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Angular Motion of Uniform Disk: Torque, Acceleration, and Rotation

Created by @nathanedwards
 at November 1st 2023, 12:40:49 am.
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#ap-physics-1
#angular-acceleration
#torque

AP Physics 1 Exam Question

A uniform disk of radius 0.5 m and mass 2 kg is initially at rest. A constant net torque of 5 Nm is applied to the disk for 4 seconds.

a) Determine the angular acceleration of the disk. b) Calculate the final angular velocity of the disk. c) What is the total angle rotated by the disk during the 4 seconds?

Answer with Step-by-Step Explanation

a) To determine the angular acceleration of the disk, we can use the equation:

τ=I⋅α \tau = I \cdot \alpha

where:

  • τ is the torque applied to the disk (5 Nm)
  • I is the moment of inertia of the disk (given as a uniform disk formula: I = (1/2) · m · r^2, where m is the mass of the disk and r is its radius)
  • α is the angular acceleration we're looking for.

First, let's calculate the moment of inertia of the disk:

I=12⋅m⋅r2 I = \frac{1}{2} \cdot m \cdot r^2
I=12⋅2 kg⋅(0.5 m)2 I = \frac{1}{2} \cdot 2 \, \text{kg} \cdot (0.5 \, \text{m})^2
I=0.5 kg⋅m2 I = 0.5 \, \text{kg} \cdot \text{m}^2

Now we can use the torque equation to solve for α:

5 Nm=0.5 kg⋅m2⋅α 5 \, \text{Nm} = 0.5 \, \text{kg} \cdot \text{m}^2 \cdot \alpha

Simplifying the equation, we get:

α=5 Nm0.5 kg⋅m2 \alpha = \frac{5 \, \text{Nm}}{0.5 \, \text{kg} \cdot \text{m}^2}
α=10 rad/s2 \alpha = 10 \, \text{rad/s}^2

Therefore, the angular acceleration of the disk is 10 rad/s^2.

b) To calculate the final angular velocity of the disk, we can use the equation:

ωf=ωi+α⋅t \omega_f = \omega_i + \alpha \cdot t

where:

  • ωf is the final angular velocity of the disk (what we're trying to find)
  • ωi is the initial angular velocity of the disk (given as 0 since it starts from rest)
  • α is the angular acceleration (calculated in part a)
  • t is the time the torque is applied (given as 4 seconds).

Given these values, we can substitute them into the equation:

ωf=0+(10 rad/s2)⋅4 s \omega_f = 0 + (10 \, \text{rad/s}^2) \cdot 4 \, \text{s}
ωf=40 rad/s \omega_f = 40 \, \text{rad/s}

Therefore, the final angular velocity of the disk is 40 rad/s.

c) To find the total angle rotated by the disk during the 4 seconds, we can use the equation:

θ=ωi⋅t+12⋅α⋅t2 \theta = \omega_i \cdot t + \frac{1}{2} \cdot \alpha \cdot t^2

where:

  • θ is the total angle rotated by the disk (what we're trying to find)
  • ωi is the initial angular velocity of the disk
  • α is the angular acceleration (calculated in part a)
  • t is the time (given as 4 seconds).

Substituting the given values:

θ=0⋅4+12⋅10 rad/s2⋅(4 s)2 \theta = 0 \cdot 4 + \frac{1}{2} \cdot 10 \, \text{rad/s}^2 \cdot (4 \, \text{s})^2
θ=80 rad \theta = 80 \, \text{rad}

Therefore, the total angle rotated by the disk during the 4 seconds is 80 rad.

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