# Thread: Walking on a merry-go-round

1. ## Walking on a merry-go-round

Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?

R[m] = 3.54;

2. Originally Posted by m_i_k_o
Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?

R[m] = 3.54;

the acceleration is centripetal.

3. Originally Posted by m_i_k_o
Suppose that you are walking outwards along a radial line on a merry-go-round platform. You walk at a constant speed of 1 m/s. The platform is rotating at the rate of 20 RPM (RPM=Revolutions Per Minute).
What is the magnitude of your acceleration (in m/s2) when you are R meters away from the centre?

R[m] = 3.54;

Write the (vector) velocity equation out as a function of time and differentiate it to give the (vector) acceleration as a function of time, then as you are at a radius R at time t=R find the required acceleration and take its absolute value.

CB

4. Originally Posted by CaptainBlack
Write the (vector) velocity equation out as a function of time and differentiate it to give the (vector) acceleration as a function of time, then as you are at a radius R at time t=R find the required acceleration and take its absolute value.

CB
Like this? (Attached pic)

5. Originally Posted by Xapphire13
Like this? (Attached pic)
Assume that $\theta=0$ at $t=0$

Then at time $t$ $(=R$) we have the velocity is:

$\bold{v}=[R \omega \sin(\omega t)+\cos(\omega t), R \omega \cos(\omega t)]+\sin(\omega t)] =$ $[t \omega \sin(\omega t)+\cos(\omega t), t \omega \cos(\omega t)]+\sin(\omega t)]$

So:

$
\dot{\bold{v}}=[t \omega^2 \cos(\omega t), 2 \omega \cos(\omega t)+t \omega^2 \sin(\omega t)]
$

Now $\omega=\frac{2 \pi}{3}$ radians/s, so when $t=3.54$ we have:

$\dot{\bold{v}}\approx[6.61158,\ 15.8338]$,

so the magnitude of the acceleration is $\approx 17.16$ m/s^2

.. check the algebra and the arithmetic

CB