# Circular Motion/Gravitational Acceleration Help Needed

• FLIHGH
In summary: All you need to do with this is combine your two formulas of Fg=(GMm)/r^2 and Fc=(mv^2)/r=mathis then cancels out the mass of the smaller object so we don't need to worry about that, then by solving the equation for a, we are able to see the acceleration due to gravity at a certain point above the Earth's surface :approve:just got to make sure that you take into account the radius of the point from Earth's centre of gravity... :approve:That will affect the Fg value...
FLIHGH
Hi, I am having problems with the following two problems:

## Homework Statement

A 1.84kg ball is swung vertically from a 0.80m cord in uniform circular motion at a speed of 2.8m/s. What is the tension in the cord at the bottom of the ball's motion?

## Homework Equations

I have tried using the centripital acceleration formula to find a, and then plugged it into the tension formula of T=ma (is that even right?), but cannot get the correct answer.

## The Attempt at a Solution

w²r = (1.84*pi)² * 0.8 = 2.67 m/s²

Tension = ma = 2*2.67 = 5.34 N

The second problem is:

## Homework Statement

The radius of the Earth is 6.38E6 m and its mass is 5.98E24 kg. What is the acceleration due to gravity at a height of 27391854.50m above the Earth's surface? (We are given that G = 6.67e-11 N*(m^2/kg^2))

## Homework Equations

I have tried using G = 6.67e-11 N*(m^2/kg^2) to find the acceleration, as well as g= -GM/r^2, but cannot come up with anything.

## The Attempt at a Solution

I have used g= -GM/r^2, and plugged in 5980000000000000000000000 for M and 6380000 for r, but nothing is coming out correct.

FLIHGH said:
Hi, I am having problems with the following two problems:

## Homework Statement

A 1.84kg ball is swung vertically from a 0.80m cord in uniform circular motion at a speed of 2.8m/s. What is the tension in the cord at the bottom of the ball's motion?

## Homework Equations

I have tried using the centripital acceleration formula to find a, and then plugged it into the tension formula of T=ma (is that even right?), but cannot get the correct answer.

## The Attempt at a Solution

w²r = (1.84*pi)² * 0.8 = 2.67 m/s²

Tension = ma = 2*2.67 = 5.34 NThanks in advance!

Hii FLIHGH ! Welcome thee to PF !

For first problem , its given that ball is swung in a "vertical" circle. You can't just apply those formulas directly.

Hint: Consider the free body diagram of the ball when it is at bottom , making an angle 0o with vertical. Then apply Newton's second law and form an equation. Resultant force contributes to centripetal acceleration.

sankalpmittal said:
Hii FLIHGH ! Welcome thee to PF !

For first problem , its given that ball is swung in a "vertical" circle. You can't just apply those formulas directly.

Hint: Consider the free body diagram of the ball when it is at bottom , making an angle 0o with vertical. Then apply Newton's second law and form an equation. Resultant force contributes to centripetal acceleration.

Thank you for the response-- what exactly is a vertical circle?

FLIHGH said:
Thank you for the response-- what exactly is a vertical circle?

Might you consider this animation of motion in vertical circle :http://www.learnerstv.com/animation/animation.php?ani= 40&cat=physics

Or see here (if your QuickTime player is not out of date) : http://phys23p.sl.psu.edu/phys_anim/mech/embederQ2.30050.html

In short , how would you solve this problem if it was a horizontal circle ? You would just put tension in string equal to centripetal force , and solve for it. In vertical circle (take a ball tied to a string and revolve it with the help of your ball and socket joint , in a plane perpendicular to ground) , gravity also plays the role. Centripetal force is resultant force of gravity and tension of string in a vertical circle. Also , its variable and depends on angle with vertical , for a given radius in vertical circle.

Now once you realize what a "vertical circle" is , get back on post #2 , and follow the hint I gave..

FLIHGH said:
The second problem is:

## Homework Statement

The radius of the Earth is 6.38E6 m and its mass is 5.98E24 kg. What is the acceleration due to gravity at a height of 27391854.50m above the Earth's surface? (We are given that G = 6.67e-11 N*(m^2/kg^2))

## Homework Equations

I have tried using G = 6.67e-11 N*(m^2/kg^2) to find the acceleration, as well as g= -GM/r^2, but cannot come up with anything.

## The Attempt at a Solution

I have used g= -GM/r^2, and plugged in 5980000000000000000000000 for M and 6380000 for r, but nothing is coming out correct.

All you need to do with this is combine your two formulas of Fg=(GMm)/r^2 and Fc=(mv^2)/r=ma
this then cancels out the mass of the smaller object so we don't need to worry about that, then by solving the equation for a, we are able to see the acceleration due to gravity at a certain point above the Earth's surface

just got to make sure that you take into account the radius of the point from Earth's centre of gravity...
That will affect the Fg value...

## 1. What is circular motion?

Circular motion is the movement of an object along a circular path. This can occur when an object is under the influence of a centripetal force, causing it to continuously change direction.

## 2. How is circular motion related to gravitational acceleration?

Circular motion and gravitational acceleration are related through Newton's law of universal gravitation. According to this law, any two objects with mass will experience a gravitational force towards each other. In circular motion, this force acts as the centripetal force that keeps an object moving along a circular path.

## 3. What is centripetal force?

Centripetal force is the force that acts towards the center of a circular path and keeps an object moving along that path. In circular motion, this force is provided by the gravitational force, tension force, or any other force that acts towards the center.

## 4. How is centripetal force calculated?

The formula for calculating centripetal force is Fc = (mv^2)/r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

## 5. Can circular motion occur without a centripetal force?

No, circular motion cannot occur without a centripetal force. Without this force, the object would move in a straight line instead of a circular path. This is because an object will continue to move in a straight line unless acted upon by a force.

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