A flywheel is rotating about a fixed axis with a kinetic energy of 360 joules and an angular speed of 30 radians per second. Determine the moment of inertia of the flywheel about its axis of rotation. The moment of inertia is a measure of how mass is distributed in a rotating object and how difficulRead more
A flywheel is rotating about a fixed axis with a kinetic energy of 360 joules and an angular speed of 30 radians per second. Determine the moment of inertia of the flywheel about its axis of rotation.
The moment of inertia is a measure of how mass is distributed in a rotating object and how difficult it is to change the rotational motion of that object. It plays a role in rotational dynamics just like the role played by mass in linear motion. When the angular speed and the moment of inertia are known, then the kinetic energy of the rotating object can be calculated.
The calculations for a given flywheel will show 0.8 kg·m² to be its moment of inertia, meaning its distribution of mass and rotational resistance matches this quantity. Understanding moments of inertia supports designing and even analyzing systems like an engine, turbines, or any mechanical flywheel for efficient safety and successful operation in general.
Thus, the moment of inertia of the flywheel is highly indicative of how it could store rotational energy and resist changes in its motion.
The time period of a planet's revolution around the Sun is governed by Kepler's Third Law, which establishes a relationship between the orbital period and the distance of the planet from the Sun. According to this law, the square of the orbital period of a planet is proportional to the cube of its aRead more
The time period of a planet’s revolution around the Sun is governed by Kepler’s Third Law, which establishes a relationship between the orbital period and the distance of the planet from the Sun. According to this law, the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. This means that as the distance of a planet from the Sun increases, its orbital period becomes significantly longer.
In this scenario, the distances of two planets from the Sun are 10¹³ meters and 10¹² meters, respectively. The ratio of their orbital periods can be determined using Kepler’s Third Law. For the first planet, which is farther from the Sun, the time period increases because the gravitational pull decreases with distance, resulting in a slower orbital speed.
Using the law, it is found that the ratio of the time periods of the two planets is 10√10. This value shows that the first planet, being ten times farther from the Sun, takes considerably longer to complete one revolution than the second planet. This result demonstrates the profound effect of distance on the orbital dynamics of celestial bodies in a solar system.
When a body of mass m is moved from the surface of the Earth to a height equal to three times the Earth's radius h = 3R, there is a change in its gravitational potential energy. Gravitational potential energy depends on the position of the body relative to the center of the Earth and decreases as thRead more
When a body of mass m is moved from the surface of the Earth to a height equal to three times the Earth’s radius h = 3R, there is a change in its gravitational potential energy. Gravitational potential energy depends on the position of the body relative to the center of the Earth and decreases as the distance from the center increases.
At the surface of the Earth, the body’s potential energy is determined by the distance R from the center. When the body is taken to a height of h = 3R, the total distance from the Earth’s center becomes 4R. The gravitational potential energy at these two points differs because potential energy is inversely proportional to the distance from the center of the Earth.
The change in gravitational potential energy is calculated as the difference between the potential energy at the surface and at the height h = 3R. After simplifying the relationship, it is found that the change in potential energy is mgR/4.
This result reflects how gravitational potential energy decreases with increasing distance from the center of the Earth. It also demonstrates the significance of height and mass in calculating energy changes during such movements, essential in space travel and satellite deployment.
The escape velocity is defined as the minimum speed by which a body has to move away from the influence of a planet's gravitational pull without further propulsion. For Earth, it would depend on how far a distance one moves away from the center of the Earth. The escape velocity is proportional to thRead more
The escape velocity is defined as the minimum speed by which a body has to move away from the influence of a planet’s gravitational pull without further propulsion. For Earth, it would depend on how far a distance one moves away from the center of the Earth. The escape velocity is proportional to the square root of the reciprocal of the radius R of the Earth at the surface.
