# Questions on Rotational Kinematics/Dynamics

• fterh
In summary, the conversation discusses questions related to physics concepts and practice problems. The main topics include tension in a rope, moments, angular momentum, and rotational dynamics. The participants also discuss the use of friction in solving problems and finding the point where the condition for rolling without slipping is met.
fterh
Some background information: I'm doing some reading up for PhO (so it's beyond what I'm supposed to learn), and so I'll post all my questions here (regarding both concepts and actual practice questions). Sorry if you feel there is a lack of effort on my part, but sometimes I'm really lost and don't know where to start.

Questions:

1. http://imgur.com/LEmOT

Since it's a single length of rope, shouldn't tension be constant throughout the rope? That's my only question (at the moment, that is :D).

fterh said:
Since it's a single length of rope, shouldn't tension be constant throughout the rope?
No. That would be true if the rope were massless (which it presumably is) and the pulleys were massless (which they are not) and frictionless.

Hi, then can you explain this: T2 * r - T1 * r = 0.5 * m * r^2 * (a/r)

I'm guessing that it means clockwise moment - anticlockwise moment = something?

But I'm not too sure.

fterh said:
Hi, then can you explain this: T2 * r - T1 * r = 0.5 * m * r^2 * (a/r)

I'm guessing that it means clockwise moment - anticlockwise moment = something?
It's applying Newton's 2nd law for rotation: The net torque = I*alpha.

Doc Al said:
It's applying Newton's 2nd law for rotation: The net torque = I*alpha.

Alright thanks a lot.

Back to the question on the pulley. The pulley with a mass if fixed, so the mass shouldn't affect the tension? The movable pulley is massless.

fterh said:
The pulley with a mass if fixed, so the mass shouldn't affect the tension?
It has mass and thus rotational inertia, so the tension in the string must be different on each side.
The movable pulley is massless.
For that pulley the tension is the same on either side.

Okay I get it, thanks! :D

Next question (2): A father of mass $m_f$ and his daughter of mass $m_d$ sit on opposite ends of a seesaw at equal distances from the pivot at the centre. The seesaw is modeled as a rigid rod of mass $M$ and length $d$ and is pivoted without friction. At a given moment, the system rotates in a vertical plane with an angular velocity $\omega$. Obtain an expression for

a) the magnitude of the system's angular momentum

Modelling the father and daughter as particles, the moment of inertia of the system is
$I = \frac{1}{12}Md^2 + \frac{1}{4}m_fd^2 + \frac{1}{4}m_dd^2 = \frac{1}{12}d^2(M + 3m_f + 3m_d)$

I get that a rod has a moment of inertia of (Md^2)/12, but how is the M.I. of the father and daughter derived? Manual derivation using calculus?

Never mind I got it :D

For anybody who needs the answer, it's because $I = r^2 \times dm = (\frac{d}{2})^2 \times m_{f} = \frac{1}{4}m_fd^2$,

fterh said:
Never mind I got it :D

Doc Al said:

It's okay, I feel great being able to solve a question completely out of my syllabus! :D

Okay, but now I have another question. :X

A uniform solid disk of radius R and mass M is set into rotation with an angular speed $\omega_i$ about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and immediately released as shown.

What is the angular speed of the disk once pure rolling takes place?

My solution:

$K_{rotational initial} = K_{rotational final} + K_{linear final}$
$(\frac{1}{2})(\frac{1}{2}MR^2)(\omega_i)^2 = (\frac{1}{2})(\frac{1}{2}MR^2)(\omega_f)^2 + (\frac{1}{2})(M)(R\omega_f)^2$
$\frac{1}{2}\omega_i^2 = \frac{1}{2}\omega_f^2 + \omega_f^2$
$\omega_f = \sqrt{\frac{1}{3}\omega_i^2}$

But the answer is $\frac{1}{3}\omega_i$. I suspect a careless mistake somewhere, but I've been through my workings and I found no mistakes.

fterh said:
My solution:

$K_{rotational initial} = K_{rotational final} + K_{linear final}$
Do not assume that kinetic energy is conserved. There is friction involved.

Doc Al said:
Do not assume that kinetic energy is conserved. There is friction involved.

But there is no mention of frictional forces in the question, so how do I go about compensating for energy losses due to friction?

