A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 83.N is applied to the rim of the wheel. The wheel has radius 0.110m . Starting from rest, the wheel has an angular speed of 14.6rev/s after 3.95s .What is the moment of inertia of the wheel?
To find the moment of inertia of the wheel, you can use the formula:
I = (τ / α)
where I is the moment of inertia, τ is the torque exerted on the wheel, and α is the angular acceleration of the wheel.
In this case, the torque can be calculated using the formula:
τ = r * F
where r is the radius of the wheel and F is the tangential force applied to the rim of the wheel.
Given:
- Radius of the wheel, r = 0.110 m
- Tangential force applied to the rim, F = 83 N
Substituting these values into the equation for torque:
τ = 0.110 m * 83 N
τ = 9.13 N⋅m
Next, the angular acceleration can be calculated using the formula:
α = Δω / Δt
where Δω is the change in angular velocity and Δt is the change in time.
Given:
- Initial angular velocity, ω_i = 0 (since the wheel starts from rest)
- Final angular velocity, ω_f = 14.6 rev/s
- Time interval, Δt = 3.95 s
Substituting these values into the equation for angular acceleration:
α = (14.6 rev/s - 0) / 3.95 s
α = 3.70 rev/s²
Finally, substituting the values of torque and angular acceleration into the formula for moment of inertia:
I = τ / α
I = 9.13 N⋅m / (3.70 rev/s²)
To convert rev/s² to rad/s², multiply by (2π rad/rev):
I = (9.13 N⋅m) / (3.70 rev/s²) * (2π rad/rev)
I = 59.1 kg⋅m²
Therefore, the moment of inertia of the wheel is 59.1 kg⋅m².