# Circular motion

1. Oct 19, 2004

### hauthuong

the sloping side is frictionless, it is spun at constant speed by rotating the wedge. show that when a mass rises up the wedge a distance L, the speed of the mass is v=sqr(gLsin0)
I got Fx=m(v^2/r) = mgsin0
v^2=g*sin0*r
can I say r=L ????
I got stuck, could you give me some hints thank you

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2. Oct 19, 2004

### Pyrrhus

Hint: Gravity is the centirpetal force because it is pulling the block towards the center.

3. Oct 19, 2004

### maverick280857

What is your specific problem with this? Tried resolving normal reaction in the right direction? Do you know what the radius of the circular path is?

4. Oct 19, 2004

### Pyrrhus

Yes r will be L.

$$mgsin\theta = m \frac{v^2}{l}$$

$$\sqrt{lgsin\theta} = v$$

5. Oct 20, 2004

### hauthuong

thank you, however, i do not quite understand why r=l. Could you please explain

6. Oct 20, 2004

### Himura Kenshin

Where you get the equation m(v^2/r)=mgsin0. I have search for many reference book but can't find this equation .Can you tell me ?

7. Oct 20, 2004

### maverick280857

In my analysis, I resolve the normal reaction in the horizontal and vertical direction. This gives,

$$N\cos\theta = mg$$
$$N\sin\theta = \frac{mv^2}{l\cos\theta}$$

I divide equation 2 by equation 1 to get

$$v^2 = gl\sin\theta$$

Cheers
Vivek

8. Oct 20, 2004

### maverick280857

mg sin theta is the component of the weight acting down the incline (resolve mg in two directions one parallel and the other perpendicular to the plane). Drawing a well labeled freebody diagram might help.

9. Oct 20, 2004

### Pyrrhus

Yes Maverick, that's a correct analysys for the way your coordinate system was put, although i prefer your way than mine, because it is using the vertical coordinate system (most used in Circular Motion) rather than the inclined one.