Could I please get an explanation that includes words? Thank you :) You are desi
ID: 1463948 • Letter: C
Question
Could I please get an explanation that includes words? Thank you :)
You are designing a new roller-coaster. The main feature of this particular design is to be a vertical circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in fact upside-down (at the top of the loop). The coaster starts at rest from a height of 80 m above the ground, speeds up as it descends to ground level, and then enters the loop, which has a radius of 20 m. Suppose a rider is sitting on a bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when the coaster is moving past the top of the loop? (You can assume that the coaster rolls on the track without friction).
Explanation / Answer
First a little overview: At the top of the Roller Coaster RC, you and your car have Potential Energy PE = mgh. As you roll down, you pick up speed and your PE is converted to KE Kinetic Energy =(1/2)mv^2. Neglecting friction, the sum of your PE + KE everywhere on the RC is equivalent to the PE at the top of the RC. When you reach the bottom of the RC where height=0, ALL of your PE has been converted to KE. Fortunately the car has brakes because your KE is quite high at the bottom.
At 30 degrees above the middle of the loop (assume that is about 2 o'clock)
The force pushing on the scale is
mV^2/r + mgcos120 Where mV^2/r is the outward force due to traveling in a circle and mgcos120 is the component of your weight summed with that scale reading. 120 is the angle between your weight and the radius of the loop terminating at your position. Notice that at the top of the loop, the angle between the radius and the force of gravity is 180 degrees meaning all your weight subtracts from mv^2/r. At the bottom of the loop the angle is 0 so all the weight adds to mv^2/r
mg = mV^2/+mgcos120 Eliminate m since it is on both sides of the equation
[g-gcos120]*20 = V^2 = 2.2657 V = 4.75m/s at 30 above the horizontal. The kinetic energy at the bottom of the loop (1/2)mVo^2 =mgh+(1/2)mV^2 at 30 degrees above the horizontal. The height h above the bottom of the loop there is
h=r+rsin30=10m Eliminate m from both sides of the equation and solve for Vo
Vo=20
At the top of the loop, all your weight subtracts from the scale reading but your velocity has also slowed. The easiest way to figure this is to calculate the KE at the bottom of the loop and then subtract the PE at the top of loop to calculate V at the top of the loop
(1/2)mVo^2 = 2.2657m. So the KE at the top of the loop is 200m - mgh = 200m-m*10*2*20/3
=200m/3 = (1/2)*m*V^2 at the top of the loop. Eliminate m from both sides
and solve for V V^2 = 400/3 => V=20/sqrt(3)m/s
NOw solve for mV^2/r + mgcos(180) = 10m = mg so the scale reads the correct amount at the top of the loop because the outward force is 2mg and the downward force is mg leaving 2mg-mg=mg