Can you give me a clue? :D

fterh said:
But there is no mention of frictional forces in the question, so how do I go about compensating for energy losses due to friction?
There must be friction, otherwise it would just sit there rotating. Just assume that the coefficient of friction is μ and see what happens.

So,

energy lost to friction = $\mu Mg \times \theta$?

Since W = F x s?

Sorry I'm wild guessing, I have no idea how to apply friction to rotational dynamics.

fterh said:
Sorry I'm wild guessing, I have no idea how to apply friction to rotational dynamics.
Sure you do. It's a force, like any other. Apply Newton's 2nd law for translation and for rotation.

Hint: The rotational speed is decreasing while the translational speed is increasing. Find the point where the condition for rolling without slipping is met.

Doc Al said:
Sure you do. It's a force, like any other. Apply Newton's 2nd law for translation and for rotation.

Hint: The rotational speed is decreasing while the translational speed is increasing. Find the point where the condition for rolling without slipping is met.

Okay... so...

$\tau = I\alpha$
$\mu MgR = \frac{1}{2}MR^2\alpha$
$\mu g = \frac{1}{2}R\alpha$
$R\alpha = 2\mu g$

$F = Ma$
$F = MR\alpha$
$F = M2\mu g$

But since F is provided for by the friction,

$\mu Mg = M2\mu g$
$1 = 2$

HUH?

Condition for pure rolling is when $\omega = v$. I could probably sub $\alpha = \frac{d\omega}{dt}$ and $a = \frac{dv}{dt}$, but why did I get 1 = 2? Where did I go wrong? :/

fterh said:
Okay... so...

$\tau = I\alpha$
$\mu MgR = \frac{1}{2}MR^2\alpha$
$\mu g = \frac{1}{2}R\alpha$
$R\alpha = 2\mu g$
Now write ω as a function of time.

Then write v as a function of time.

Solve for when the condition for rolling without slipping is met. (You can't assume it's met for all times; that was your error.)

Doc Al said:
Now write ω as a function of time.

Then write v as a function of time.

Solve for when the condition for rolling without slipping is met. (You can't assume it's met for all times; that was your error.)

So my formula $\tau = \mu MgR$ is correct?

$\omega_f = \omega_i + \alpha t$, $v = at$

When the disk is rolling without sleeping, i.e. when $\omega_f = v$,

$\omega_i + \alpha t = at$
$\omega_i + \frac{2\mu g}{R}t = \frac{\mu Mg}{M}t$
$\omega_i = \frac{\mu Mg}{M}t - \frac{2\mu g}{R}t = \mu gt (1 - \frac{2}{R})$

So from here, do I find an expression for t, then apply to $\omega_f = \omega_i + \alpha t$?

fterh said:
So my formula $\tau = \mu MgR$ is correct?
Yes.

$\omega_f = \omega_i + \alpha t$, $v = at$
Yes, but don't forget that alpha is negative, since the rotational speed is decreasing.

When the disk is rolling without sleeping, i.e. when $\omega_f = v$,
Almost: $R\omega_f = v$

fterh said:
So my formula $\tau = \mu MgR$ is correct?

$\omega_f = \omega_i + \alpha t$, $v = at$

When the disk is rolling without sleeping, i.e. when $R\omega_f = v$,

$R\omega_i - R\alpha t = at$
$R\omega_i - R\frac{2\mu g}{R}t = \frac{\mu Mg}{M}t$
$R\omega_i - 2\mu gt = \mu gt$
$R\omega_i = 3\mu gt$
$t = \frac{R\omega_i}{3\mu g}$

$\omega_f = \omega_i - \alpha t$
$\omega_f = \omega_i - \frac{2\mu g}{R} \times \frac{R\omega_i}{3\mu g}$
$\omega_f = \omega_i - \frac{2}{3}\omega_i$
$\omega_f = \frac{1}{3}\omega_i$

Hardly a eureka moment, but still :D

Thanks a lot!

Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

$R\omega_i = 3\mu gt$
$\omega_i = \frac{3\mu gt}{R}$
$\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t$
$\omega_f = \frac{\mu gt}{R}$

It's not what the question wants, but is it correct as well?

fterh said:
Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

$R\omega_i = 3\mu gt$
$\omega_i = \frac{3\mu gt}{R}$
$\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t$
$\omega_f = \frac{\mu gt}{R}$

It's not what the question wants, but is it correct as well?

