Uniform Circular Motion Notes

Period (T)

The time it takes to complete one full revolution in a circular motion.

Constant speed

When an object's speed remains unchanged over time.

Uniform Circular Motion

Motion in a circular path where the speed is constant but the direction changes.

Centripetal force (Fc)

The force that acts on an object moving in a circular path, directed towards the center of the circular path.

Gravitational field strength (g)

The force of gravity acting on an object per unit mass.

Tension (F_T)

The force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting at either end.

Net force (Fnet)

The overall force acting on an object, taking into account all individual forces.

Mass (m)

A measure of the amount of matter in an object, usually measured in kilograms.

Balanced forces

Forces that are equal in size and opposite in direction, resulting in no change in motion.

Instantaneous velocity (V)

The velocity of an object at a specific instant in time.

Kinetic energy (KE)

The energy of an object due to its motion, calculated as 1/2 mv² where m is mass and v is velocity.

Acceleration (a)

The rate of change of velocity of an object, typically measured in meters per second squared (m/s²).

Frictional force (Ff)

The force that opposes the relative motion of solid surfaces, fluid layers, or material elements sliding against each other.

Centripetal acceleration (a_c)

The acceleration that is directed towards the center of the circle, keeping an object moving in its circular path; also given by ac = v²/r.

Radius (r)

The distance from the center of a circle to any point on its circumference.

Period of uniform circular motion (T)

The time taken to complete one full revolution around the circle.

Frequency (f)

The number of complete revolutions per unit time, typically measured in hertz (Hz), where f = 1/T.

Linear speed (v)

The distance travelled per unit time along the circular path is calculated as v = 2πr/T. Where T is the period.

Newton's law of universal gravitation

A law stating that every point mass attracts every other point mass in the universe with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

F_{g =} Gmm/r²

This force is known as the gravitational force (F_g), where G is the gravitational constant.

Perfectly uniform circular motion

Motion that occurs when an object travels along a circular path with constant radius and constant speed, maintaining a uniform circular motion.

Centripetal Force Formula

F_c = mv²/r, where F_c is centripetal force, m is mass, v is the linear speed, and r is the radius.

Acceleration Formula

a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

Kinetic Energy Formula

KE = 1/2 mv², where KE is kinetic energy, m is mass, and v is velocity.

Linear Speed Formula

v = 2πr/T, where v is linear speed, r is the radius, and T is the period of the motion.

Centripetal Acceleration Formula

a_c = v²/r, where ac is centripetal acceleration, v is linear speed, and r is the radius.

Newton's Second Law Formula

F_net = ma, where F_net is net force, m is mass, and a is acceleration.

Uniform Circular Motion

A type of motion where an object moves along a circular path at a constant speed, requiring a net centripetal force directed towards the center of the circle, causing a continuous change in the object's direction.

Centripetal Force (Fc)

The necessary net force that keeps an object in uniform circular motion. It acts perpendicular to the object's velocity and towards the center of the circle, calculated using the formula Fc = mv²/r, where m is mass, v is linear speed, and r is the radius.

Factors Affecting Centripetal Force

The centripetal force required for uniform circular motion is influenced by the object's mass, the linear speed of the motion, and the radius of the circular path, demonstrating that an increase in linear speed or mass results in a corresponding increase in centripetal force.

Centripetal Acceleration (a_c)

The acceleration experienced by an object moving in a circular path, directed towards the center of the path. It can be expressed as a_c = v²/r, where v is linear velocity and r is the radius of the curvature.

Gravitational Field Strength (g)

The gravitational force per unit mass at a given point in space, typically measured in newtons per kilogram (N/kg), influences the weight of objects and acts as the centripetal force in planetary motion. Formula: g = F_g/m, where F_g is the gravitational force and m is mass.

Applications of Centripetal Force

Centripetal force concepts are critical in diverse fields, including engineering (designing curved roads), physics (understanding satellites in orbit), and sports (analyzing motion of athletes), highlighting its relevance across various practical and theoretical applications.

Calculate the gravitational field strength on an object with a mass of 10 kg that experiences a gravitational force of 98 N. What is the gravitational field strength (g)?

To find g, use the formula g = F/m. Here, F = 98 N and m = 10 kg. So, g = 98 N / 10 kg = 9.8 N/kg.

An astronaut on the surface of the Moon weighs 60 N. What is the gravitational field strength (g) on the Moon if the astronaut has a mass of 10 kg?

Using the formula g = F_g/m, where F_g = 60 N and m = 10 kg, we find g = 60 N / 10 kg = 6 N/kg.

If the gravitational field strength on Mars is approximately 3.7 N/kg, what would be the weight of an object with a mass of 15 kg on Mars?

