A Spacecraft Which Can Be Considered to Be a Hollow Sphere With a Diameter

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All High School Physics Resources

Correct answer:

80N

Explanation:

In order to get the spacecraft spinning, the rockets must supply a torque to the border of the spacecraft.

$\tau = I\alpha = Fr$

Nosotros can calculate the moment of inertia of the spacecraft and the 4 rockets along the edge.

$I = I_{spacecraft} + 4I_{rocket}$

The spacecraft tin can be considered a compatible disk.

$I_{spacecraft} = \frac{1}{2}m_{spacecraft}r^2$

The rocket can exist calculated

$I_{rocket} = m_{rocket}r^2$

So the total moment of inertia

$I = \frac{1}{2}m_{spacecraft}r^2 + 4(m_{rocket}r^2)$

$I = \frac{1}{2}(4000kg)(5m)^2 + 4(400kg)(5m)^2$

I = 50000kgm^2 + 40000kgm^2

I = 90000kgm^2

We can also calculate the angular acceleration of the rocket

$= \frac{\delta \omega}{t}$

Since the spacecraft starts from rest the initial angular velocity is .

The last angular velocity needs to be converted to radians per second.

$\frac{40 rotations}{minute} * \frac{2\pi rad}{rotation} * \frac{1min}{60 seconds} = \frac{4\pi}{3} rad/s$

We also need to convert the 4 minutes to seconds

$4min * \frac{60seconds}{1minute} = 240seconds$

$\alpha = \frac{(\frac{4\pi}{3} rad/s) - 0}{240s}$

$\alpha = 0.02rad/s^2$

Therefore the total torque applied by the rockets is

$\tau = I\alpha$

$\tau = (90000kgm^2)(0.02rad/s^2)$

$\tau = 1570Nm$

Each rocket contributes to the torque. So to determine the torque contributed by one rocket we would divide this by iv

$\frac{1570Nm}{4} = 393Nm$

We can now determine the force applied by one rocket through the equation

$\tau = Fr$

393N = F(5m)

F = 79N

Nosotros can approximate that to about 80 N

A merry-go-round has a mass of 1500kg and radius of . How much net work is required to accelerate information technology from rest to a ration rate of revolution per seconds? Presume information technology is a solid cylinder.

Correct answer:

14800J

Explanation:

We know that the piece of work-kinetic free energy theorem states that the work washed is equal to the change of kinetic free energy. In rotational terms this ways that

$W = \delta KE$

$W = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

In this case the initial angular velocity is 0rad/sec .

We tin can convert our final athwart velocity to radians per second.

$\frac{1rev}{8s}*\frac{2\pi rad}{1rev} = \frac{\pi}{4}rad/s$

We as well can calculate the moment of inertia of the merry-go-round assuming that it is a uniform solid disk.

$I = \frac{1}{2}mr^2$

$I = \frac{1}{2}(1500kg)(8m)^2$

I = 48000kgm^2

We tin can put this into our work equation now.

$W = \frac{1}{2}(48000kgm^2)(\frac{\pi}{4}rad/s)^2 - 0$

W = 14800J

An auto engine slows down from 4000rpm to 2000rpm in . Calculate its angular acceleration.

Correct answer:

-52.34rad/s^2

Caption:

The first thing we demand to practice is convert our velocities to radians to per second.

$\frac{4000rotations}{minute} * \frac{2\pi rad}{1 rotation} * \frac{1minute}{60 seconds} = 418.67rad/s$

$\frac{2000rotations}{minute} * \frac{2\pi rad}{1 rotation} * \frac{1minute}{60 seconds} = 209.33rad/s$

Nosotros can now find the angular dispatch through the equation

$\alpha = \frac{\omega_f - \omega_i}{\delta t}$

$\alpha = \frac{209.33rad/s - 418.67rad/s}{4s}$

$\alpha = -52.34rad/s^2$

What is the angular momentum of a 0.30kg ball revolving on the end of a thin string in a circle of radius 1.50m at an athwart speed of 10.5rad/s ?

Right answer:

7.1kgm^2/s

Caption:

The equation for angular momentum is equal to the moment of inertia multiplied past the angular speed.

$L = I \omega$

The moment of inertia of an object is equal to the mass times the radius squared of the object.

I = mr^2

Nosotros can substitute this into our athwart momentum equation.

$L = mr^2 \omega$

Now nosotros can substitute in our values.

L = (0.30kg)(1.5m)^2(10.5rad/s)

L = 7.1kgm^2/s

An ice skater performs a fast spin by pulling in her outstretched arms close to her body. What happens to her angular momentum most the centrality of rotation?

Possible Answers:

Information technology increases

It changes only information technology is incommunicable to tell which manner

It does non change

It decreases

Correct reply:

It does not alter

Explanation:

According to the constabulary of conservation of momentum, the momentum of a organisation does non modify. Therefore in the example, the angular momentum of the ice skater is constant. When she pulls her arms in, she is reducing her moment of inertia which causes her angular velocity to increase

Several objects roll without slipping down an income of vertical top H, all starting from rest. The objects are a battery (solid cylinder), a frictionless box, a nuptials band (hoop), an empty soup tin can, and a marble (solid sphere). In what club do they reach the bottom of the incline?

