angular momentum

[Mango12]

Part 1: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 2: 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 honestly have no idea how to start solving this.

 

[Deleted member 4993]

Part 1: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 2: 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 honestly have no idea how to start solving this.

The rotational energy = 1/2 * I * w²

Does that ring a bell^?

If it does not - you need more help than we can provide.

 

[Mango12]

The rotational energy = 1/2 * I * w²

Does that ring a bell^?

If it does not - you need more help than we can provide.

Yes, it does actually! Thank you. So..with that info i successfully solved for part 1, but I am still confused on what part 2 is asking me to do and how I can solve it.