# Questions on Rotational Kinematics/Dynamics

#### Doc Al

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

#### fterh

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.

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

#### fterh

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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?

#### Doc Al

Mentor
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.

#### Doc Al

Mentor
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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.

#### fterh

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?

#### mj345

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..........

#### Doc Al

Mentor
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.

#### fterh

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?

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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. ><

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#### fterh

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?

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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... :)

#### Doc Al

Mentor
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?
Yes, you'll need the perpendicular axis theorem. To make use of the hint, try to express the moment of inertia about a diagonal in terms of Ix and Iy. (Consider how you'd have to rotate the x and y axes to align with the diagonal. It's a bit tricky.)

#### Doc Al

Mentor
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. ><
No energy is lost to friction--conservation of energy can be used. Consider the impulse given by the floor and how that effects the angular and linear motion of the ball.

#### fterh

No energy is lost to friction--conservation of energy can be used. Consider the impulse given by the floor and how that effects the angular and linear motion of the ball.
But doesn't the fact that after some time the ball will stop bouncing and spinning mean that energy is continually being converted into other forms due to friction?

#### Doc Al

Mentor
But doesn't the fact that after some time the ball will stop bouncing and spinning mean that energy is continually being converted into other forms due to friction?
In a real collision between ball and floor, sure, the ball will eventually stop bouncing. But this is meant as an idealized problem and the 'perfectly elastic impact' means energy is conserved.

#### fterh

In a real collision between ball and floor, sure, the ball will eventually stop bouncing. But this is meant as an idealized problem and the 'perfectly elastic impact' means energy is conserved.
Okay. I've given some thought to this question and I have a question.

If the backspin causes the ball to bounce back towards the left after landing at Q, won't it continue moving to the left rather than bounce back to the right towards Q (after landing at P)?

#### fterh

I need some help manipulating equations.

$m_1 a = m_1 g sin\alpha - \frac{aI}{r^2} - F$
$m_2 a = m_2 g sin\alpha - \mu m_2 g cos\alpha + F$

From the above 2, I am able to get:

$a = g \cdot \frac{(m_1 + m_2) sin\alpha - \mu m_2 cos\alpha}{m_1 + m_2 + \frac{I}{r^2}} ...(1)$

But how do I find an expression for F that doesn't include $a$?

#### fterh

Okay. I've given some thought to this question and I have a question.

If the backspin causes the ball to bounce back towards the left after landing at Q, won't it continue moving to the left rather than bounce back to the right towards Q (after landing at P)?
I need some help manipulating equations.

$m_1 a = m_1 g sin\alpha - \frac{aI}{r^2} - F$
$m_2 a = m_2 g sin\alpha - \mu m_2 g cos\alpha + F$

From the above 2, I am able to get:

$a = g \cdot \frac{(m_1 + m_2) sin\alpha - \mu m_2 cos\alpha}{m_1 + m_2 + \frac{I}{r^2}} ...(1)$

But how do I find an expression for F that doesn't include $a$?
Bump for help! :)

#### Doc Al

Mentor
I need some help manipulating equations.

$m_1 a = m_1 g sin\alpha - \frac{aI}{r^2} - F$
$m_2 a = m_2 g sin\alpha - \mu m_2 g cos\alpha + F$

From the above 2, I am able to get:

$a = g \cdot \frac{(m_1 + m_2) sin\alpha - \mu m_2 cos\alpha}{m_1 + m_2 + \frac{I}{r^2}} ...(1)$

But how do I find an expression for F that doesn't include $a$?
What problem are you trying to solve here?

If it's the bouncing ball, consider using conservation laws and the fact that the collision is elastic.

#### fterh

What problem are you trying to solve here?

If it's the bouncing ball, consider using conservation laws and the fact that the collision is elastic.
It's actually a problem from IPhO 1968.