Question

Consider a bicycle wheel that initially is not rotating. A block of mass m

is attached to the wheel via a string and is allowed to fall a distance h. Assume that the wheel has a moment of inertia I about its rotation axis. Consider the case that the string tied to the block is wrapped around the outside of the wheel, which has a radius rA. Find ωA , the angular speed of the wheel after the block has fallen a distance, for this case.
Express ωA in terms of m , g, h, rA and I

Answer 1

To calculate the angular speed ωA of the wheel after the block falls a distance h, use the conservation of energy equating potential and kinetic energy, ωA = √((2mgh) / (I + mrA2)).

Step-by-step explanation:

To find the angular speed ωA of a bicycle wheel after a block of mass m falls a distance h, we can use the principle of conservation of energy. The potential energy of the block is transformed into kinetic energy, both rotational (of the wheel) and translational (of the block). Initially, the system has potential energy given by mgh, and finally, it has rotational kinetic energy given by ½Iω2. Assuming no other forces are doing work (such as friction), these two energy quantities can be equated:

mgh = ½Iω2

Solving for ω (the angular speed), we get:

ω = √(2mgh / I)

However, note that as the string unwinds and the wheel rotates, the block also has translational kinetic energy. The translational kinetic energy is given by ½mv2, and since v = ωrA, this translates into another term of ½m(ωrA)2 in energy.

So the corrected equation becomes:

mgh = ½Iω2 + ½m(ωrA)2

This yields:

ωA = √((2mgh) / (I + mrA2))