MHB Angular momentum and a basketball

[Mango12]

Putting spin on the ball improves it’s stability (the spin imparts angular momentum and it takes an outside torque to change angular momentum), but at a cost. As you know, for a given input of energy, spin consumes some of that energy and leaves less for translation.

Part A: So, How much more energy must a player expend in order to make the same shot, but impart a reasonable spin of 3.00 revolutions per second? A standard basketball has a mass of 0.500 kg and a diameter of 25.4 cm. Your answer should be approximately 1.00J of energy.

Part B: What percentage is that of the kinetic energy of the original shot taken without spin?And, does it seem that controlling the amount of spin could make a big enough difference to miss or make a shot?

~I have no idea how to start this. Can anyone help?

 

[I like Serena]

Putting spin on the ball improves it’s stability (the spin imparts angular momentum and it takes an outside torque to change angular momentum), but at a cost. As you know, for a given input of energy, spin consumes some of that energy and leaves less for translation.

Part A: So, How much more energy must a player expend in order to make the same shot, but impart a reasonable spin of 3.00 revolutions per second? A standard basketball has a mass of 0.500 kg and a diameter of 25.4 cm. Your answer should be approximately 1.00J of energy.

Hi Mango12! (Smile)

Starting with A, the rotational energy involved is given by $\frac 12 I \omega^2$, where $I$ is the moment of inertia of a basketball of mass $m$ and $\omega$ is the angular velocity corresponding to the required spin.
Do you perchance have a formula for that moment of inertia?
And can you calculate the required angular velocity?

 

[Mango12]

Hi Mango12! (Smile)

Starting with A, the rotational energy involved is given by $\frac 12 I \omega^2$, where $I$ is the moment of inertia of a solid ball of mass $m$ and $\omega$ is the angular velocity corresponding to the required spin.
Do you perchance have a formula for that moment of inertia?
And can you calculate the required angular velocity?

moment of inertia for a solid sphere is 2/5MR^2, which equals 32.3

the formula for angular velocity is w=v/r. I think I have to do something with radians, but I'm not sure.

 

[I like Serena]

moment of inertia for a solid sphere is 2/5MR^2, which equals 32.3

the formula for angular velocity is w=v/r. I think I have to do something with radians, but I'm not sure.

Actually, I have just realized that a basketball is not a solid sphere but a hollow sphere.
The moment of inertia for a hollow sphere is $\frac 23 m r^2$.

And with 3 revolutions per second, we have an angular velocity of $\omega = 3 \times 2\pi \text{ rad/s}$, since we have $2\pi \text{ rad}$ in one revolution.

 

[Mango12]

Actually, I have just realized that a basketball is not a solid sphere but a hollow sphere.
The moment of inertia for a hollow sphere is $\frac 23 m r^2$.

And with 3 revolutions per second, we have an angular velocity of $\omega = 3 \times 2\pi \text{ rad/s}$, since we have $2\pi \text{ rad}$ in one revolution.

okay, so when I tried to find rotational energy I did 1/2(Iw^2) but I don't know if I did it right...

1/2(2/3)mr^(2w^2)
1/2 (53.8)w^2
1/2(53.8)(18.8)
=505.7

 

[I like Serena]

okay, so when I tried to find rotational energy I did 1/2(Iw^2) but I don't know if I did it right...

1/2(2/3)mr^(2w^2)
1/2 (53.8)w^2
1/2(53.8)(18.8)
=505.7

That doesn't look quite right...

Let's redo that:
$$
\frac 12 I\omega^2 = \frac 12 \left(\frac 23 m r^2\right)\omega^2 = \frac 12 \left(\frac 23 \cdot 0.5 \cdot \left(\frac 12 \cdot 0.254\right)^2\right) \cdot (3\cdot 2\pi)^2
= 0.955 \text{ J}
$$

 

[Mango12]

That doesn't look quite right...

Let's redo that:
$$
\frac 12 I\omega^2 = \frac 12 \left(\frac 23 m r^2\right)\omega^2 = \frac 12 \left(\frac 23 \cdot 0.5 \cdot \left(\frac 12 \cdot 0.254\right)^2\right) \cdot (3\cdot 2\pi)^2
= 0.955 \text{ J}
$$

That number looks a lot better. But is that the final answer for part A?

 

[Mango12]

That number looks a lot better. But is that the final answer for part A?

And how do I go about solving part B? (I really appreciate all the help by the way) :)

 

[I like Serena]

That number looks a lot better. But is that the final answer for part A?

Yes.

Part B: What percentage is that of the kinetic energy of the original shot taken without spin?And, does it seem that controlling the amount of spin could make a big enough difference to miss or make a shot?

And how do I go about solving part B? (I really appreciate all the help by the way) :)

There seems to be some information missing here.
Can you find anything about the kinetic energy of the original shot?
Is that perhaps part of a previous question?

 

[Mango12]

Yes.

There seems to be some information missing here.
Can you find anything about the kinetic energy of the original shot?
Is that perhaps part of a previous question?

Well, when the ball is originally shot, it is traveling at 7.67m/s, so..

1/2mv^2= (.5)(.5)(7.67^2)= 14.7

 

[I like Serena]

Well, when the ball is originally shot, it is traveling at 7.67m/s, so..

1/2mv^2= (.5)(.5)(7.67^2)= 14.7

So...?

 

[Mango12]

So...?

Would I subtract the .995 and the 14.7? But wait...it says it wants to know the percentage of KE from the original shot, so maybe I should divide them?

 

[I like Serena]

Would I subtract the .995 and the 14.7? But wait...it says it wants to know the percentage of KE from the original shot, so maybe I should divide them?

Yep. Let's divide them.

 

[Mango12]

Yep. Let's divide them.

Or do I need to add .955 to the 14.7? (to get the total energy, since the .955 is for the spin and the 14.7 is for a normal shot?)

but if we are just dividing then .955/14.7=.065=6.5%

 

[I like Serena]

Or do I need to add .955 to the 14.7? (to get the total energy, since the .955 is for the spin and the 14.7 is for a normal shot?)

but if we are just dividing then .955/14.7=.065=6.5%

I'm assuming the original shot is without spin, in which case this is just fine.
So 6.5% of the shot is needed to give it the proper spin.
That doesn't sound like too much does it?
So if we can spin it without affecting our aim, that's probably worthwhile to stabilize the shot.

 

[Mango12]

I'm assuming the original shot is without spin, in which case this is just fine.
So 6.5% of the shot is needed to give it the proper spin.
That doesn't sound like too much does it?
So if we can spin it without affecting our aim, that's probably worthwhile to stabilize the shot.

Wow, I actually understand this now! I really appreciate you taking the time to go through all of this with me :)