| **Bicycle science problem** | jaybird
*Nov 4, 2003 4:12 PM* | | Here is a problem that I recently encountered in an engineering class. See if you can get it... I'll post the answer later.
Given: A mountain biker crests a hill at 10.0 mph, then coasts down the other side, traveling 100 ft down a 20% incline (average rise/run). The cyclist and bike weigh a total of 200 lb, and 1300 ft·lb of the cyclist's energy is lost to friction on the way down the hill.
Required: The cyclist's speed (mph) at the bottom of the hill. |
| **re: Bicycle science problem** | Spiderman
*Nov 4, 2003 4:29 PM* | | It doesn't matter how much energy he is losing because he is coasting down?? |
| **What's this got to do with Lance and Sandra? n/m** | AaronL
*Nov 4, 2003 4:36 PM* | | |
| **re: Bicycle science problem** | TimA
*Nov 4, 2003 5:26 PM* | | 39.3 MPH, Neglecting rotational weight and wind resistance of course. This also assumes the 100 ft referred to the height of the hill, not the distance travelled. I wasn't sure which way was meant and this assumption was a bit easier.
BTW: Yes I am bored at work. |
| **re: Bicycle science problem** | torquecal
*Nov 4, 2003 8:09 PM* | | Assuming he stops at the bottom of the hill for a rest break his speed is zero at the bottom. |
| **re: Bicycle science problem** | EvilDeer
*Nov 5, 2003 6:16 AM* | | 10mph = 10 x 5280 / 3600 = 14.667 ft/s
Hill Height = 14.79 ft (for travelling distance of 100ft)
Final Energy = Initial kinetic energy + Potential energy - Losses
I'm assuming that you mean 1300 ft^2.lb/s^2 not 1300 ft.lb since you say energy.
0.5mV^2 =0.5mv^2 + mgh - losses
0.5(100)V^2 = 0.5(200)(14.67)^2 + (200)(32.2)(14.79) - 1300 = 115468 ft^2.lb/s^2
Final Speed = 33.9 ft/s = 23.2 mph
As a bonus question, calculate the effect on final speed of a 1lb beer bottle thrown at 20mph at the back of the cyclist from a pickup truck moving at 30mph ;) |
| **re: Bicycle science problem** | jaybird
*Nov 6, 2003 5:45 AM* | | You got the closest... The actual height of the the hill is 19.6ft. You have to calculate the angle of incline which comes out to be 11.3deg by finding arctan of 20/100 since the rise over run is 20% or 20/100. Then take the sin of that angle and multiply by 100 to get the rise with a hypotenuse of 100.
the actual answer is 22.2 mph. |
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