# Homework Help: Mechanical engineering Swing of Death

1. Oct 29, 2012

### gl0ck

Hi there,
again maybe stupid question but I have no idea about this exercise.
or maybe a little clue. i know that v=r?
but with the given 40° when i convert it to a rmp it is very small value 0.111111
and the r is 3m or 7m..
Please if someone can help me..
also if someone can help me with the FBD of A wheel being hauled up a slope by a wire attached to its axle.

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2. Oct 29, 2012

### CWatters

You should show your working before asking for help.

For example how did you arrive at two values? Neither match the answer I got.

3. Oct 29, 2012

### gl0ck

I just wondered if the r is equal to the wire attached or the whole body of the swing?
that is how i find 3m and 7m because the string is 3m long and the whole body is 7m

4. Oct 30, 2012

### CWatters

Neither.

With the wire hanging down vertically the radius would be 4m. With the ride spinning around at very high speed the wires would fly out and up until almost horizontal and the radius would be approx 4 + 3=7.

However the problem asks what speed is necessary for them to be at a 40 degree angle to the vertical. You can use basic trig to work out the exact radius. It's 4M + ?M in the diagram...

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Last edited: Oct 30, 2012
5. Oct 30, 2012

### CWatters

Re your PM. Best I reply on thread so anyone else with similar question sees the follow up..

No the radius would be..

= 4M + ?M see my diagram.
= 4 + 3Sin(40)
= 4 + 1.93
= 5.93m

Since it's not moving vertically the vertical components must sum to zero.

Lets take downwards as +ve

-T*cos(40) + m*g = 0
so
T = m*g/cos(40)...........................................(Eq 1)

Now for the horizontal..

In order for it to move in a circle the wire must provide centripetal force = mv2/r towards the centre so..

T*sin(40) = m*v2/r
T = m*v2/(r*sin(40)) ...................(Eq 2)

over to you