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Katyanochek1 [597]
3 years ago
5

A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

the hoop is released from rest, the string unwinds and the hoop descends. How does the tension in the string (T) compare to the weight of the hoop
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R ,      R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I  α

= I  a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

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The two cars will eventually have the same velocity and it occurs between 4 and 5. Velocity is normally defined as displacement over time. This shows that a change in the value of either time or displacement will affect the value of the velocity. Thus, at a given time between the points 4 and 5, the two cars have the same value of velocity.

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At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu
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Answer:

a) 0.159

b) 0.84

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Let the vertical electric field component = E_{v}

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I = \frac{E_{m} ^{2} }{2c \mu}..............................(1)

E_{m} is the resolution of the vertical and horizontal components, E_{h} and    E_{v}

E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}..................(2)

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Put equation (2) into equation (3)

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After the glasses were put on the horizontal component vanishes, i.e. E_{h} = 0

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Divide equation (5) by equation (4)

\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}...............................(6)

But E_{h} = 2.3E_{v}......................(7)

Insert equation (7) into (6)

\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2}  }{(2.3E_{v})^{2}   + E_{v} ^{2}   } \\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{5.29E_{v}^{2}   + E_{v} ^{2}   }\\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{6.29E_{v}^{2}  }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\

\frac{I_{2} }{I_{1} }= 0.159

b) When the sunbather lies on his side, the vertical component vanishes, i.e E_{v} = 0

\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}

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Answer:

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