Math Problem Statement

A large stone disk is viewed from above and is initially at rest. The disk has a mass of 145.0 kg and a radius of 1.800 m. A constant force of 40.0 N is applied tangent to the edge of the disk for 56.0 s, causing the disk to spin around the z-axis. Calculate the angular acceleration, final angular velocity, and translational speed at the edge of the stone after 56.0 s.

Solution

Let's go through each part of the problem to see where errors might be occurring and confirm the calculations.

Given Information

  • Mass of disk, m=145.0kgm = 145.0 \, \text{kg}
  • Radius of disk, r=1.8mr = 1.8 \, \text{m}
  • Force applied, F=40.0NF = 40.0 \, \text{N}
  • Time of application, t=56.0st = 56.0 \, \text{s}

Part (a): Angular Acceleration α\alpha

The angular acceleration can be calculated using the formula: α=τI\alpha = \frac{\tau}{I} where:

  • τ\tau is the torque, given by τ=F×r\tau = F \times r.
  • II is the moment of inertia of a solid disk, I=12mr2I = \frac{1}{2} m r^2.
  1. Calculate Torque τ\tau: τ=F×r=40.0N×1.8m=72.0Nm\tau = F \times r = 40.0 \, \text{N} \times 1.8 \, \text{m} = 72.0 \, \text{Nm}

  2. Calculate Moment of Inertia II: I=12mr2=12×145.0kg×(1.8m)2=234.9kgm2I = \frac{1}{2} m r^2 = \frac{1}{2} \times 145.0 \, \text{kg} \times (1.8 \, \text{m})^2 = 234.9 \, \text{kg} \cdot \text{m}^2

  3. Calculate Angular Acceleration α\alpha: α=τI=72.0234.90.3065rad/s2\alpha = \frac{\tau}{I} = \frac{72.0}{234.9} \approx 0.3065 \, \text{rad/s}^2 This matches the answer provided in the solution, 0.3065rad/s2k^0.3065 \, \text{rad/s}^2 \, \hat{k}.

Part (b): Final Angular Velocity ω\omega

The final angular velocity ω\omega can be found using the formula: ω=α×t\omega = \alpha \times t where α=0.3065rad/s2\alpha = 0.3065 \, \text{rad/s}^2 and t=56.0st = 56.0 \, \text{s}.

  1. Calculate ω\omega: ω=0.3065×56.017.564rad/s\omega = 0.3065 \times 56.0 \approx 17.564 \, \text{rad/s}

However, this answer is marked incorrect. The reason might be due to rounding errors or requiring more precision in intermediate steps.

Part (c): Translational Speed vv at the Edge

The translational speed at the edge can be calculated using: v=ω×rv = \omega \times r

  1. Using ω17.564rad/s\omega \approx 17.564 \, \text{rad/s}: v=17.564×1.831.6152m/sv = 17.564 \times 1.8 \approx 31.6152 \, \text{m/s}

This answer is also close but might require more precision. To improve accuracy, consider using all intermediate steps with more decimal places.

Would you like detailed calculations with extended precision, or help on any particular section?

Related Questions

  1. How does changing the force affect the angular acceleration?
  2. What would happen to the final angular velocity if the mass of the disk increased?
  3. How would a change in the radius affect the moment of inertia and torque?
  4. Why might rounding errors impact the final answer?
  5. Can we calculate the kinetic energy of the rotating disk after 56 seconds?

Tip:

Always carry extra significant figures in intermediate calculations to minimize rounding errors in the final answer.

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Math Problem Analysis

Mathematical Concepts

Rotational Motion
Angular Acceleration
Angular Velocity
Torque
Moment of Inertia
Translational Speed

Formulas

Torque (τ) = Force (F) * Radius (r)
Moment of Inertia (I) for a disk = (1/2) * Mass (m) * Radius^2
Angular Acceleration (α) = Torque / Moment of Inertia
Angular Velocity (ω) = Angular Acceleration * Time
Translational Speed (v) = Angular Velocity * Radius

Theorems

Newton's Second Law for Rotation

Suitable Grade Level

College Physics