Ask question

Ask question

All categories

3 years ago

11

A circular disk is rotating about a central axis with an angular velocity which is increasing at a constant rate. Point 1 on the

disk is located at a distance r1 from the center of the disk and point 2 on the disk is located at a distance r2 from the center of the disk. If r2 = 2r1, which of the following ratios is the largest? 1 and 2 refer to being at points 1 and 2 in the subscripts below, all values occurring at the same time.
a. ω2/ω1
b. α2/α1
c. ac2/ac1
d. All the ratios are 1 so none is larger.

1 answer:

antiseptic1488 [7]3 years ago

8 0

The largest ratio of the circular disk at the two points is ac2/ac1.

The given parameters:

Location of point 1 = r1
Location of point 2 = r2 = 2r1

<h3>What is angular speed?</h3>

The angular speed of an object is the rate of change of angular displacement of the object.

The angular speed of each disk is given as;

$\omega = 2\pi N$

where;

N is the number of revolutions

thus, the angular speed is independent of radius of the circular disk

ω2/ω1 = 1

the centripetal acceleration of each disk is calculated as follows;

$a_c = \frac{v^2}{r} \\\\a_c_1 = \frac{v^2}{r_1} \\\\a_c_2 = \frac{v^2}{2r_1} \\\\\frac{a_c_2}{a_c_1} = \frac{2r_1}{v^2} \times \frac{v^2}{r_1} = 2$

the angular acceleration of the disk is calculated as follows;

$\alpha = \frac{\omega }{t}$

the angular acceleration is independent of the radius of the disk.

α2/α1 = 1

Thus, the largest ratio of the circular disk at the two points is ac2/ac1.

Learn more about centripetal acceleration here: brainly.com/question/79801

You might be interested in

In boxing, the use of 16-ounce gloves rather than 12-ounce gloves reduces the chance of injury because the force is distributed

mr Goodwill [35]

Answer:

true

Explanation:

Here we have assumed that increasing the mass of a glove will increase the surface area.

Injury is caused by the application of pressure at a point on the body. The application of pressure takes place via the area of the gloves. Pressure is given by

$P=\dfrac{F}{A}$

where

F = Force

A = Area to which the force is applied

So, a bigger glove will increase the surface area and reduce the pressure resulting in a lower chance of injury.

Hence, the statement is true.

5 0

3 years ago

A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio

Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

8 0

3 years ago

The magnetic field lines of a bar magnet spread out from the north end to the south end. South end to the north end. Edges to th

Gwar [14]

<h3><u>Answer;</u></h3>

the north end to the south end.

<h3><u>Explanation;</u></h3>

Magnetic field lines from a bar magnet form lines that are closed. The direction of magnetic field is taken to be outward from the North pole of the magnet and in to the South pole of the magnet.
A magnetic field refers to the area surrounding a magnet where a force is exerted on certain objects. These lines are spread out of the north end of the magnet.
The magnetic field lines resemble a bubble.

3 0

3 years ago

Read 2 more answers

A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)

frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

$\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow \text{Time taken by light}=2.8583\times 10^{-6}$

Time taken by light = 2.8583×10⁻⁶ seconds

3 0

3 years ago

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

pantera1 [17]

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.

3 0

3 years ago