Oh, and in addition to the above 2 questions, part c of the question is:

Assume the coefficient of friction between the disc and surface is $\mu$. What is the time interval after setting the disk down before pure rolling motion begins?

So based on this, I'm assuming that that I don't need to find t in part a? Could it be a case of "over-doing", when you do too much and too complicated for a relatively simple question?

fterh said:
Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

$R\omega_i = 3\mu gt$
$\omega_i = \frac{3\mu gt}{R}$
$\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t$
$\omega_f = \frac{\mu gt}{R}$

It's not what the question wants, but is it correct as well?
I don't know what problem you are trying to solve here, so I can't comment on whether it's correct or not.

Doc Al said:
I don't know what problem you are trying to solve here, so I can't comment on whether it's correct or not.

I meant is it a valid derivation? Does it hold true?

fterh said:
Oh, and in addition to the above 2 questions, part c of the question is:

Assume the coefficient of friction between the disc and surface is $\mu$. What is the time interval after setting the disk down before pure rolling motion begins?

So based on this, I'm assuming that that I don't need to find t in part a? Could it be a case of "over-doing", when you do too much and too complicated for a relatively simple question?
In solving part a you wrote an expression for the speed as a function of time. You solved for the final speed, so now you can just plug in that speed and solve for the time. (In solving for the final speed you didn't need to solve for the time explicitly, but for part c you do.)

fterh said:
I meant is it a valid derivation? Does it hold true?
A derivation of what? What problem does that equation represent?

Doc Al said:
In solving part a you wrote an expression for the speed as a function of time. You solved for the final speed, so now you can just plug in that speed and solve for the time. (In solving for the final speed you didn't need to solve for the time explicitly, but for part c you do.)
But for part a, in solving for the final circular speed, I needed to substitute $t = \frac{R\omega_i}{3\mu g}$ into $\omega_f = \omega_i - \alpha t$, which would give me the answer for part c already.

Doc Al said:
A derivation of what? What problem does that equation represent?
I meant is this equation $\omega_f = \frac{\mu gt}{R}$ true?

--------------------------------------------------

From Wikipedia, $L = I\omega$ describes a body's angular momentum. $L = r \times p$ on the other hand describes a particle's angular momentum w.r.t. some point of origin.

Applying to this question (http://books.google.com/books?id=oSQNvb5WzJgC&printsec=frontcover#v=onepage&q&f=false page 59 problem 26), how come, in the solution, these two are equated?

i.e. $\frac{I(u+v)}{r} = P(h-r)$. Since they are two different things how come they can be equated to each other?

fterh said:
But for part a, in solving for the final circular speed, I needed to substitute $t = \frac{R\omega_i}{3\mu g}$ into $\omega_f = \omega_i - \alpha t$, which would give me the answer for part c already.
Right.

I meant is this equation $\omega_f = \frac{\mu gt}{R}$ true?
Yes, that's true.

fterh said:
--------------------------------------------------

From Wikipedia, $L = I\omega$ describes a body's angular momentum. $L = r \times p$ on the other hand describes a particle's angular momentum w.r.t. some point of origin.

Applying to this question (http://books.google.com/books?id=oSQNvb5WzJgC&printsec=frontcover#v=onepage&q&f=false page 59 problem 26), how come, in the solution, these two are equated?

i.e. $\frac{I(u+v)}{r} = P(h-r)$. Since they are two different things how come they can be equated to each other?
They are applying the impulse = change in momentum principle. Realize that the force exerted on the ball exerts both a linear impulse and and angular impulse--it's still the same force.

Doc Al said:
They are applying the impulse = change in momentum principle. Realize that the force exerted on the ball exerts both a linear impulse and and angular impulse--it's still the same force.

I think I get it, thanks.

Another thing though. Since the force exerted by the cushion on the ball is not tangential, initially, I felt that the tangential component of the impulse needed to be determined (then multiplied by r for $\delta L$.

But then after thinking about it, I realized that since the impulse we are concerned about in the question is only the horizontal component, thus the P in the solution refers only to the horizontal impulse, and the vertical component of the impulse can be disregarded. Thus there is no need to resolve for the tangential component of the horizontal impulse (which wouldn't make sense).

