Circular Motion/Gravitational Acceleration Help Needed

[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 centripetal 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.

Thanks in advance!

 

[sankalpmittal]

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 centripetal 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 0^o with vertical. Then apply Newton's second law and form an equation. Resultant force contributes to centripetal acceleration.

 

[FLIHGH]

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 0^o 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?

 

[sankalpmittal]

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..

 

[StaceyPurcher]

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.

Thanks in advance!

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

 

[StaceyPurcher]

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...
So just add the radius of the Earth to your given radius value