Question

I have a question, and I need to know why my answer is incorrect. (Phys)

Answer 1

In the game of tether-ball, a 1.25 m rope connects a
M= 0.780 kg ball to the top of a vertical pole so that
the ball can spin around the pole as shown in the figure below.
http://i.imgur.com/Tb0DjOD.jpg ((Copy paste, don't click, just in case...))

What's the speed of the ball as it rotates around the pole
when the angle θ of the rope is 36.0º with the vertical?

Answer 2

The force of Tension is:
\(\color{blue}{ T= mg\cos\theta}\)

The x-component of tension is equivalent to the centripetal force:
\(\color{blue}{F_c = T\sin\theta }\) \(\\[0.8em]\)
\(\color{blue}{F_c= mg\cos\theta \sin\theta}\) \(\\[0.8em]\)
\(\color{blue}{mv^2/R= mg\cos\theta \sin\theta}\) \(\\[0.8em]\)
\(\color{blue}{v^2/R= g\cos\theta \sin\theta}\) \(\\[0.8em]\)
\(\color{blue}{v^2/(1.25\times \sin \theta )= g\cos\theta \sin\theta}\) \(\\[0.8em]\)
\(\color{blue}{v^2= 1.25g\cos\theta \sin^2\theta}\) \(\\[0.8em]\)
\(\color{blue}{|v|= \sqrt{ 1.25g\cos\theta \sin^2\theta}}\) \(\\[0.8em]\)
\(\color{blue}{|v|= \sqrt{ 1.25\times 9.80 \cos\theta \sin^2\theta}}\) \(\\[0.8em]\)
\(\color{blue}{|v|= \sqrt{ 1.25\times 9.80 \cos(36) \sin^2(36)}=1.85 }\) \(\\[0.8em]\)

Answer 3

That answer was incorrect.
So, the centripetal force is not equivalent to the tension in the x?

Answer 4

@dumbcow

Answer 5

ok sorry for wait, your equation for Tension is wrong, its not "mg cos"


\[T^2 = T^2 \sin^2 \theta + m^2 g^2\]
\[T \cos \theta = mg\]
\[T = \frac{mg}{\cos \theta}\]

Answer 6

Oh, so in practice I should be dividing by cos(theta) instead of multiplying.
I see, nice triangle. My teacher didn't explain anything in class (she was talking about her headache... that is more than just reasonable. Thanks, I will try this:)

Answer 7

thx, and yes the centripetal force is always the force of tension in the x

Answer 8

Is my operation with radius correct?

I mean, it is still right that,
R=(1.25 × sin(theta))

yes?

Answer 9

Well, it makes perfect sense to me to be that way.
Just clarifying.

Answer 10

yes your radius is correct

Answer 11

Nice!

Answer 12

\(\color{blue}{F_c = T\sin\theta }\)

\(\color{blue}{\displaystyle F_c = \left(Mg\sec\theta\right)\sin\theta }\)

\(\color{blue}{\displaystyle MV^2/R = Mg\sec\theta\sin\theta }\)

\(\color{blue}{\displaystyle V^2/R = g\sec\theta\sin\theta }\)

\(\color{blue}{\displaystyle V = \sqrt{R g\sec\theta\sin\theta} }\)

\(\color{blue}{\displaystyle V = \sqrt{(1.25\sin\theta) g\sec\theta\sin\theta} }\)

\(\color{blue}{\displaystyle V = \sqrt{1.25 g\sec\theta\sin^2\theta} }\)

\(\color{blue}{\displaystyle V = \sqrt{1.25(9.80)\sec(36)\sin^2(36)} }\)

Answer 13

V=2.28722

Answer 14

Yes, yes, yes!!!!

Answer 15

:)

Answer 16

:)

Answer 17

v = 2.3 m/s is correct