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Igoryamba
3 years ago
14

a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height

h=3v2/4g
Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′  and mass ′

M′  is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

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Answer:

The change of the momentum of the ball is -19.8\, \frac{mkg}{s}

Explanation:

We should find \varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}} (1)with \overrightarrow{p_{i}} the initial momentum and \overrightarrow{p_{f}} the final momentum. Linear momentum is defined as \overrightarrow{p}=m\overrightarrow{v}, using that on (1):

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It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall \overrightarrow{v_{i}}=+2.2\, \frac{m}{s} and \overrightarrow{v_{f}}=-2.2\, \frac{m}{s} so (2) becomes:

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
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Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

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(a)

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so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

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