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sveta [45]
3 years ago
6

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

sphere attached to the end opposite the pivot. This arrangement is a good approximation to a simple pendulum (period = 0.61 s), because the mass of the sphere (lead) is much greater than the mass of the rod (aluminum). When the sphere is removed, the pendulum no longer is a simple pendulum, but is then a physical pendulum. What is the period of the physical pendulum?
Physics
1 answer:
gavmur [86]3 years ago
8 0

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

T_{simple = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / I ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and I  is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

I = \frac{1}{3}mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/I) OR T = 2π√(I/mgL)

so we can use I = \frac{1}{3}mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =  \frac{1}{2}D.

now, substituting these equations, the period becomes;

T = 2π/√(I/mgL) OR T = 2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } } OR T = 2π√(2D/3g )  ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω_{simple = 2π/T_{simple OR  ω_{simple = √(g/D) OR  ω_{simple = 2π√( D/g )  

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × T_{simple

we substitute in value of T_{simple

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

 

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Read 2 more answers
A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas
Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0
3 years ago
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