Ask question

Ask question

All categories

3 years ago

13

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha

s a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

1 answer:

sineoko [7]3 years ago

7 0

Answer:

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

$T=2\pi \sqrt{\frac{I}{Mgd}}$ (1)

Where:

I is the moment of inertia
M is the mass of the pendulum
d is the distance from the center of mass to the pivot
g is the gravity

Let's solve the equation (1) for I

$T=2\pi \sqrt{\frac{I}{Mgd}}$

$I=Mgd(\frac{T}{2\pi})^{2}$

Before find I, we need to remember that

$T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s$

Now, the moment of inertia will be:

$I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}$

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

I hope it helps you!

You might be interested in

A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

3 years ago

Which of the following are true? Select all that apply. The net electric field at any location inside a block of copper is zero

Agata [3.3K]

Answer:

1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.

2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.

3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.

Explanation:

1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.

2) The charged particles on a conductor in equilibrium are at rest.

6 0

3 years ago

You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont

choli [55]

Answer:

$490.5\ \text{N/m}$

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

x = Displacement of spring = $78-50=28\ \text{cm}$

k = Spring constant

The force balance of the system is given by

$kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}$

The spring constant for that spring is $490.5\ \text{N/m}$ .

6 0

3 years ago

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne

mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

6 0

3 years ago