disk, $I=\frac{1}{2}mR^2$ . Then substitute this relation into the equation and solve the equation for t. $t=-\frac{\omega_{i}Rm}{3F}$

Substitute the equation into $\omega=\omega_{i}+\frac{FR}{I}t$ , we have $\omega=\omega_{i}+\frac{FR}{I} (-\frac{\omega_i Rm}{3F})$ .

Simplify the equation, we have $\omega=\frac {1}{3} \omega_{i}$ . Hence, the ratio is 0.333 (to 3 significant figures).

Initially, the disc has only angular velocity $\omega_i$ . For pure rolling motion to start, $\omega_f.R = v$ which is brought about by $frictional force, f$ . This implies the friction force will be such that:

1. angular acceleration, $\alpha$ produced by it opposes $\omega_i$ .
2. it produces acceleration, $a$ in forward direction.

Therefore, $v$ = $0 + a.\tau$ $,$ where $a$ = $\frac {f}{M}$

$\omega_f = \omega_i - \alpha.\tau$ $;$

$f.R = I_cm.\alpha$ where $I_cm = \frac {M}{2R^2}$ $\Rightarrow$ $\alpha = \frac {2f}{MR}$

Since $\omega_f.R = v$ $\omega_i.R - \alpha.R.\tau = a.\tau$

$\Rightarrow$ $\omega_i = \frac {f.\tau}{MR} + \frac {2f.\tau}{MR}$ Therefore, $\omega_i = \frac {3f.\tau}{MR}$ and $\omega_f = \frac {f.\tau}{MR}$ .

Hence, the answer $\frac {\omega_f}{\omega_i} = \frac {1}{3} = \boxed {0.33}$

initially angular momentum of the body about a point on the ground was L=mr^2 (initial angular velocity) ....(1) Since friction acts along the ground it produces no torque So, conserving angular momentum about the point Final angular momentum= 3mr^2 (final angular velocity) ....(2) Equating eqn. (1) and (2) Required ratio=1/3=.333

Let the center of mass (which is also the geometric center) of the disc be O. Choose the table surface/earth as the frame of reference for the linear motion of O, and O as the center of rotation. Assume the disk in initially rotating clockwise.

The velocity of O as a function of t is v(t); the angular speed around O is W(t). Let the time pure rotation without slipping starts be t=T. v(0)=0 v(T)=W(T) R (1)

While sliding, the net external force on the disk is a constant friction f. Because at the point of contact P with the table, the relative motion of the disk against the table is to the left (clockwise), f points to the right, which is the direction of the center of mass O's motion. Newton II yield: a=f/M (a constant), v(t)=at = ft/M (2)

Let Tq be the net torque regarding O; Tq= fR, counterclockwise, consequently reducing the angular speed of the disk. Let I be the Rotational moment of inertia of the solid disk; I=(1/2)MR^2
Angular accleration alpha(t) = Tq/I = 2f/(MR) is a constant. Hence W(t)= W(0) - 2ft/(MR) (3)

Use eqautions (1), (2), and (3) at t=T, W(T) = W(0)-2fT/(MR) = V(T)/R = fT/(MR) Simplify into 3fT/(MR)=W(0), namely 3W(T)=W(0). Hence the proportion W(T)/W(0) = 1/3

Note that this proportion varies with the moment of inertia I of the object. For a hollow disk (a circle), I=MR^2, W(T)/W(0) = 1/2. For a solid sphere, I = (2/5)MR^2, the answer would be 2/7. For I=kMR^2, where k is a positive real coefficient describing the distribution of mass of the object, W(T)/W(0) = 1/(1/k +1).

since -FR = Iα, where I denotes the moment of inertia of the spinning disk and α denotes its angular acceleration

α = (ωf - ωi) / t

F = m(vf - vi) / t

vf = ωf*R

as there was initially no linear momentum,

vi = 0

from the above equations we can obtain,

(- (mωf*R^2)/t ) = 1/2 * mR^2 (ωf - ωi)/t

-2 ωf = ωf - ωi

3ωf = ωi

ωf/ωi = 1/3

Initial Angular momentum = I (w i) Final angular momentum = I (w f) + m(w_f)r^2 where I = (1/2)mr^2 (moment of inertia)

Equating them we get (1/2)w i = (1/2)w f + w_f

or (w f)/(w i) = 1/3 = 0.333

I am sorry that I cannot add images.I will try to add them in comments as soon as I can. Since Lokesh Sharma went to the extent to making a 'note' to know the solution to this question, I am going to give a detailed answer.

In physics problems, it is always important to be clear about conventions and much more important is to stick to them. Assume that the disk is spinning in the clockwise direction. I have chosen clockwise direction (for rotational part) as positive and right direction (for linear part) as positive.

LINEAR MOTION

We know that $f=\mu N$ .

Balancing forces in the y-direction:

$N=Mg \Rightarrow f=\mu Mg$ .

Initially the disk has no linear momentum $\Rightarrow\ u$ i.e. initial velovity $=0$ . Since the only linear force acting on the disk is the frictional force, we have

$f=Ma \Rightarrow \mu Mg=Ma \Rightarrow a=\mu g$ .

(Once you see the diagram, it will become more clear as to why frictional force causes the disk to accelerate and not decelerate).

