# Circular Motion

1. Feb 7, 2010

### lemon

1. The problem statement, all variables and given/known data
Ignoring the motion of the Earth around the Sun and the motion of the Sun through space, find:

a) the angular velocity
b) the linear speed
c) the centripetal acceleration

of a body resting on the ground at the equator.

What would the length of a day be if the angular speed of the Earth's rotation on its axis were to increase until the body became effectively weightless? Radius of Earth = 6.4 x 106m

2. Relevant equations

1: ω=θ/t
2: v=2πr/T and v=ωr
3: a=v2/r and a=rω²

3. The attempt at a solution

a) ω=θ/t
Hmmm!
We have 24 hours for one complete rotation of the Earth = 24x60x60=86400s
One complete rotation = 360° = 2π(rads)

b) v=ωr
v=(73x10-6)x(6.4x106)=467.2m/s

c) a=rω2
a=(6.4x106)x(73x10-6)2=0.0341056m s-1
=34.1x10-3m s-1

In order to answer the last part of the question, I guess we must consider what angular speed would cause the body to become effectively weightless. We don't have a mass of the body (kg) and therefore, can't calculate its weight.
Much appreciated as always

2. Feb 7, 2010

### kuruman

You don't need the mass of the body. If the Earth rotates so fast that a 70 kg father becomes weightless, so does his child who has a mass of only 20 kg. Bodies become weightless when the acceleration of gravity is equal to the centripetal acceleration.

The other parts of the question look OK although I did not perform the numerical calculations.

3. Feb 7, 2010

### ehild

Could you solve the problem if the body had 1 kg mass?

ehild

4. Feb 7, 2010

### lemon

When the acceleration of gravity is equal to the centripetal acceleration:
Centripetal acceleration = 10m s-2

a=rω², so ω=±√a/r = √10m s-2/6.4 x 106m
= 1.25x10-3rad s-1 (is that unit correct?)

Therefore, 2π x 1.25x10-3 = 7.9x10-3s
A really quick day!

Is that right?

5. Feb 7, 2010

### kuruman

The unit is correct.

What equation is this? Please write it symbolically so that I can see where it is coming from.

6. Feb 8, 2010

### ehild

No.

$$\omega=\frac{2\pi}{T}=1.25 \cdot 10^{-3} s^{-1}\rightarrow T=?$$

ehild

7. Feb 8, 2010

### lemon

This is the new centripetal acceleration:
ω=±√a/r = √10m s-2/(6.4 x 106m)

I think I may have messed up there and multiplied 2π by ω (2π x 1.25x10-3 = 7.9x10-3s). Instead I guess I should have divided by ω.
Using ω=2π/T would give: 2π/(1.25x10-3rad s-1=1600π or 5026.5s (1d.p.)

8. Feb 9, 2010

### ehild

looks better, doesn't it?

ehild

9. Feb 17, 2010

### lemon

Looks really good to me. Still a very quick day.
Glad I can get more sunshine than that.
Thanks everyone