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rewona [7]
4 years ago
7

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent ho

op about that axis, if the hoop has the same mass M and a radius k given by The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Physics
1 answer:
faltersainse [42]4 years ago
6 0

Answer:

Explanation:

Let mass of cylinder be M

Moment of inertia of cylinder

= 1/2 M R² r is radius of cylinder

If radius of equivalent  hoop be k

Mk² = 1/2 x MR²

k = R / √2

1.2 / 1.414

Radius of gyration = 0.848 m

b )

moment of inertia of spherical shell

= 2 / 3 M R²

Moment of inertia of equivalent hoop

Mk²

So

Mk² = 2 / 3 M R²

k = √2/3 x R

= .816 X 1.2

Radius of gyration = .98 m

c )

Moment of inertia of solid sphere

= 2/5 M R²

Moment of inertia of equivalent hoop

= Mk²

Mk² = 2/5 M R²

k √ 2/5 R

Radius of gyration = .63 R

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<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

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Generally, the equation for Rate of flow of Liquid is  mathematically given as

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Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

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In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

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5 0
2 years ago
Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing
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Specific heat capacity is an intensive property and does not depend on sample size.

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v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

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initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

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u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

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