The balance wheel of a watch oscillates with angular amplitude 1.7π rad and period 0.64 s. Find (a) the maximum angular speed of the wheel, (b) the angular speed of the wheel at displacement 1.7π/2 rad, and (c) the magnitude of the angular acceleration at displacement 1.7π/4 rad.
It undergoes simple harmonic motion with an equation for angular displacement, a(t), that can be written
a(t) = A sin (2*pi*t/P)
where A is the amplitude and p is the period. I chose time zero such that the phase angle is zero at that time, to simplify the equation.
(a) w = A*(2*pi/P) cos(2*pi*t/P)
w(max) = 2*pi*(1.7 pi)/(0.46)
(b) When the displacement is half the amplitude,
2*pi*t/P = pi/6 radians
At that time, w is cos (pi/6) = (sqrt3)/2 times the amplitude
(c) When the displacement is 1/4 the amplitude, the angular argument
2*pi*t/P = 0.253 radians
The angular acceleration at any time is
alpha (t) = -A*(2*pi/P)^2*sin(2*pi*t/P)
= (A/4)(2*pi/P)^2
To solve this problem, we'll use the formulas for angular amplitude, angular speed, and angular acceleration.
(a) The maximum angular speed of the wheel is given by the formula:
Maximum angular speed = (2π) / Period
Given that the period is 0.64 s, we can substitute this value into the formula:
Maximum angular speed = (2π) / 0.64 = 3.125π rad/s
So, the maximum angular speed of the wheel is 3.125π rad/s.
(b) The angular speed of the wheel at any displacement can be calculated using the formula:
Angular speed = (2π) / Period
Given that the displacement is 1.7π/2 rad, we can substitute this value into the formula:
Angular speed = (2π) / 0.64 = 3.125π rad/s
So, the angular speed of the wheel at a displacement of 1.7π/2 rad is 3.125π rad/s.
(c) The magnitude of the angular acceleration can be calculated using the formula:
Angular acceleration = (Maximum angular speed^2) / Angular amplitude
Given that the displacement is 1.7π/4 rad, we can substitute the maximum angular speed (3.125π rad/s) and the angular amplitude (1.7π rad) into the formula:
Angular acceleration = (3.125π^2) / (1.7π) = (9.765625π^2) / (1.7π) = 5.75961538π rad/s^2
So, the magnitude of the angular acceleration at a displacement of 1.7π/4 rad is approximately 5.759π rad/s^2.
To find the answers to these questions, we need to understand the relationship between angular displacement, angular speed, and angular acceleration.
Angular displacement (θ) is the change in angle from the starting position. It is measured in radians (rad).
Angular speed (ω) is the rate at which the angle changes with respect to time. It is given by the formula:
ω = θ / t
where θ is the angular displacement and t is the time taken.
Angular acceleration (α) is the rate at which the angular speed changes with respect to time. It is given by the formula:
α = ω / t
where ω is the angular speed and t is the time taken.
Now let's solve each part of the question.
(a) To find the maximum angular speed, we need to find the angular displacement and period. The maximum angular speed occurs at the extreme points of the oscillation, where the angular displacement is 1.7π rad. The period is given as 0.64 s.
We can use the formula for angular speed:
ω = θ / t
Substituting the values, we have:
ω = (1.7π rad) / (0.64 s)
ω ≈ 2.66 rad/s
So, the maximum angular speed of the wheel is approximately 2.66 rad/s.
(b) To find the angular speed at a displacement of 1.7π/2 rad, we can again use the formula for angular speed:
ω = θ / t
Substituting the values, we have:
ω = (1.7π/2 rad) / (0.64 s)
ω ≈ 2.49 rad/s
So, the angular speed of the wheel at a displacement of 1.7π/2 rad is approximately 2.49 rad/s.
(c) To find the magnitude of the angular acceleration at a displacement of 1.7π/4 rad, we need to find the angular speed and period. The angular displacement is given as 1.7π/4 rad, and the period is given as 0.64 s.
We can use the formula for angular acceleration:
α = ω / t
First, let's find the angular speed using the formula mentioned earlier:
ω = θ / t
Substituting the values, we have:
ω = (1.7π/4 rad) / (0.64 s)
ω ≈ 1.33 rad/s
Now, let's use the formula for angular acceleration:
α = ω / t
Substituting the values, we have:
α = (1.33 rad/s) / (0.64 s)
α ≈ 2.08 rad/s^2
So, the magnitude of the angular acceleration at a displacement of 1.7π/4 rad is approximately 2.08 rad/s^2.
for part a: is it the 2*PI*(1.7PI)/.46 OR IS IT 2*pi* (1.7pi/.46)? AND FOR PART B:
DO YOU MULTIPLY THE AMPLITUDE BY (SQRT3)/2? AND I UNDERSTAND PART C