Projectile, and uniform circular motion questions

[reverse]

Projectile, and uniform circular motion questions please help!

can anyone please help me.. i keep getting the wrong answers over and over.. you don't have to solve the problem or give me the results.. explanation of how to do them would really be appreciated.. thanks in advance.

1) A projectile's horizontal range on level ground is R=(V0)^2sin2(theta)/g. At what launch angle or angles will the projectile land at half of its maximum possible range.

2) A well-lubricated bicycle wheel spins a long time before stopping. Suppose a wheel initially rotating at 150 takes 62 to stop. If the angular acceleration is constant, how many revolutions does the wheel make while stopping?

3)A 300-m-tall tower is built on the equator. How much faster does a point at the top of the tower move than a point at the bottom?

4)A magnetic computer disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at 600 rad/ s^2 for 1/2 s, then coasts at a steady angular velocity for another 1/2 s . a) What is the speed of the dot at t = 1.0 s?
b)Through how many revolutions has it turned?


thanks advance for any help.

 

[Mattowander]

For the first question : Vo squared times sin 2 theta. This means that whatever angle you plug in, it will be doubled. Think about which angle will give you the maximum possible sin value (1).

 

[reverse]

For the first question : Vo squared times sin 2 theta. This means that whatever angle you plug in, it will be doubled. Think about which angle will give you the maximum possible sin value (1).

ya i did that.. and i got 45 degrees, is that right tho??.. because it wasnt the right answer..

and if taking the half of it would be 22.5 degrees??

 

[Mattowander]

I apologize. I misread the first question. Instead of looking for the angle that will give you the maximum possible sin value, look for the angle that will give you half of that. :)

 

[reverse]

I apologize. I misread the first question. Instead of looking for the angle that will give you the maximum possible sin value, look for the angle that will give you half of that. :)

no problem..
ok so i got 30 degrees and 150 degrees..?

 

[Mattowander]

It's most likely going to be 30 degrees although I suppose 150 degrees would work too :)

You tell me, is that the right answer?

 

[reverse]

It's most likely going to be 30 degrees although I suppose 150 degrees would work too :)

You tell me, is that the right answer?

no :( i don't know why??

 

[Mattowander]

Because of the 2 in the sin 2 theta. Does that hint help you? What's the angle?

 

[reverse]

Because of the 2 in the sin 2 theta. Does that hint help you? What's the angle?

i divided and multiplied 30 by 2 and both answers were wrong

 

[Mattowander]

That's weird. I'm not sure why that answer would be wrong.

 

[fmr172]

I figured it out! Well I didn't figure it out, I got the right answer. I cheated by graphing y=sin2x, and traced the curve to about .5 on the y-axis on both sides of 45[tex]\circ[/tex]. I'm still not sure on the logic though, I can see how the period is cut in half by the 2x. And I can see how the shorter period combined with the same max height make the curve "steeper." But I don't understand why it's not at the halfway point of the period. Is it because its the halfway point of the tangents slope?