A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel h and observes that drops of water fly off tangentially. She measures the heights reached by drops mov- ing vertically (Fig. P7.8). A drop that breaks loose from the tire on one turn rises vertically 54.0 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The radius of the wheel is 0.381 m. (a) Why does the first drop rise higher than the second drop? (b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel’s angular acceleration (assuming it to be constant).

a) The first drop rises higher than the second drop because the speed of the tire has reduced from what it was during the first turn to a lower value during the second turn. This will most likely be due to how bicycle tires are set up, plenty frictional elements to tamper and reduce the speed of the bike until it is pedalled again.

Note that the speed with which the drops of water rise are both equal to the corresponding tangential speeds of the tire at those points in time. And since the tangential speed of the tire reduces in between turns, the height travelled by the drops too, reduces.

b) To calculate the angular acceleration for the two cases.

The kinetic energy of the drops of water while on the tire is converted to the energy used to attain the respective heights that they attain.

(1/2)mv² = mgh

v = √(2gh)

For the first drop

h₁ = 54.0 cm = 0.54 m

r = 0.381 m

v₁ = √(2gh₁)

v₁ = √(2×9.8×0.54) = 3.253 m/s

w₁ = (v₁/r)

w₁ = (3.253/0.381) = 8.538 rad/s

For the second drop

h₂ = 51.0 cm = 0.51 m

r = 0.381 m

v₂ = √(2gh₂)

v₂ = √(2×9.8×0.51) = 3.162 m/s

w₂ = (v₂/r)

w₂ = (3.162/0.381) = 8.300 rad/s

Using the equations of angular motion,

w₂² = w₁² + 2αθ

θ = 2π

8.3² = 8.538² + 4π (α)

α = -0.32 rad/s²

Negative because it is angular deceleration

Hope this Helps!!!