Possible Answers:

Empty Soup Can, Wedding Band, Marble, Bombardment, Box

Box, Marble, Battery, Empty Soup Tin can, Hymeneals Band

Nuptials Band, Empty Soup Can, Bombardment, Marble, Box

Hymeneals Ring, Box, Empty Soup Can, Marble, Battery

Marble, Empty Soup Tin can, Bombardment, Box, Wedding ceremony Band

Correct answer:

Box, Marble, Bombardment, Empty Soup Can, Wedding ceremony Band

Explanation:

We can use conservation of energy to compare the gravitational potential energy at the time of the hill to the rotational and kinetic energy at the lesser of the colina.

The box would exist the fastest as all of the gravitational potential energy would convert to translational energy.

The round objects would share the gravitational potential free energy between translational and rotational kinetic energies.

$mgh = \frac{1}{2} mv^2 + \frac{1}{2} Iw^2$

The moment of inertia is equal to a numerical factor () times the mass and radius squared. Since the mass is the same in each term, the speed does non depend on .

$mgh = \frac{1}{2} mv^2 + \frac{1}{2} xmr^2w^2$

$gh = \frac{1}{2}v^2 + \frac{1}{2} xr^2w^2$

Additionally we tin can substitute athwart speed for translational velocity using the equation

w=v/r

$gh = \frac{1}{2} v^2 + \frac{1}{2}xr^2(v/r)^2$

The radius cancels out and we are left with

$gh = \frac{1}{2}v^2 + \frac{1}{2} xv^2$

Therefore the velocity is purely dependent on the numerical factor () in the moment of inertia and the height from which it was released. Since all of these objects were released from the same height, nosotros can examine the moment of inertia for each to decide which volition exist the fasters.

Hoop (wedding ceremony band) = mr^2

Hollow cylinder (empty can) = $\frac{1}{2}m(r1^2 + r_2^2)$

Solid cylinder (Battery) = $\frac{1}{2} mr^2$

Solid sphere (Marble) = $\frac{1}{2} mr^2$

From this nosotros can meet that the marble will reach the bottom at the fastest velocity every bit information technology has the smallest numerical factor. This will be followed by the bombardment, the empty can and the wedding ceremony ring.

A potter's bicycle is rotating around a vertical axis through its heart a frequency of 1.5 rev/s . The wheel tin be considered a uniform deejay of mass 5.0kg and bore 0.40m . The potter then throws a 2.5kg clamper of dirt, approximately shaped as a flat disk of radius 6.0cm , onto the center of the wheel. What is the angular velocity of the wheel after the dirt sticks to it?

Correct answer:

6.85rad/s

Explanation:

Nosotros can use the conservation of angular momentum in social club to solve this problem. The law of conservation of angular momentum states that the momentum before the collision must equal to the momentum afterwards the standoff. Angular momentum is calculated with the equation

$L = I \omega$

Before the standoff we only have the potter's cycle rotating.

We know that the moment of inertia of the cycle can be considered as a uniform deejay.

$I = \frac{1}{2}mr^2$

$I = \frac{1}{2}(5kg)(0.4m)^2$

I = 0.4kgm^2

We tin convert the velocity of the bicycle to rad/s

$1.5rev/s * \frac{2\pi rad}{1rev} = 3\pi rad/s$

We can now summate the momentum earlier the collision.

$L = (0.4kgm^2)(3\pi rad/s)$

$L = 1.2\pi kgm^2/s$

Now information technology is time to analyze the momentum subsequently the standoff. At this point nosotros have added a slice of clay which is now moving at the aforementioned angular velocity as the pottery.

$L = (I_{wheel} + I_{clay})\omega$

We know that the moment of inertia of the clay can be considered as a compatible deejay.

$I = \frac{1}{2}mr^2$

$I = \frac{1}{2}(2.5kg)(0.06m)^2$

I = 0.15kgm^2

We know the angular momentum at the beginning equals the angular momentum at the end.

$1.2\pi kgm^2/s = (0.4kgm^2 +0.15kgm^2) \omega$

Nosotros can at present solve for the angular velocity

$1.2\pi kgm^2/s = (0.55kgm^2) \omega$

$\omega = 6.85rad/s$

Determine the moment of inertia of a 11kg sphere of radius 0.5m when the axis of rotation is through its eye.

Correct answer:

1.375kgm^2

Explanation:

A wheel tin can exist looked at equally a compatible disk. We can then look up the equation for the moment of inertia of a solid cylinder.The equation is

$I = \frac{1}{2}mr^2$

We can now solve for the moment of inertia.

$I = \frac{1}{2}(11kg)(0.5m)^2$

I = 1.375kgm^2

Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an centrality through the center?

Possible Answers:

Both the aforementioned

Correct Answer

The hollow ane

The solid one

Correct respond:

The hollow ane

Explanation:

Since both spheres have the same radius and the same mass, we need to look at the equations for the moment of inertia of a solid sphere and a hollow sphere.

A solid sphere $I = \frac{2}{5}MR^2$

A hollow sphere $I = \frac{2}{3}MR^2$

If both of these have the same mass and radius, the only difference is the constant that is being multiplied by MR^2

In this example the hollow sphere has a larger constant and therefore would have the larger moment of inertia.

This too conceptually makes sense since all the mass is distributed forth the outside of the sphere significant it all has a larger radius. A solid sphere has mass that is both close to the center and farther away, pregnant that it would have a reduced moment of inertia.

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