Question

You push a 560 mm radius bicycle wheel, which is approximately hoop shaped, along a flat region until it reaches a speed of 5.01 m/s. It then goes up a smooth hill of irregular shape and height 0.6 m before falling off the vertical face on the opposite side.
What is its speed at the top of the hill?

Answer 1

Answer:

The speed at the top of the hill is 4.38 m/s

Explanation:

Here we have total Kinetic = KE (translational) +KE (rotational)

=0.5 m v² + 0.5m·r²v²/r² = m·v²

Therefore at height 0.6 m we have

0.6 mg = mv²

When v = 5.01 m/s maximum height is

m·g·h=m·5.01²

or h = 2.56 m

Therefore at 0.6 m we have 2.56 - 0.6 more height energy to climb

which gives

1.96·m·g = m v₂²

or v₂² = 19.22

v₂ = 4.38 m/s.