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4 years ago

A circular hoop of diameter d hangs on a nail. what is the period of its oscillations at small amplitude?

Physics

1 answer:

Misha Larkins [42]4 years ago

4 0

m = mass of the circular hoop

r = radius of the hoop

I = moment of inertia of the hoop

moment of inertia of the hoop about the center of hoop is given as

I = m r²

k = distance of the point of suspension from center of mass = r

using parallel axis theorem

I' = moment of inertia of hoop about the point of suspension

I' = I + m k²

I' = m r² + m k²

I' = m r² + m r²

I' = 2 m r²

Time period of oscillation for the hoop is given as

T = 2π sqrt(I'/mgk)

T = 2π sqrt(2 m r²/(mgr))

T = 2π sqrt(2 r/g)

since 2r = diameter = d

T = 2π sqrt(d/g)

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Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$

Donde:

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

$A_{t}$ - Área transversal del orificio, medido en metros cuadrados.

$\dot V_{out}$ - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que $v_{out} = 2.426\,\frac{m}{s}$ y $A_{t} = 5\,cm^{2}$ , el caudal de salida del chorro es:

$\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)$

$\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}$

El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

5 0

3 years ago

A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of t

ivolga24 [154]

Answer:

5572.8 N

Explanation:

Applying,

F = ma.............. Equation 1

Where F = Force, m = mass of the car, a = acceleration.

We can find a by applying,

v² = u²+2as............. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, = distance.

From the question,

Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m

Substitute these value into equation 2

0² = 1.7²+2×0.21×a

a = -1.7²/(2×0.21)

a = -2.89/0.42

a = -6.88 m/s²

Also given: m = 810 kg

Substitute these value into equation 1

F = 810(-6.88)

F = -5572.8 N

Hence the force on the bumber is 5572.8 N

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3 years ago

While spinning in a centrifuge a 70.0 KG astronaut experiences an acceleration of 2.00g, or twice the acceleration due to gravit

alisha [4.7K]

Centripetal acceleration is the force acting on the body who's moving in a circular motion. It is still a force, which according to Newton's Second Law of Motion,

F = ma
F = (70 kg)(2)(9.81 m/s²)
F = 1373.4 N

<em>Thus, the answer is B.</em>

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4 years ago

Consider the numbers 23.68 and 4.12. The sum of these numbers has ____ significant figures, and the product of these numbers has

damaskus [11]

Answer:multiplying will give us 7 significant figures and addition will give us 3 significant figures

Explanation:

After multiplying the two numbers they resulting value will give a value in its 4 decimal places because both given values are in 2 decimal places. The 4 dp is gotten by the addition of the decimal places of both given numbers (2+2) and

The result of its addition will give us a value in its 1dp and 3 significant figures since the addition of 23.68 and 4.12 will give us 27.8

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3 years ago

If continuous spectrum light from a star passes through a cool, low-density gas on its way to your telescope and spectroscope, _

natima [27]

Answer: a dark (absorption) line

Explanation:

This is as a result of absorption of electromagnetic a specific wavelength. The pattern followed by such lines is characteristic of specific atoms in the path of radiation.

6 0

3 years ago