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 260 N applied to its edge causes the wheel to have an angular acceleration of 0.795 rad/s2.

(a) What is the moment of inertia of the wheel?
1 kg · m2

(b) What is the mass of the wheel?
2 kg

(c) If the wheel starts from rest, what is its angular velocity after 4.70 s have elapsed, assuming the force is acting during that time?
3 rad/s

To solve this problem, we need to use the formulas related to rotational motion.

(a) The moment of inertia (I) of a solid cylinder can be calculated using the formula:

I = (1/2) * m * r^2

where m is the mass of the cylinder and r is the radius.

To find the moment of inertia, we need to use the given values of radius and the angular acceleration. We can rearrange the formula to solve for I:

I = (2 * a) / r^2

where a is the angular acceleration.

Substituting the values:

I = (2 * 0.795) / (0.330^2)
I = 2.996 / 0.1089
I ≈ 27.53 kg · m^2

Therefore, the moment of inertia of the wheel is approximately 27.53 kg · m^2.

(b) The mass (m) of the wheel can be calculated using the formula:

m = I / (r^2)

Substituting the values:

m = 27.53 / 0.330^2
m = 27.53 / 0.1089
m ≈ 252.85 kg

Therefore, the mass of the wheel is approximately 252.85 kg.

(c) To find the angular velocity (ω) of the wheel after a given time, we can use the formula:

ω = ω0 + (α * t)

where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.

In this case, the wheel starts from rest, so the initial angular velocity (ω0) is zero. Substituting the given values:

ω = 0 + (0.795 * 4.70)
ω ≈ 3.73 rad/s

Therefore, the angular velocity of the wheel after 4.70 seconds is approximately 3.73 rad/s.