What is the stopping distance for a round steel bar using friction braking?

[capterdi]

Hi,

Please help me with this problem:

Suppose a round steel bar of 25 mm diameter and 72 m lengh, which is traveling at 3.7 m/sec. I need to know which is the distance needed for the bar to brake by frictioning against a flat steel surface. Coefficient of friction is 0.3 and density of steel 7.85 ton/m3.

Thank you.

 

[Topher925]

Do you need help finding an answer or do you just want someone to do this for you?

I would start by determining the mass of the rod and then finding the Lagrangian of the system.

 

[capterdi]

I gladly would like someone to solve this problem for me, so I can see which are the formulas involved, and also the solution.

Thanks

 

[minger]

It seems to me that you can just find the weight of the bar. Assuming the bar is horizontal, the weight will also be the normal force. Multiply said normal force by the coefficient of friction to obtain the friction force. Divide by the mass to get your acceleration. Divide the velocity by the acceleration to get the time needed.

 

[capterdi]

Minger,

Ok...let me try and see what I get...

Thanks

 

[uart]

The stopping time is,

t = \frac{v}{\mu g}And the stopping distance is,

d = v t - 0.5 \mu g t^2If you don't need to know the time then you can calculate the stopping distance directly from the equation,

d = \frac{v^2}{2 \mu g}

Where \mu is the coefficient of friction, v is the initial velocity and g is the gravitational aceleration aceleration (approx 9.8 m/s^2).

 

[capterdi]

uart,

Ok...I can understand those equations, and have all needed data to solve them. Thanks a lot.

 

[capterdi]

Sorry...didn't notice before: The mass isn´'t involved in the equations? I suppose it's not the same distance and time for a mass of 1 kg than for 1,000kg.

 

[minger]

You should have derived these using my approach. You would have arrived at the same point, but with the knowledge of where they come from. The friction force again is the coefficient times the normal force:
F_f = \mu m g
This will be the only force acting on the bar, so we can relate it with the famous equation:
F_f = \mu m g = m a
Of course the masses will cancel out and you get the relationship between acceleration and coefficient of friction. From here, substitute the definition for acceleration:
\mu g = \frac{v}{t}
and of course simply solve for time. So, theoretically in an ideal world, yes, both a 1kg and a 1000kg bar moving at the same initial velocity will stop at the same time.

 

[uart]

Sorry...didn't notice before: The mass isn´'t involved in the equations? I suppose it's not the same distance and time for a mass of 1 kg than for 1,000kg.

Yes it actually is the same, the mass cancels out in the calculations. Think of it like this, the 1000kg object will experience 1000 times the frictional force compared with the 1kg object, but the deceleration (a=F/m) will be the same.

 

[capterdi]

Right.

Thanks minger & uart for your help.