Math Problem Statement

The figure shows a uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of 1.6 cm and a mass of 20 grams and is initially at rest. Starting at time t = 0, two forces are to be applied tangentially to the rim as indicated, so that at time t = 1.2 s the disk has an angular velocity of 280 rad/s counterclockwise. Force F1 has a magnitude of 0.096 N. What is the magnitude of F2?

Solution

To solve this problem, let's analyze the information provided and calculate the magnitude of F2F_2.

Given Data:

  1. Radius of disk (RR) = 1.6 cm = 0.016 m
  2. Mass of disk (mm) = 20 grams = 0.02 kg
  3. Initial angular velocity (ω0\omega_0) = 0 (since it starts from rest)
  4. Final angular velocity (ω\omega) = 280 rad/s (counterclockwise)
  5. Time interval (tt) = 1.2 s
  6. Force F1F_1 = 0.096 N

The question requires us to find the magnitude of the force F2F_2.

Approach:

  1. Calculate the moment of inertia (I) of the disk: For a uniform disk rotating around its center, the moment of inertia is: I=12mR2I = \frac{1}{2} m R^2

  2. Calculate the angular acceleration (α\alpha): Using the formula for angular acceleration: α=ωω0t\alpha = \frac{\omega - \omega_0}{t}

  3. Calculate the net torque (τ\tau): Torque (τ\tau) is related to angular acceleration by: τ=Iα\tau = I \cdot \alpha

  4. Set up the equation for net torque with F1F_1 and F2F_2: Since the forces are applied tangentially, the torque due to each force is: τ=F1RF2R\tau = F_1 \cdot R - F_2 \cdot R Since we know τ\tau, we can solve for F2F_2.

Let's go through the calculations:

  1. Moment of inertia: I=12×0.02kg×(0.016m)2=2.56×106kgm2I = \frac{1}{2} \times 0.02 \, \text{kg} \times (0.016 \, \text{m})^2 = 2.56 \times 10^{-6} \, \text{kg} \cdot \text{m}^2

  2. Angular acceleration: α=280rad/s01.2s=233.33rad/s2\alpha = \frac{280 \, \text{rad/s} - 0}{1.2 \, \text{s}} = 233.33 \, \text{rad/s}^2

  3. Net torque: τ=Iα=(2.56×106kgm2)×233.33rad/s2=0.000597Nm\tau = I \cdot \alpha = (2.56 \times 10^{-6} \, \text{kg} \cdot \text{m}^2) \times 233.33 \, \text{rad/s}^2 = 0.000597 \, \text{N} \cdot \text{m}

  4. Torque equation with F1F_1 and F2F_2: τ=F1RF2R\tau = F_1 \cdot R - F_2 \cdot R Substituting τ=0.000597Nm\tau = 0.000597 \, \text{N} \cdot \text{m}, F1=0.096NF_1 = 0.096 \, \text{N}, and R=0.016mR = 0.016 \, \text{m}: 0.000597=(0.096)(0.016)F2(0.016)0.000597 = (0.096) \cdot (0.016) - F_2 \cdot (0.016)

  5. Solve for F2F_2: F2=(0.0960.016)0.0005970.016F_2 = \frac{(0.096 \cdot 0.016) - 0.000597}{0.016} F2=0.0015360.0005970.016=0.0009390.016=0.0587NF_2 = \frac{0.001536 - 0.000597}{0.016} = \frac{0.000939}{0.016} = 0.0587 \, \text{N}

Answer:

The magnitude of F2F_2 is approximately 0.0587N0.0587 \, \text{N}.

Would you like a further breakdown of any part of this solution?

Here are some related questions you might consider:

  1. How would the solution change if the disk had a different mass?
  2. What if the disk’s radius were larger? How would F2F_2 be affected?
  3. How does the direction of the forces impact the angular velocity?
  4. Can this approach be applied to non-uniform objects, like a ring?
  5. What role does the moment of inertia play in rotational dynamics?

Tip: Always double-check unit conversions, especially when switching from grams to kilograms or centimeters to meters.

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Math Problem Analysis

Mathematical Concepts

Rotational Motion
Angular Acceleration
Torque
Moment of Inertia

Formulas

Moment of Inertia for a disk I = (1/2) * m * R^2
Angular acceleration α = (ω - ω0) / t
Torque τ = I * α
Net torque equation τ = F1 * R - F2 * R

Theorems

Newton's Second Law for Rotation (τ = I * α)

Suitable Grade Level

Undergraduate Physics