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January 27, 2020, 09:05:41 pm

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#### bigtick

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##### pendulum
« on: May 31, 2008, 02:23:06 pm »
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The bob of a 1m long pendulum moves at sqrt(10) m/s when the pendulum makes 30 deg with the vertical. Find the magnitude of the bob's accel.

#### /0

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##### Re: pendulum
« Reply #1 on: May 31, 2008, 09:58:26 pm »
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The only forces acting on the bob are gravity $F_g$ and tension $T$.

If you draw a diagram in which thee bob is displaced by $\theta = \frac{\pi}{6}$ with the vertical, the restoring force will be $F_g\sin{\theta}$ ($F_g\cos{\theta}$ is canceled by the tension). Hence the acceleration is $a=\frac{F_g\sin{\theta}}{m}$.

$\Rightarrow v = \int \frac{F_g\sin{\theta}}{m} d\theta=\frac{-F_g \cos{\theta}}{m}$

When $\theta = \frac{\pi}{6}$, $v=\sqrt{10}$

$\Rightarrow \frac{9.8 \times \frac{\sqrt{3}}{2}}{m} = \sqrt{10}$

$m = 2.68$

$\Rightarrow a = \frac{-9.8 \sin{\frac{\pi}{6}}}{2.68}=-1.828 \ ms^{-2}$

#### Mao

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##### Re: pendulum
« Reply #2 on: May 31, 2008, 10:02:46 pm »
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I dont exactly know kinematics all that well yet, but

$\frac{dv}{d\theta} = a$?!

you cant integrate like that man

the force acting on the bob at any moment is $F_g = m\cdot g$ and tension [opposite of gravity resolved into components parallel and orthogonal to the $\theta$) $F_T = \cos(\theta) \cdot m \cdot g$

hence, the restoring force, as already found, is
\begin{align*}
F_{net} &= \sin(\theta) \cdot m\cdot g \\
m\cdot a_{net} &= \sin(\theta) \cdot m\cdot g \\
a_{net} &= g\cdot \sin(\theta) \\
a_{\theta = 30^o} &= \frac{g}{2} \approx 5ms^{-2}\\
\end{align*}

i think....? [velocity wasnt even used...]
that seemed a bit too simple for the information provided...
« Last Edit: June 02, 2008, 08:40:36 pm by Mao »
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#### /0

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##### Re: pendulum
« Reply #3 on: May 31, 2008, 10:06:07 pm »
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#### bigtick

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##### Re: pendulum
« Reply #4 on: June 02, 2008, 06:12:01 pm »
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The given answer is a=11.1 m/s2.

#### mark_alec

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##### Re: pendulum
« Reply #5 on: June 02, 2008, 09:03:19 pm »
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Wrong. The acceleration due to gravity will have a maximum value of  $9.8 ms^{-2}$ when the arm of the pendulum is horizontal. It will be less at all other points.

#### Neobeo

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##### Re: pendulum
« Reply #6 on: June 02, 2008, 09:19:10 pm »
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« Last Edit: June 02, 2008, 09:34:03 pm by Neobeo »
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#### mark_alec

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##### Re: pendulum
« Reply #7 on: June 02, 2008, 09:51:39 pm »
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Question is abusing meaning of pendulum. Please slap whoever wrote it.

#### bigtick

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##### Re: pendulum
« Reply #8 on: June 02, 2008, 10:05:31 pm »
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What is a simple pendulum?

#### mark_alec

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##### Re: pendulum
« Reply #9 on: June 02, 2008, 10:24:35 pm »
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A simple pendulum could be understood as mass on the end of a rope/pole that is free to oscillate due to the force of gravity. In the situation they gave, the 'pendulum' is being powered, like a metronome, so to model it circular motion needs to be taken into account.

#### /0

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##### Re: pendulum
« Reply #10 on: June 02, 2008, 10:29:29 pm »
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A simple pendulum could be understood as mass on the end of a rope/pole that is free to oscillate due to the force of gravity. In the situation they gave, the 'pendulum' is being powered, like a metronome, so to model it circular motion needs to be taken into account.

Wait... in 'physics' land, neglecting air resistance and any other forces, the bob would rise to the same height due to conservation of energy, wouldn't it? So it doesn't need to be 'powered'?

#### mark_alec

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##### Re: pendulum
« Reply #11 on: June 02, 2008, 10:58:48 pm »
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physics' land, neglecting air resistance and any other forces, the bob would rise to the same height due to conservation of energy, wouldn't it? So it doesn't need to be 'powered'?
It would in order to have the velocity that was given.

#### /0

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##### Re: pendulum
« Reply #12 on: June 03, 2008, 02:19:23 am »
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but if you dropped the bob from a certain height (under a suitable gravitational force) wouldn't it eventually reach that velocity anyway?

#### bigtick

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##### Re: pendulum
« Reply #13 on: June 03, 2008, 05:30:15 pm »
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find the angle with the vertical when the bob is released.