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3 years ago

7

A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the

rim?

1 answer:

stepan [7]3 years ago

3 0

Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

$I = mr^{2}\\$

where,

I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²

r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2

r = 0.65 m/2 = 0.325 m

m = Mass of the Bicycle Rim = ?

Therefore,

$0.21\ kg.m^{2} = m(0.325\ m)^{2}\\m = \frac{0.21\ kg.m^{2}}{(0.325\ m)^{2}}\\$

<u>m = 1.99 kg = 2 kg</u>

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Complete Question

The diagram of this question is shown on the first uploaded image

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Explanation:

The free body diagram for the diagram in the question is shown

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From the question we are told that

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The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

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Substituting value

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$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

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$PE_B = m_b gh$

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$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

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$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

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