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AfilCa [17]
3 years ago
8

You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict

ion for the cylinder on the ramp is μs. If the cylinder rolls downhill without slipping, what is the magnitude of the friction force that the ramp exerts on the cylinder? Express your answer in terms of the variables β, M and acceleration due to gravity g.
Physics
1 answer:
zlopas [31]3 years ago
6 0

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f  , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / (  R² / k² + 1)

Putting the given values

f = Mgsinβ /(  R² / k² + 1)

for cylinder ,  R² / k² = 2

f =  Mgsinβ /3

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A spatially challenged goldfish swims along the x-axis only. Its initial position is 7.8 m. After swimming back and forth a whil
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<em>The fish displacement was -3.4 m</em>

Explanation:

<u>Distance and Displacement</u>

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dtotal=d1+d2+d3+...+dn

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We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

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The fish displacement was -3.4 m

3 0
3 years ago
Write two different unit in which mass is measured.​
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2 years ago
A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
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To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

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PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

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This expression is equivalent to,

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Replacing we have,

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Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
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