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kirill115 [55]
3 years ago
5

A cylinder with moment of inertia I about its center of mass, mass m, and radius r has a string wrapped around it which is tied

to the ceiling (Figure 1) . The cylinder's vertical position as a function of time is y(t). At time t=0 the cylinder is released from rest at a height h above the ground.
A) The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate ?and v, the velocity of the center of mass of the cylinder?
B) Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity.
C) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?
D) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?
Physics
1 answer:
NARA [144]3 years ago
5 0

Answer:

Explanation:

Let us study the downward  movement of cylinder which accelerates as well as rotates .

A)

If v be the linear downward velocity of cm of cylinder and ω be angular velocity of cylinder

v = ωr , when there is no slippage of string around cylinder.

B &C )

Total kinetic energy = Rotational + linear

= 1/2 Iω² + 1/2 m v²

1/2 x1/2 mr²ω² +1/2 m v²

= 1/4  mv² +1/2 m v²

= 3/4 m v²

For downward acceleration ,

mg - T = ma where T is tension in string.

Rotational movement

Torque = T x r

Tr = I α , I is moment of inertia and α is angular acceleration .

= I a/r

T = I a / r² , Putting this value of T in earlier equation

mg - I a / r²  = ma

a (I  / r² +m )= mg

a = mg /  (I  / r² +m )

For cylinders

I = .5 mr²

a = g / (.5 +1)

= g / 1.5

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Answer:

a

Explanation:

because they are easily reduced and oxidised

3 0
3 years ago
According to Newton’s law of universal gravitation, which statements are true?
andreyandreev [35.5K]

Before we solve this, we should know this fact:

According to Newton's Law of Gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two objects. It can be shown by this:

F ∝ \frac{Mm}{ {d}^{2} }

Now, let us check all the options.

A. As we move to higher altitudes, the force of gravity on us decreases.

<em>This </em><em>statement </em><em>is </em><em>true.</em>

The force of gravity is inversely proportional to the square of distance from the centre of the earth. If, we go up the surface of the earth, the distance from the centre of the earth increases and hence the value of force of gravity decrease. So, force of gravity decreases with altitude.

B. As we move to higher altitudes, the force of gravity on us increases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

We have already got the result in option A. that the force of gravity decreases with altitude. It never increases with altitude.

C. As we gain mass, the force of gravity on us decreases.

<em>This </em><em>statement</em><em> </em><em>is </em><em>false.</em>

The force of gravity is directly proportional to the product of the masses. So, if increase our mass, then the force of gravity will also increase and if we decrease our mass, then the force of gravity decreases.

D. As we gain mass, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true.</em>

As mentioned earlier in option C., the force of gravity is directly proportional to the product of the masses of the earth and another object. So, as we gain mass, the force of gravity on us increases.

E. As we move faster, the force of gravity on us increases.

<em>This </em><em>statement</em><em> is</em><em> </em><em>true</em><em>.</em>

Here, we have to consider a different formula. According to Newton's Second Law,

F = ma, where F is the force, m is the mass and a is the acceleration.

In other words,

F ∝ a, i.e., force is directly proportional to acceleration.

We know, acceleration is the rate of change of velocity of an body within a time period.

So, if speed is increased, then acceleration will also be greater, which results in the increase of force. So, as we move faster, the force of gravity on us increases.

<u>Answers:</u>

A: As we move to higher altitudes, the force of gravity on us decreases.

D: As we gain mass, the force of gravity on us increases.

E: As we move faster, the force of gravity on us increases.

Hope you could understand.

If you have any query, feel free to ask.

7 0
2 years ago
5. A car accelerates from 0 to 72 km/hour in 8.0 seconds. What is the car's acceleration?
MA_775_DIABLO [31]

Answer:

2.5 m/s

Explanation:

There are calculators online that can help you easily calculate the accerlation.

8 0
2 years ago
In a given chemical reaction the energy of the products is less than the energy of the reactants. Which statement is true for th
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Answer:

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7 0
3 years ago
Normalize the equations
tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k

Part b)

\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1

\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}

7 0
3 years ago
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