Acarnival clown rides a motorcycle down a ramp and then up and around a large, vertical loop. if the loop has a radius of 18 m, what is the slowest speed the rider can have at the top of the loop so that the motorcycle stays in contact with the track and avoids falling? m/s

Answer59.2kg; /

I'm having some difficulty seeing the diagram clearly.but the question says there's a net inward force on the object.so we know right away,  from newton's 2nd law of motion, thatthat baby is accelerating.  it doesn't even matter which way thenet force is pointing.  if there's any net force, then either theobject's speed or else its direction must be changing.

To compare speeds speeds of wagons we will use second newton's law. it states: f = m * ahere we can see that force depends on the acceleration. if we know acceleration for both wagons we can compare them to make conclusion of speeds of wagons.we have: m1 = 10lbm2 = 20lbf1 = 2*f2also: m2 = 2*m1for bill we have equation: f1 = m1 * a1for tom we have equation: f2 = m2 * a2after inserting identities we have: 2*f2 =m1 * a1f2 = 2*m1 * a2in order to make conclusions about speeds we will solve these equations for a and find their ratio: a1 = 2f2 / m1a2 = f2 / 2m1a1 / a2 = (2f2/m1) / (f2/2m1)a1 / a2 = (2f2*2m1) / (m1*f2)a1 / a2 = (4f2m1) / (m1f2)a1 / a2 = 4this ratio is bigger than 1 and this means that a1 is greater than a2. a1 is 4 times greater than a2. speed of bill's wagon is 4 times greater than speed of tom's wagon.