61) A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s. A very light bolt is attached to extend the axle in length, and a string is attached to the axle at a distance of 0.10 m from the wheel. Initially the axle of the spinning wheel is horizontal, and the wheel is suspended only from the string. We can ignore the mass of the axle and spokes. At what rate will the wheel process about the vertical

The rate the wheel will process about the vertical is 2.86 RPM

Explanation:

Given;

radius of the bicycle wheel, R = 0.36 m

mass of the wheel, m = 3.2 kg

angular velocity, ω = 4 rev/s

The rate at which the wheel will process about the vertical is given by;

Ф = τ/L

Where;

τ is the torque

L is the angular momentum

τ = Fr

τ = mgr = 3.2 x 9.8 x 0.1 = 3.126 N.m

L = Iω =  MR²ω

L = 3.2 x (0.36)² x (4 x 2π)

L = 10.4244 kg.m²/s

Ф = τ/L

Ф = (3.126) / (10.4244)

Ф = 0.29987 rad/s

Ф = 0.29987 rad/s  x (60 / 2π)

Ф = 2.86 RPM

Therefore, the rate the wheel will process about the vertical is 2.86 RPM

mistakes are ok

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever