A cyclist is on a rolling stationary trainer pedaling such that his wheels are traveling at 30
km/h. He has an unfortunate incident and ends up falling over 90° on his side onto a large
patch of ice. Given the cyclist and bicycle have an 4.5 kg*m^2, find the angular velocity
that the cyclist will have on the ice after he is on his side? Assume there is negligible mass in
the hub and the spokes of the wheels.
To find the angular velocity of the cyclist after falling on his side, we can apply the principle of conservation of angular momentum. According to this principle, the total angular momentum before the fall will be equal to the total angular momentum after the fall.
The formula for angular momentum is given as:
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
Given:
Moment of Inertia of cyclist and bicycle = 4.5 kg*m^2 (I)
Angular velocity of the wheels before the fall = 30 km/h = (30 * 1000) m / (60 * 60) s = 8.33 m/s (ω₁)
Since the cyclist and bicycle are considered as a single object, we need to calculate the initial angular momentum of the system before the fall. Assuming the cyclist is upright, the initial angular momentum is:
Initial Angular Momentum (L₁) = I * ω₁
To find the angular velocity (ω₂) of the cyclist on the ice after falling, we can rearrange the formula for angular momentum:
L₁ = L₂, where L₂ is the final angular momentum on the ice.
Therefore, I * ω₁ = I * ω₂
Simplifying the equation, we find:
ω₂ = ω₁
So, the angular velocity of the cyclist on the ice after falling will be equal to the angular velocity of the wheels before the fall, which is 8.33 m/s.
Please note that in real-world scenarios, factors such as friction and other forces may affect the actual angular velocity after the fall. This calculation assumes ideal conditions with no external forces acting on the system.