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NikAS [45]
2 years ago
6

A wagon wheel consists of 8 spokes of uniform diameter, each of mass ms and length L cm. The outer ring has a mass mring. What i

s the moment of inertia of the wheel
Physics
1 answer:
just olya [345]2 years ago
6 0

Answer:

The moment of inertial of the wheel,  I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2

Explanation:

Given;

8 spokes of uniform diameter

mass of each spoke, = M_s

length of each spoke, = L

mass of outer ring, = M_r

The moment of inertial of the wheel will be calculated as;

I = 8I_{spoke} + I_{ring}

where;

I_{spoke is the moment of inertia of each spoke

I_{ring is the moment of inertia of the rim

Moment of inertia of each spoke =\frac{1}{3}M_sL^2

Moment of inertial of the wheel

I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2

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Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

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Answer:

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C₂ = ∈A/2d.............................(2a)

From equations (1) and (2), the relationship between C₁ and C₂ is:

C₂ = C₁/2....................................(2b)

The original  Energy stored in the capacitor is given by :

E₁ = Q²/2C₁...............................(3)

On doubling the separation between the capacitance plates

E₂ = Q²/2C₂...............................(4a)

E₂ = Q²/2  *  1/C₂........................(4b)

Putting equation (2b) into (4b)

E₂ = Q²/2  *  2/C₁

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