Question

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial angular speed of 2.7 rad/s at its equilibrium position and oscillates with a frequency of 2.5 Hz, what is its maximum angular displacement?

Answer 1

To find the maximum angular displacement of the torsional pendulum, you can use the formula for the maximum angular displacement and the formula for the moment of inertia. Substitute the given values into the equations and solve for the maximum angular displacement.

Step-by-step explanation:

In this problem, we can use the formula for the frequency of a torsional pendulum:

f = (1 / 2π) * √(k / I)

Where f represents the frequency of the oscillation, k represents the torsional constant of the wire or spring, and I represents the moment of inertia of the disk. Rearrange the equation to solve for I:

I = (4π² * m * r²) / k

Substitute the given values:

I = (4π² * 0.45 kg * (0.035 m)²) / k

Now we can calculate the maximum angular displacement:

θmax = √(2 * E / k)

Where θmax represents the maximum angular displacement and E represents the mechanical energy of the system. Rearrange the equation to solve for E:

E = (1 / 2) * k * θmax²

Substitute the given values:

E = (1 / 2) * k * (θmax)²

Answer 2

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

`\omega = 2\pi f`

`\omega = 2\pi (2.5)`

`\omega = 5\pi rad/s`

The angular displacement is given as the form:

`\theta (t) = \theta_0 cos(\omega t)`

In the equlibrium we have to
`t=0, \theta(t) = \theta_0` and in the given position we have to

`\theta(t) = \theta_0 cos(5\pi t)`

Derived the expression we will have the equivalent to angular velocity

`(d\theta)/(dt) = 2.7rad/s`

Replacing,

`\theta_0(sin(5\pi t))5\pi = 2.7`

Finally

`\theta_0 = (2.7)/(5\pi)rad = 9.848\°`

Therefore the maximum angular displacement is 9.848°