Now consider a platform located at a height equal to the radius of the Earth (R) above its surface. This makes the total distance from the Earth’s center to the platform (2R). Since escape velocity decreases with an increase in distance from the planet’s center, the escape velocity from this platform is less than that from the Earth’s surface.
This, knowing that escape velocity varies in inverse proportion to the distance from the center’s square root, makes escape velocity at a distance of 2R a fraction of the surface escape velocity. The factor f linking between the two velocities can then be calculated as follows, 1/√2 . Therefore, if the escape velocity from earth’s surface is some quantity ‘a’, then from this ‘platform’ it becomes an amount ‘a divided by f’.
This shows that gravity depends on distance, indicating that the escape velocity at altitude varies.
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet's influence. The conventional value of escape velocity from Earth's surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees wiRead more
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet’s influence. The conventional value of escape velocity from Earth’s surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees with the vertical, then the whole speed needed for escape becomes a different story.
The required launch velocity, at an angle, would be the summation of the components of velocity. The vertical component needs to equal the escape velocity so that the body can rise infinitely against the gravity of Earth. Now, for an angle of 60 degrees, we find that in relation between the required launch velocity and the escape velocity, it shows that the necessary velocity for escape is increasing.
Specifically, when launched at an angle of 60 degrees, the total velocity needed becomes approximately 11√3 km/s. This is because the vertical component of the launch velocity has to compensate for the angle at which the body is launched. Consequently, a launch speed of 11√3 km/s guarantees that the vertical component will match the required escape velocity of 11.2 km/s and the body will break free from Earth’s gravitational influence. This is how launch angle affects the dynamics of reaching escape velocity.
A flywheel rotating about fixed axis has a kinetic energy of 360 joule when its angular speed is 30 radian/sec. The moment of inertia of the wheel about the axis of rotation is
A flywheel is rotating about a fixed axis with a kinetic energy of 360 joules and an angular speed of 30 radians per second. Determine the moment of inertia of the flywheel about its axis of rotation. The moment of inertia is a measure of how mass is distributed in a rotating object and how difficulRead more
A flywheel is rotating about a fixed axis with a kinetic energy of 360 joules and an angular speed of 30 radians per second. Determine the moment of inertia of the flywheel about its axis of rotation.
The moment of inertia is a measure of how mass is distributed in a rotating object and how difficult it is to change the rotational motion of that object. It plays a role in rotational dynamics just like the role played by mass in linear motion. When the angular speed and the moment of inertia are known, then the kinetic energy of the rotating object can be calculated.
The calculations for a given flywheel will show 0.8 kg·m² to be its moment of inertia, meaning its distribution of mass and rotational resistance matches this quantity. Understanding moments of inertia supports designing and even analyzing systems like an engine, turbines, or any mechanical flywheel for efficient safety and successful operation in general.
Thus, the moment of inertia of the flywheel is highly indicative of how it could store rotational energy and resist changes in its motion.
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See lessThe distance of two planets from the sun are 10¹³ m and 10¹² m respectively. The ratio of time periods of the planets is
The time period of a planet's revolution around the Sun is governed by Kepler's Third Law, which establishes a relationship between the orbital period and the distance of the planet from the Sun. According to this law, the square of the orbital period of a planet is proportional to the cube of its aRead more
The time period of a planet’s revolution around the Sun is governed by Kepler’s Third Law, which establishes a relationship between the orbital period and the distance of the planet from the Sun. According to this law, the square of the orbital period of a planet is proportional to the cube of its average distance from the Sun. This means that as the distance of a planet from the Sun increases, its orbital period becomes significantly longer.
In this scenario, the distances of two planets from the Sun are 10¹³ meters and 10¹² meters, respectively. The ratio of their orbital periods can be determined using Kepler’s Third Law. For the first planet, which is farther from the Sun, the time period increases because the gravitational pull decreases with distance, resulting in a slower orbital speed.
