# 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 said:
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.)

fterh said:
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.

Doc Al said:
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?

fterh said:
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.

Doc Al said:
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)?

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 said:
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 said:
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! :)

fterh said:
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.

Doc Al said:
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.

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