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GalinKa [24]
3 years ago
6

A torsion pendulum consists of an irregularly-shaped object of mass 29.0 kg suspended vertically by a wire of torson constant 1.

14 Nm through its center of mass. If this pendulum oscillates through 98 cycles in 74.0 s, find the rotational inertia of the object.
Physics
1 answer:
GarryVolchara [31]3 years ago
4 0

Answer:

I=0.0503\ kg-m^2

Explanation:

Given that,

Mass of the object, m = 29 kg

Torsion constant of the wire, K = 1.14 N-m

Number of cycles, n = 98

Time, t = 74 s

To find,

The rotational inertia of the object.

Solution,

Relationship between the moment of inertia, time period and the torsion constant of the spring is given by :

T=2\pi\sqrt{\dfrac{I}{K}}

Where I is the moment of inertia

K is spring constant

Let T Is the time period of oscillation, such that,

T=\dfrac{98}{74}=1.32\ s

I=\dfrac{T^2K}{4\pi ^2}

I=\dfrac{(1.32)^2\times 1.14}{4\pi ^2}

I=0.0503\ kg-m^2

So, the rotational inertia of the object is 0.0503\ kg-m^2.

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Answer:

The differences between the Lincoln, Johnson, and Congress Reconstruction plans include: Although both the Lincoln and Johnson plan were open to readmission of the southern States, Congress claimed that the two administrations were too lenient and sought stiffer punishment for the States

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3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

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