Let the final velocity of the disk be $v$ .Since the disk would be in pure rolling motion, we can say that $v=R\omega_{f}$ Using laws of motion, we have

$v-u=at \Rightarrow v-0=\mu gt \Rightarrow v=\mu gt \Rightarrow R\omega_{f}=\mu gt$ ....................(1)

ROTATIONAL MOTION

We know that $\tau= R \times F\Rightarrow \tau=-Rf$ (Remember the sign conventions)

$\tau=-Rf\Rightarrow \tau=-R\mu Mg$ ..............(2)

Also, $\tau=I\alpha\Rightarrow \tau=\frac{MR^{2}\alpha}{2}$ ............(3)

(2)=(3) gives us

$-R\mu Mg=\frac{MR^{2}\alpha}{2}\Rightarrow \alpha=-\frac{2\mu g}{R}$

Using simple law of motion,

$\omega_{f}-\omega_{i}=\alpha t$

$\Rightarrow \omega_{f}-\omega_{i}=-\frac{2\mu g}{R}t$ ..............(4)

Dividing Equation (4) by (1) gives

$\frac{\omega_{f}-\omega_{i}}{R\omega_{f}}=-\frac{2\mu gt}{R \mu gt}$

$\Rightarrow 1-\frac{\omega_{i}}{\omega_{f}}=-2$

$\Rightarrow \frac{\omega_{f}}{\omega_{i}}=\frac{1}{3}=\boxed{0.334}$

The disk initially starts out with rotational velocity $v_{R,i} = \omega_i R$ . Due to the frictional force, there will be a loss in rotational velocity and an increase in translation velocity. Therefore, the final rotational velocity will be of the form $v_{R,f} = v_{R,i} - a_R t$ , and the final translational velocity will be $v_T = a_T t$ . Sinc the criterion for rolling without slipping requires the translational and rotational velocities to be equal, $v_{R,i} - a_R t = a_T t$ .

Finding $a_T$ is simple (it's just $\mu g$ ), which makes $v_T$ just $\mu g t$ . To find $a_R$ , we set up an equation that expresses the net torque:

$\tau = (\mu M g) (R) = I \alpha$

$\mu MgR = \left(\frac{1}{2} MR^2 \right) \frac{a_R}{R}$

$2 \mu g = a_R$

Therefore, $v_{R,i} = a_R t + a_T t = 3 \mu g t$ , so $\omega_f / \omega_i = v_T / v_{R,i} = \boxed{\frac{1}{3}}$ .

The torque produced by the friction is written as $T = I \alpha = \mu M g R$ , being $\mu$ the friction coefficient and $\alpha$ the angular acceleration. Therefore $\alpha = \frac{\mu M g R}{I}$ . But the friction is what is causing the center of mass to change its motion. Then we also know that $f_{a} = \mu M g = M a_{CM} \rightarrow a_{CM} = \mu g$ . When the disc stops slipping, it is true that $v_{CM} = \omega_{CM} R = \mu g t$ is its speed, where $\omega_{CM}$ is its final angular velocity and $t$ is the time the disc took to stop slipping. Using this last equation, we have $(\omega_{i} - \alpha t) R = \mu g t \rightarrow t = \frac{\omega_{i} R}{ \alpha R + \mu g}$ Now, this equation together with the equation for the final speed of the not-anymore-slipping-disc give us $\omega_{f} = \frac{v_{f}}{R} \rightarrow \frac{\omega_{f}}{\omega_{i}} = \frac{\mu g \omega_{i}}{\alpha R + \mu g} = \frac{1}{\frac{\alpha R}{\mu g} + 1}$ But remember that $\alpha = \frac{\mu M g R}{I}$ and $I$ is the moment of inertia of the disc which is $I = \frac{1}{2} M R^{2}$ Therefore $\alpha = \frac{2 \mu g}{R}$ and going back to the quotient of angular velocities we have $\frac{\omega_{f}}{\omega_{i}} = \frac{1}{3}$

When doing problems like this, always think if you can apply the laws of angular momentum and energy conservation or not. Angular momentum is conserved about some point only when there is no net torque about that point. Energy will not be conserved because a friction, a dissipative force acts on the disk as soon as it touches the ground and starts rolling with slipping. After it starts rolling without slipping, friction is no longer dissipative.

Draw the free body diagram of the disk when it is on the ground. Friction should act at it's bottommost point forwards.(Since disk is slipping, it will oppose the disc's rotational motion). The weight mg and Normal force N will also act through the centre of the disk.

Is there any point about which torque is zero? If we take torque about the center of the disk, the torques due to $mg$ and $N$ will be zero because they pass through the centre. But there will be a torque due to friction, equal to $fR$ .

Now consider the bottommost point. There will be no torque here due to either $mg$ , $N$ or friction, because they all pass through this point. Angular momentum will be conserved about this point.

The initial angular momentum of the disk will be $Iw_{i}$ , where I is the moment of inertia about the disk abt the center of mass. Yes, we take moment of inertia about the center of mass, even though angular momentum is conserved about the bottom point. That's because it comes out that way in the derivation of angular momentum of a rotating body.

The final angular momentum will be $Iw_{f}+mvR$ . The disk starts moving forward with a linear velocity, as soon as it touches the ground. Then friction decreases angular velocity and increases linear velocity until the relation $v=wR$ is reached, the one required for pure rolling.

So, $Iw_{i}=Iw_{f}+mvR$

Putting $I=\frac{mR^{2}}{2}$ and $v=wR$ , we get $\frac{w_{f}}{w_{i}}=\frac{1}{3}$