What are the Components of Acceleration in a Vertical Circle?

[jmwachtel]

Homework Statement

One end of a cord is fixed and a small 0.250 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.50 m as shown in the figure below. When θ = 20.0°, the speed of the object is 7.70 m/s. At this instant, find each of the following.

(a) the tension in the string

(b) the tangential and radial components of acceleration

ar =

at =

(c) the total acceleration

Homework Equations

I have been trying to use ar = mgcos(20)
at = mgsin(20)

Also for the Tension, I was using T= mg(v^2/Rg + cos(20)

The Attempt at a Solution

My numbers are not coming out right, and I am afraid I don't no understad this problem in the slightest. Please help me. Thank You.

 

[jmwachtel]

Any help would be appriciated. I am new to this, please help.

 

[physixguru]

THIS SHOULD HELP YA AT ITS BEST...::

Tension=mg*cos theta

FOR Tangential accln...
TAKE angular displacement as=(pi/180)*20 radian
v=7.70 m/s
Omega=v/R
Use omega^2=0+2*alpha*(pi/180)*20 radian

Calculate alpha...
then calculate t using omega=0+alpha*t

then Tangential accln=v/t

Total accln={at^2+ac^2}^1/2

Radial accln=v^2/r

 

[WiFO215]

Newton's laws will always rescue us.

Tension=mg*cos theta

That is incorrect. That is only so when the object/particle is at the extremes of oscillation and at such points v=0 but it is not so in this case.

I have been trying to use ar = mgcos(20)

That is not right. Radial accelaration is T - mgcos(20) = m(v^2)/R for it is centripetally accelarating.( Fnet= ma! You've got to consider all the forces.)

T= mg(v^2/Rg + cos(20)

Thats right. How come you have used T-mgcos(20)=m(v^2)/R here? Tension should come out right.

Feel free to ask anything else if you still feel unsure about something.

 

[Oerg]

That is not right. Radial accelaration is T - mgcos(20) = m(v^2)/R for it is centripetally accelarating.( Fnet= ma! You've got to consider all the forces.)

I believe you meant centripetal acceleration.

Radial acceleration comes from the other component of the weight

 

[WiFO215]

Huh? It is radially accelerating inwards, yes? Then should radial accn. not equal Centripetal accn.?

 

[jmwachtel]

You said the tension part is incorrect, what is the correct way to find the tension?

 

[Oerg]

Huh? It is radially accelerating inwards, yes? Then should radial accn. not equal Centripetal accn.?

Radial acceleration means the acceleration that is linear to its movement,

Centripetal acceleration means the acceleration that is perpendicular to its movement.

EDIT:
OK so sorry, radial acceleration is centripetal acceleration. The acceleration that is linear to movement is known as tangenial acceleration

 

[jmwachtel]

I figured out the tension and the tagnet acceleration, but I need to figure out the radial now. How might I do that, I don't understand all the stuff that guro did. I used at = gsin(20).

 

[jmwachtel]

Ok, I have solved all of the problem except for one piece:

atotal = 23.9 m/s2 inward and below the cord at °

What does below the cord mean? I know it's asking for an angle.

 

[Oerg]

Hi Jim,For an object moving in a citcle we know that the resultant acceleration is given as the centripetal force of acceleration. In the question, the tension and the weight component add up to this centripetal acceleration. Can you figure this out now?

 

[jmwachtel]

It's Joe. Yes I did! Thank You! I took the tan inverse of at/ar and that have me 8.04. Thank You!