Math Problem Statement

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 210 N applied to its edge causes the wheel to have an angular acceleration of 0.854 rad/s2. (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 6.00 s have elapsed, assuming the force is acting during that time?

Solution

Given Data:

  • Radius of the grinding wheel, r=0.330mr = 0.330 \, \text{m}
  • Tangential force, Ft=210NF_t = 210 \, \text{N}
  • Angular acceleration, α=0.854rad/s2\alpha = 0.854 \, \text{rad/s}^2
  • Time, t=6.00st = 6.00 \, \text{s}

Relevant Equations:

  1. Torque: τ=Iα\tau = I \alpha
  2. Relationship between torque and force: τ=Ftr\tau = F_t r
  3. Moment of inertia for a solid cylinder: I=12Mr2I = \frac{1}{2} M r^2
  4. Angular velocity: ω=ω0+αt\omega = \omega_0 + \alpha t, where ω0\omega_0 is the initial angular velocity.

(a) Moment of Inertia

From τ=Iα\tau = I \alpha and τ=Ftr\tau = F_t r, we can write: Iα=Ftr    I=FtrαI \alpha = F_t r \quad \implies \quad I = \frac{F_t r}{\alpha}

Substitute the values: I=2100.3300.854=81.15kg\cdotpm2I = \frac{210 \cdot 0.330}{0.854} = 81.15 \, \text{kg·m}^2

So, the moment of inertia is: I=81.15kg\cdotpm2I = 81.15 \, \text{kg·m}^2


(b) Mass of the Wheel

For a solid cylinder, I=12Mr2I = \frac{1}{2} M r^2. Rearranging for MM: M=2Ir2M = \frac{2I}{r^2}

Substitute I=81.15kg\cdotpm2I = 81.15 \, \text{kg·m}^2 and r=0.330mr = 0.330 \, \text{m}: M=281.15(0.330)2=162.30.1089=1490.58kgM = \frac{2 \cdot 81.15}{(0.330)^2} = \frac{162.3}{0.1089} = 1490.58 \, \text{kg}

So, the mass is: M=1490.58kgM = 1490.58 \, \text{kg}


(c) Angular Velocity after 6.00 seconds

From the angular velocity equation: ω=ω0+αt\omega = \omega_0 + \alpha t

Since the wheel starts from rest (ω0=0\omega_0 = 0), the equation becomes: ω=αt\omega = \alpha t

Substitute α=0.854rad/s2\alpha = 0.854 \, \text{rad/s}^2 and t=6.00st = 6.00 \, \text{s}: ω=0.8546.00=5.124rad/s\omega = 0.854 \cdot 6.00 = 5.124 \, \text{rad/s}

So, the angular velocity after 6 seconds is: ω=5.124rad/s\omega = 5.124 \, \text{rad/s}


Final Answers:

(a) Moment of inertia: 81.15kg\cdotpm281.15 \, \text{kg·m}^2
(b) Mass: 1490.58kg1490.58 \, \text{kg}
(c) Angular velocity after 6.00 seconds: 5.124rad/s5.124 \, \text{rad/s}


Would you like a step-by-step breakdown or have any additional questions?
Here are some related questions for practice:

  1. How would the moment of inertia change if the wheel's radius doubled?
  2. What is the tangential velocity of a point on the edge after 6 seconds?
  3. What would happen to the angular acceleration if the applied force is halved?
  4. How much work is done by the tangential force over 6 seconds?
  5. If the wheel were hollow, how would its mass and moment of inertia change?

Tip: Always confirm the units when calculating quantities like angular velocity and moment of inertia to avoid mistakes!

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Math Problem Analysis

Mathematical Concepts

Physics
Rotational Motion
Moment of Inertia
Angular Kinematics

Formulas

Torque: τ = Iα
Relationship between torque and force: τ = F_t r
Moment of inertia for a solid cylinder: I = (1/2) M r²
Angular velocity: ω = ω₀ + αt

Theorems

Newton's Second Law for Rotation

Suitable Grade Level

Grades 11-12