# Homework Help: Wall of death problem

1. Apr 7, 2009

### ab3477

1. The problem statement, all variables and given/known data
- FBD of description of forces

- Mass of the rider ( is to decide ) = 70 kg
- Dimensions of the Ride ( is to decide )
Height = 10 meters

Givens:
gmax= 4 g
Speed of bike = 80 Km/h
Mass of Bike = 180 kg
mu= vertical coefficient of friction = 1.00

Find:
Minimum speed to keep rider on wall
Maximum based on gmax

2. Relevant equations
Fc= m.ac = m. 4pi^2* Radius/ T^2

3. The attempt at a solution
mtotal= 250 kg ( 180 + 70 )
speed of bike in meters = 22.22 m/s
mT*g = direction= downwards= 2452.5 N [Down]

So far this is what I have done for attempt. I'm not asking to do this question.
1) direction of forces..Fg would be down. Normal force?? and what provides the centripetal force.
2) If you can help me do this question.

2. Apr 8, 2009

### Slepton

Can you type the exact question?

3. Apr 8, 2009

### lanedance

I assuming its a bkie riding horizontally around the internal surface of a cylinder?

If this the case... the frictional force must balance the gravitational force in the vertical direction for the rider to stay up on the wall

Whilst the reaction force from the wall (horizontally) is exactly that required to change the rider & bike direction motion... ie it is the centripetal force

4. Apr 8, 2009

### ab3477

yes, it is a vertical cylinder or surface. ( 90 degrees of angle )
I understood that the reaction force is causing motion that is centripetal force.
Weight force would Down. What way would be the reaction force and normal force? and frictional force!
I'm guessing normal force will be perpendicular to surface.
Plus, I need help in finding these:
-Minimum speed to keep rider on wall
-Maximum based on gmax (4 g )

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5. Apr 8, 2009

### lanedance

I would never use south for vertically down, just confusing

Force Balance on Rider & Bike

Vertically
Down - Gravitational force
Up - Frictional force

Horizontally
Toward centre of cylinder - Normal force (perp to surface as you said)

For the rider to not accelerate up or down, the net force in the vertical dierction must balance to zero.

The maximum allowable frictional force F, for a given normal force N, is F = mu.N, where mu is the coefficient of friction. If the gravitataional force is greater than the max allowable frictional force the rider will slip down. Note the firctional force can be less than this, but mu.N is the upper limit for a given normal force.

From this information you should be able to find the minimum speed... ie when the speed is just enough, that the normal force provides sufficient frictional force to balance the gravitational force

the max allowable speed should come straight from the centripetal force for circular motion

6. Apr 9, 2009

### ab3477

here's the try...
A) V= Sqrt[(g*r)/mu)] (( equations found on this website with similar problem))
= sqrt[(9.81*10)/(1.00)]
= 9.91 m/s or 35.66 Km/h
Therefore, minimum of 35.6 km/h speed is required to stay on the wall.

B) Max based on Gmax (4 g)
4xg (9.81 ) = 39.24
Fc=250*39.24 = 9810 N

Gforce/Fc = (mv2)/r
V= sqrt[(Fc.r)/(m)]
= sqrt[(9810*10)/250)]
= 19.798 m/s or 71.24 km/h...will give the maximum of 4g.

One question arising now, why the mass does not matter to calculate the minimum speed as it does for max (based on gmax)

Thank you for help!!

7. Apr 16, 2009

### lanedance

for the minimum speed you have
$$Frictional Force = F = \mu.N = \mu.\frac{m v^2}{r} = mg = Gravitational Force$$
so the m's cancel out

I'm not totally sure what you have done in the 2nd case, I think the numbers are correct, but you may have just carried m in and cancelled later without realising...

$$a_{max} = 4g = \frac{F_c}{m} = \frac{\frac{mv^2}{r}}{m} = \frac{v^2}{r}$$