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Alexeev081 [22]

3 years ago

7

Calculate the moment of inertia and the rotational kinetic energy for the following objects spinning about a central axis (in un

its of Joules): a solid sphere with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a thin spherical shell with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a solid cylinder with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a hoop with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec.

1 answer:

Helga [31]3 years ago

6 0

Answer:

a) K = 6.25 10⁻⁴ J , b) K = 10.41 10⁻⁴ J , c) K = 7.81 10⁻⁴ J , d) K = 15.625 10⁻⁴ J

Explanation:

The moment of inertia is

I = ∫ r² dm

For high symmetry bodies it is tabulated.

The rotational kinetic energy is

K = ½ I w²

Let's calculate the values of the given bodies

a) solid sphere

I = 2/5 M R²

I = 2/5 0.200 0.05²

I = 2 10⁻⁴ kg m²

K = ½ 2 10⁻⁴ 2.5²

K = 6.25 10⁻⁴ J

b) spherical shell

I = 2/3 M R²

I = 2/3 0.200 0.05²

I = 3.33 10⁻⁴ kg m²

K = ½ 3.33 10-4 2.5²

K = 10.41 10⁻⁴ J

c) solid cylinder

I = ½ M R²

I = ½ 0.200 0.05²

I = 2.5 10⁻⁴ kg m²

K = ½ 2.5 10-4 2.5²

K = 7.81 10⁻⁴ J

d) hoop

I = M R²

I = 0.200 0.05²

I = 5 10⁻⁴ kg m²

K = ½ 5 10⁻⁴ 2.5²

K = 15.625 10⁻⁴ J

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Answer:

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Explanation:

Given,

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If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

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Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

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Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

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A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

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Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

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L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

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a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

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So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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