Is my explanation correct? I'm still kind of unsure >< Can you explain why there isn't a need to resolve for tangential impulse?

dude keep it short and simple...
Force is the rate of change of MOMENTUM
Torque is the rate of change of ANGULAR MOMENTUM

let friction force which acts to casue pure rolling be F
F acts to accelerate the Disc to a velocity say V
F.R.t=change in angualr momentum= -{Iw(f)-Iw(i)} (negatuve sign as the torque acts to retard the angualr velocity)...i

sine F acts to accelerate the disc it implies F.t=Mv...ii
for pure rolling V=Rw...iii
using i ,ii and iii we have the w(f)=w(i)/3...

fterh said:
I think I get it, thanks.

Another thing though. Since the force exerted by the cushion on the ball is not tangential, initially, I felt that the tangential component of the impulse needed to be determined (then multiplied by r for $\delta L$.

But then after thinking about it, I realized that since the impulse we are concerned about in the question is only the horizontal component, thus the P in the solution refers only to the horizontal impulse, and the vertical component of the impulse can be disregarded. Thus there is no need to resolve for the tangential component of the horizontal impulse (which wouldn't make sense).

Is my explanation correct? I'm still kind of unsure >< Can you explain why there isn't a need to resolve for tangential impulse?
Realize that Torque = r X F = r*F*sinθ. Whether you take the force (F) times the perpendicular distance to the axis (r*sinθ), which is what they did here, or you take the tangential component of the force (F*sinθ) times the radius (r), you end up with the same resulting torque.

Okay thanks, I get it! :D

A couple more questions:

A thin rectangular plate of length a, width b and mass M has a moment of inertia $I = \frac{1}{12}M(a^2 + b^2)$ about an axis through its center and perpendicular to its plane. What is the moment of inertia of the plate about an axis in the plane of the plate and forming a diagonal of the rectangle? [Hint: $I_x = \frac{1}{12}Ma^2$ and $I_y = \frac{1}{12}Mb^2$]

Do I use perpendicular axis theorem then apply sort of like pythagoras theorem?

------------------------------

A small elastic ball (radius r) is projected with velocity $v_1$ and backspin $\omega_1$ at 45 deg to the horizontal. The ball is observed to repeatedly bounce back and forth between P and Q and always rising at an angle of 45 deg from the rough floor. Assuming perfectly elastic impact, determine

a) the backspin required, in terms of r and $v_1$, and
b) the minimum coefficient of friction between the ball and the floor.

Okay, I'm totally stumped on this question. All I know is that "perfectly elastic impact" means that speed of approach/"hit" equals speed of separation/"bounce". In this case conservation of energy cannot be used, because energy is lost to friction. Conservation of momentum and angular momentum cannot be used too, due to friction. So I don't know what can be used to solve this. ><

Last edited:
fterh said:
Okay thanks, I get it! :D

A couple more questions:

A thin rectangular plate of length a, width b and mass M has a moment of inertia $I = \frac{1}{12}M(a^2 + b^2)$ about an axis through its center and perpendicular to its plane. What is the moment of inertia of the plate about an axis in the plane of the plate and forming a diagonal of the rectangle? [Hint: $I_x = \frac{1}{12}Ma^2$ and $I_y = \frac{1}{12}Mb^2$]

Do I use perpendicular axis theorem then apply sort of like pythagoras theorem?

------------------------------

A small elastic ball (radius r) is projected with velocity $v_1$ and backspin $\omega_1$ at 45 deg to the horizontal. The ball is observed to repeatedly bounce back and forth between P and Q and always rising at an angle of 45 deg from the rough floor. Assuming perfectly elastic impact, determine

a) the backspin required, in terms of r and $v_1$, and
b) the minimum coefficient of friction between the ball and the floor.

Okay, I'm totally stumped on this question. All I know is that "perfectly elastic impact" means that speed of approach/"hit" equals speed of separation/"bounce". In this case conservation of energy cannot be used, because energy is lost to friction. Conservation of momentum and angular momentum cannot be used too, due to friction. So I don't know what can be used to solve this. ><

Bump... :)

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