MHB A torsional pendulum - determine angular speed

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SUMMARY

The discussion focuses on determining the angular speed of a torsional pendulum, specifically a horizontal uniform metal disk suspended by a wire. The angular displacement is modeled by the equation θ = 0.2 cos(π(t - 0.5)). By applying calculus, the derivative of angular displacement with respect to time, denoted as ω(t), is calculated as ω(t) = -0.2π sin(π(t - 0.5)). At t = 3.1 seconds, the angular speed is found to be approximately -0.6 radians per second, indicating a decrease in angular displacement.

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A torsional pendulum made by suspending a horizontal uniform metal disk by a wire from its center. If the disk is rotated and then released, it willexecute simple (angular) harmonic motion. Suppose at t seconds the angular displacement of \theta radians from the initial position is given by the equation \theta = 0.2 cos \pi(t-0.5). Determine to the nearest tenth of a radian per second, how fast the angle is changing at 3.1 sec.
 
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Hello and welcome to MHB! (Wave)

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Okay, that's our canned response to new users who have posted a question without showing any work or thoughts. It serves us well, but let's consider, we are given:

$$\theta=0.2\cos\left(\pi(t-0.5)\right)$$

Now, we are being asked how at what rate the angular displacement $\theta$ is changing with respect to time $t$. What concept from calculus allows us to determine the instantaneous rate of change of one variable with respect to another?
 
To follow up, we can use the derivative to find the time rate of change of the angular displacement, which is typically denoted by $\omega$.

$$\omega(t)=\d{\theta}{t}=-0.2\pi\sin\left(\pi(t-0.5)\right)$$

Hence:

$$\omega(3.1)=-0.2\pi\sin\left(\pi(3.1-0.5)\right)=-0.2\pi\sin(2.6\pi)\approx-0.6$$

So, we find that at time $t=3.1\text{ s}$, the angular displacement is decreasing at a rate of about 0.6 radians per second.
 
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