Question

a bicyclist rides downhill at 9.0 m/s, when suddenly, at a vertical height of 60 meters relative to the bottom of the hill, their 2.25 kg front wheel comes off and rolls without slipping downhill. since most of the mass of the wheel is on its perimeter, you can model it as a thin hoop with a 0.40 m radius. a. determine how fast the wheel is moving when it reaches the bottom of the hill. answer: 25.9 m/s

Answer 1

The wheel is moving at 25.9 m/s when it reaches the bottom of the hill.

To determine the speed of the wheel when it reaches the bottom of the hill, we can use the conservation of energy principle.

This principle states that the total energy of a closed system remains constant. In this case, the closed system is the wheel.

Initially, the wheel has only potential energy due to its height above the bottom of the hill.

As it rolls down the hill, this potential energy is converted into kinetic energy.

At the bottom of the hill, the wheel has only kinetic energy.

We can use the following equations to relate potential energy (PE), kinetic energy (KE), and height (h):

PE = mgh

KE = 1/2mv^2

where:

m is the mass of the object

g is the acceleration due to gravity (9.8 m/s^2)

v is the velocity of the object

Setting the initial PE equal to the final KE, we get:

mgh = 1/2mv^2

Solving for v, we get:

v = sqrt(2gh)

Plugging in the values for h (60 m) and m (2.25 kg), we get:

v = sqrt(2 * 9.8 m/s^2 * 60 m)

v = 25.9 m/s

Therefore, the wheel is moving at 25.9 m/s when it reaches the bottom of the hill.