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 200 N applied to its edge causes the wheel to have an angular acceleration of 1.048 rad/s2.
(a) What is the moment of inertia of the wheel? HINT: Remember that there are two equations for torque. τ = r Fsinθ and τ = I α.
kg·m2
(b) What is the mass of the wheel? HINT: You will need to refer to the specific formula that determines the moment of inertia for a solid disk, or solid cylinder, in terms of its mass and radius.
kg
(c) If the wheel starts from rest, what is its angular velocity after 4.40 s have elapsed, assuming the force is acting during that time? HINT: You will want to use the equations for uniform angular acceleration.
rad/s
To answer these questions, we will need to use some basic equations related to torque and angular acceleration.
(a) To find the moment of inertia, we can use the equation τ = I α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
In this case, we are given the torque (τ = rFsinθ) and the angular acceleration (α = 1.048 rad/s^2). We need to calculate the moment of inertia.
First, we need to find the radius of the wheel. In the given information, the radius is given as 0.330 m.
Now, we can substitute these values into the equation τ = I α:
τ = rFsinθ
I α = rFsinθ
Rearranging the equation, we get:
I = τ / α
Substituting the given values:
I = (0.330 m * 200 N * sin90°) / 1.048 rad/s^2
Calculating the equation:
I ≈ 62.72 kg·m^2
So, the moment of inertia of the wheel is approximately 62.72 kg·m^2.
(b) To find the mass of the wheel, we can use the formula for the moment of inertia of a solid cylinder, which is I = (1/2) m r^2, where m is the mass and r is the radius.
Rearranging the equation, we get:
m = 2I / r^2
Substituting the given values:
m = 2 * 62.72 kg·m^2 / (0.330 m)^2
Calculating the equation:
m ≈ 1194.18 kg
So, the mass of the wheel is approximately 1194.18 kg.
(c) To find the angular velocity of the wheel after 4.40 s, we can use the equation for angular velocity, ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity (0 rad/s as the wheel starts from rest), α is the angular acceleration, and t is the time.
Substituting the given values:
ω = 0 + (1.048 rad/s^2)(4.40 s)
Calculating the equation:
ω ≈ 4.5976 rad/s
So, the angular velocity of the wheel after 4.40 s is approximately 4.5976 rad/s.