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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 250 N applied to the edge causes the wheel to have an angular acceleration of 0.940 rad/s^2.

(a) What is the moment of inertia of the wheel?
(b) What is the mass of the wheel?
(c) If the wheel starts from rest, what is its angular velocity after 5.00 s have elapsed, assuming the force is acting during that time?

Respuesta :

onyebuchinnaji

Answer:

(a) I = 87.766 kg.m²

(b) m = 1611.864 kg

(c) ω = 4.7 rad/s

Explanation:

Given;

radius of the wheel, r = 0.330 m

applied force, F = 250 N

angular acceleration, α = 0.940 rad/s²

Part (a) the moment of inertia of the wheel

τ = F x r = Iα

I = (F x r) / α

where;

I is the moment of inertia

I = (250 x 0.33) / 0.94

I = 87.766 kg.m²

Part (b) the mass of the wheel

I = ¹/₂mr²

2I = mr²

m = (2I) / r²

where;

m is the mass of the wheel

m = (2 x 87.766) / (0.33²)

m = 1611.864 kg

Part (c) angular velocity after 5.00 s

ω = ω₀ + αt

where;

ω is the angular velocity after 5.00 s

ω₀ is the initial angular velocity = 0

ω = 0 + (0.94 x 5)

ω = 0 + 4.7

ω = 4.7 rad/s

adefioyelaoye

The radius of the wheel, r = 0.330 m

Force, F = 250 N

Angular acceleration, α = 0.940 rad/s²

  • The moment of inertia of the wheel

  τ = F x r = Iα

where   τ is torque, F is force, r is radius, I is inertia and α is angular acceleration.

     I = (F x r)

             α

I = (250 x 0.33)

          0.94

I = 87.766 kg.m²

  • The mass of the wheel

I = ¹/₂mr²

2I = mr²

m = (2I)

       r²

m = (2 x 87.766)

           (0.33²)

m = 1611.86kg

  • Angular velocity after 5.00 s

          ω = ω₀ + αt

In this equation, ω is the angular velocity after 5.00 s

ω₀ is the initial angular velocity which is 0

ω = 0 + (0.94 x 5)

ω = 0 + 4.7

ω = 4.7 rad/s

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