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kakasveta [241]
3 years ago
10

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the ho

op's plane at an edge.
Physics
1 answer:
mote1985 [20]3 years ago
7 0

Answer:

 I = 2 MR²

Explanation:

Given that

Radius of the hollow ring ( hoop ) = R

The mass of the hoop = M

We know that mass moment of inertia of a hoop about its center is given as

Io= M R²

By using theorem  ,mass moment of inertia at distance d from center is given as

I= Io + m d²

Here ,M= m  ,d =R

Now by putting the values in the above equation we get

I =  M R² +  M R²

I = 2 MR²

Therefore the mass moment of inertia will be  2 M R².

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I need it due in 10 mins ​
ExtremeBDS [4]

Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

Velocity = Angular velocity × radius = 2×1.2 = 2.4 m/s

Centripetal force= mv²/r = 3 × 2.4×2.4/1.2 = 3 × 2.4 × 2

= 14.4 N

7 0
3 years ago
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad
Anna71 [15]

Answer:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

a_{r} = \omega^{2}\cdot R

Where:

\omega - Angular speed, measured in radians per second.

R - Radius of rotation, measured in meters.

The angular speed is first determined:

\omega = \frac{\pi}{30}\cdot \dot n

Where \dot n is the angular speed, measured in revolutions per minute.

If \dot n = 3000\,rpm, the angular speed measured in radians per second is:

\omega = \frac{\pi}{30}\cdot (3000\,rpm)

\omega \approx 314.159\,\frac{rad}{s}

Now, if \omega = 314.159\,\frac{rad}{s} and R = 0.1\,m, the resultant acceleration is then:

a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)

a_{r} = 9869.588\,\frac{m}{s^{2}}

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

6 0
3 years ago
You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g
TiliK225 [7]

Answer:

Only 9% weaker

Explanation:

Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.

4 0
3 years ago
Read 2 more answers
How many times larger than a centigram is a dekagram
Umnica [9.8K]

Answer:

A dekagram is thousand (1000) times larger than a centigram.

Explanation:

→ [1 dekagram = 1,000 centigrams]

→ 1 dekagram = 10 grams

→ 10 grams = 100 decigrams

→ 100 decigrams = 1,000 centigrams

3 0
3 years ago
FREE BRAINLEIAST FOR FIRST GOGOGOGO!!!!heererer
Kamila [148]

Answer:

ME PLS

Explanation:

6 0
3 years ago
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