Using the law, it is found that the ratio of the time periods of the two planets is 10√10. This value shows that the first planet, being ten times farther from the Sun, takes considerably longer to complete one revolution than the second planet. This result demonstrates the profound effect of distance on the orbital dynamics of celestial bodies in a solar system.
See lessA body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3 R. Then change in gravitational
When a body of mass m is moved from the surface of the Earth to a height equal to three times the Earth's radius h = 3R, there is a change in its gravitational potential energy. Gravitational potential energy depends on the position of the body relative to the center of the Earth and decreases as thRead more
When a body of mass m is moved from the surface of the Earth to a height equal to three times the Earth’s radius h = 3R, there is a change in its gravitational potential energy. Gravitational potential energy depends on the position of the body relative to the center of the Earth and decreases as the distance from the center increases.
At the surface of the Earth, the body’s potential energy is determined by the distance R from the center. When the body is taken to a height of h = 3R, the total distance from the Earth’s center becomes 4R. The gravitational potential energy at these two points differs because potential energy is inversely proportional to the distance from the center of the Earth.
The change in gravitational potential energy is calculated as the difference between the potential energy at the surface and at the height h = 3R. After simplifying the relationship, it is found that the change in potential energy is mgR/4.
This result reflects how gravitational potential energy decreases with increasing distance from the center of the Earth. It also demonstrates the significance of height and mass in calculating energy changes during such movements, essential in space travel and satellite deployment.
See lessThe earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the earth. The value of f is
The escape velocity is defined as the minimum speed by which a body has to move away from the influence of a planet's gravitational pull without further propulsion. For Earth, it would depend on how far a distance one moves away from the center of the Earth. The escape velocity is proportional to thRead more
The escape velocity is defined as the minimum speed by which a body has to move away from the influence of a planet’s gravitational pull without further propulsion. For Earth, it would depend on how far a distance one moves away from the center of the Earth. The escape velocity is proportional to the square root of the reciprocal of the radius R of the Earth at the surface.
Now consider a platform located at a height equal to the radius of the Earth (R) above its surface. This makes the total distance from the Earth’s center to the platform (2R). Since escape velocity decreases with an increase in distance from the planet’s center, the escape velocity from this platform is less than that from the Earth’s surface.
This, knowing that escape velocity varies in inverse proportion to the distance from the center’s square root, makes escape velocity at a distance of 2R a fraction of the surface escape velocity. The factor f linking between the two velocities can then be calculated as follows, 1/√2 . Therefore, if the escape velocity from earth’s surface is some quantity ‘a’, then from this ‘platform’ it becomes an amount ‘a divided by f’.
This shows that gravity depends on distance, indicating that the escape velocity at altitude varies.
See lessFor a satellite velocity from earth is 11.2 km/s. If a body is to be launched at an angle of 60ᵒ with the vertical, then escape velocity will be
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet's influence. The conventional value of escape velocity from Earth's surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees wiRead more
While ejecting a body from the Earth, escape velocity becomes very crucial to overcome gravitational force so that it becomes free from the planet’s influence. The conventional value of escape velocity from Earth’s surface is about 11.2 km/s. But if the launch angle is considered to be 60 degrees with the vertical, then the whole speed needed for escape becomes a different story.
The required launch velocity, at an angle, would be the summation of the components of velocity. The vertical component needs to equal the escape velocity so that the body can rise infinitely against the gravity of Earth. Now, for an angle of 60 degrees, we find that in relation between the required launch velocity and the escape velocity, it shows that the necessary velocity for escape is increasing.
Specifically, when launched at an angle of 60 degrees, the total velocity needed becomes approximately 11√3 km/s. This is because the vertical component of the launch velocity has to compensate for the angle at which the body is launched. Consequently, a launch speed of 11√3 km/s guarantees that the vertical component will match the required escape velocity of 11.2 km/s and the body will break free from Earth’s gravitational influence. This is how launch angle affects the dynamics of reaching escape velocity.
See less