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frosja888 [35]
3 years ago
10

A holiday ornament in the shape of a hollow sphere with mass 0.010 kg and radius 0.055 m is hung from a tree limb by a small loo

p of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)
Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

0.608 s

Explanation:

Given that

m = 0.01 kg

r = 0.055 m

the period of a pendulum is primarily stated as

T = 2π √(I/mgd), where

I = moment of inertia

m = mass of pendulum

g = acceleration due to gravity

d = radius

moment of inertia, I is given as

I = ⅔MR² + MR²

I = 5/2 MR²

also, going forward, we assume d = R

next, we substitute each into the equation for period, i.e 2π√(I/mgd)

T = 2π √[(5/3MR²) / MgR]

T = 2π √[(5/3R) / g]

T = 2π √(5R/3g)

Next, we plug in the values of each, and we have

T = 2π √[(5 * 0.055) / (3 * 9.8)]

T = 2π √(0.275/29.4)

T = 2π √0.00935

T = 2π * 0.0967

T = 2 * 3.142 * 0.0967

T = 0.608 s

Therefore, the period of the hollow sphere is 0.608 s

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F = m x a          --------------(i)

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F = m v²/ r             --------------------(ii)

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F = qvBsinθ          -------------(iii)

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<em>Combine equations (ii) and (iii) as follows;</em>

m (v² / r) = qvBsinθ         [divide both side by v]

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v = 1.48 x 10⁷m/s

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θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

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T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

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T = 0.245 / 1.48 x 10⁷

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