Weight can be calculated using the formula Weight = mg. Here, m = 15 kg and g = 3.7 N/kg. So, Weight = 15 kg * 3.7 N/kg = 55.5 N.

A satellite orbits the Earth at an altitude where the gravitational field strength is 8.7 N/kg. Calculate the gravitational force acting on a 500 kg satellite.

Use the formula F = mg. Here, m = 500 kg and g = 8.7 N/kg. Thus, F = 500 kg * 8.7 N/kg = 4350 N.

On an alien planet, an object experiences a gravitational force of 40 N while having a mass of 5 kg. What is the gravitational field strength (g) on that planet?

Apply the formula g = F/m. Here, F_g = 40 N and m = 5 kg. So, g = 40 N / 5 kg = 8 N/kg.

If the gravitational field strength on Jupiter is 24.79 N/kg, how much would a 70 kg object weigh on Jupiter?

Weight can be calculated as Weight = mg. Here, m = 70 kg and g = 24.79 N/kg. Therefore, Weight=F_g = 70 kg * 24.79 N/kg = 1735.3 N.

A 25 kg object experiences a gravitational force of 245 N on Earth. What is the gravitational field strength (g) experienced by the object?

Use g = F/m. Here, F_g = 245 N and m = 25 kg. Thus, g = 245 N / 25 kg = 9.8 N/kg.

Calculating the coefficient of friction for a car turning a corner

To find the coefficient of friction needed to keep a car moving in a circle without sliding, use the equation μ = F_c/ F_n, where F_c is the centripetal force and F_n is the normal force acting on the car. F_n = F_g and can be calculated as Fn = mg, where m is the mass of the car and g is the gravitational field strength. Fc can be determined using Fc = mv²/r, where v is the velocity of the car and r is the radius of the turn. μ = v²/gr,

Centripetal force (F_c) during a turn

The net force required to keep an object moving in circular motion, expressed as F_c= mv²/r, where m is mass, v is linear speed, and r is radius of the turn.

Normal force (F_n)

The perpendicular force exerted by a surface on an object in contact with it, typically equal to the weight of the object when on a flat surface: F_n= mg, where m is mass and g is gravitational field strength.

A car of mass 1000 kg turns a corner with a radius of 50 m at a speed of 20 m/s. Compute the necessary friction coefficient.

First, find F_c = mv²/r = 1000 kg * (20 m/s)² / 50 m = 8000 N. Assuming F_n = mg = 1000 kg * 9.8 N/kg = 9800 N, then μ = F_c / F_n = 8000 N / 9800 N = 0.816.

What is gravitational field strength (g) at a given point in space?

The gravitational force per unit mass, influencing the weight of objects and acting as the centripetal force in planetary motion. Formula: g = F_g/m, where F_g is the gravitational force and m is mass.

Calculating the period (T) of a satellite around a planet

Calculating the period (T) of a satellite around a planet
Definition: The period of a satellite is the time it takes to complete one full orbit around a planet. It can be calculated using the formula GMT² = 4π²R³, where r is the distance from the center of the planet to the satellite, G is the gravitational constant, and M is the mass of the planet.

Solve for the mass of a planet given the period T of a Sattelite

The period of a satellite is the time it takes to complete one full orbit around a planet. The mass of the planet M can be calculated using the formula GMT² = 4π²R³, where r is the distance from the center of the planet to the satellite, G is the gravitational constant, and M is the mass of the planet.

What is the equation for the work done on a satellite?

The work done on a satellite is defined as the change in its gravitational potential energy as it moves in the gravitational field of a planet. It can be expressed as W = ΔPE = PE final - PE initial, where U is the gravitational potential energy given by U = -GMm/r.

W = -GMm(1/r_f- 1/r_i) = GMm(1/r_i - 1/r_f), where ri is the initial distance from the planet's center and rf is the final distance.

What is the gravitational potential energy of a satellite(any object) in space ?

The gravitational potential energy (PE) of a satellite in space is the energy an object possesses due to its position in a gravitational field, expressed as PE = -GMm/r, where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the distance from the centre of the planet to the satellite.

How do you find the speed of a falling meteorite as it speeds towards a planet?

To find the speed of a falling meteorite, you can use the conservation of mechanical energy principle. The total mechanical energy (kinetic + potential) remains constant, or ΔKE = -ΔPE, so

1/2mv_f² - 1/2mv_i² = - GMm(1/r_i - 1/r_f),

What is the equation for ∆PE for large distances? ,

The equation for ∆PE for large distances is approximated as ∆PE = GMm(1/r_i - 1/r_f), where r_f is the final distance, and r_i is the initial distance from the planet's center. This expression simplifies to reflect changes in gravitational potential energy for objects far from the planet.