Tangential and Radial components of acceleration and tension

[Eohlas]

Homework Statement

One end of a cord is fixed and a small 0.500 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m as shown. (I don't have access to a camera right now, apologies.) When the angle is 20.0 degrees, the speed of the object is 8.00 m/s. At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration, and (c) the total acceleration. (d)Is the answer to this question changed if the object is swinging down toward its lowest point instead of swinging up? Explain.

Known data:
mass is 0.500 kg
Radius is 2.00
Angle is 20.0 degrees
Speed of object when it is 20.0 degrees to the right of the vertical is 8.00 m/s

Homework Equations

Radial acceleration = (V²)/radius)
mg = 4.9N
mgsinθ = 1.68 N
mgcosθ = 4.60 N

The Attempt at a Solution

I solved it earlier today, but now that I've taken a second look, I'm starting to think I didn't solve it correctly.

I set up the equations of motion for both the y and x components. (I've set up the coordinate system so the y-axis is along the cord)
ƩFy: T - mgcosθ - ma_r = 0 (originally, I did not include the centripetal force)
ƩFx: -mgsinθ= -ma (the force is acting on the object in the opposite direction)
A) Find tension in the cord
T = mgcosθ + ma_r

A_r is V²/R so, (8.00^2)/2.00 --> 32.0 m/s^2 ( seems excessive to me)
T = mgcosθ + ma_r --> 4.60N + 16.0N = 20.6N

B) Find tangential and radial components of acceleration

Already found A_r in part A. I think tangential acceleration is obtained from -mgsinθ = -ma, which when I cancel out the mass, is -gsinθ= -a which yields -3.35 m/s^2.

C) the total acceleration is √( (A_tangential)² + A_r²) ) which yields 32.2 m/s^2

D) The answers to the previous questions will not change if the object was swinging in the opposite direction, because the magnitudes of the tension, radial acceleration, and tangential acceleration would still be the same. In other words, they are independent of the direction of motion.

Thank you for taking the time to read this.

 

[ehild]

Homework Statement

One end of a cord is fixed and a small 0.500 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m as shown. (I don't have access to a camera right now, apologies.) When the angle is 20.0 degrees, the speed of the object is 8.00 m/s. At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration, and (c) the total acceleration. (d)Is the answer to this question changed if the object is swinging down toward its lowest point instead of swinging up? Explain.

Known data:
mass is 0.500 kg
Radius is 2.00
Angle is 20.0 degrees
Speed of object when it is 20.0 degrees to the right of the vertical is 8.00 m/s

Homework Equations

Radial acceleration = (V²)/radius)
mg = 4.9N
mgsinθ = 1.68 N
mgcosθ = 4.60 N

The Attempt at a Solution

I solved it earlier today, but now that I've taken a second look, I'm starting to think I didn't solve it correctly.

I set up the equations of motion for both the y and x components. (I've set up the coordinate system so the y-axis is along the cord)
ƩFy: T - mgcosθ - ma_r = 0 (originally, I did not include the centripetal force)
ƩFx: -mgsinθ= -ma (the force is acting on the object in the opposite direction)
A) Find tension in the cord
T = mgcosθ + ma_r

A_r is V²/R so, (8.00^2)/2.00 --> 32.0 m/s^2 ( seems excessive to me)
T = mgcosθ + ma_r --> 4.60N + 16.0N = 20.6N

B) Find tangential and radial components of acceleration

Already found A_r in part A. I think tangential acceleration is obtained from -mgsinθ = -ma, which when I cancel out the mass, is -gsinθ= -a which yields -3.35 m/s^2.

C) the total acceleration is √( (A_tangential)² + A_r²) ) which yields 32.2 m/s^2

D) The answers to the previous questions will not change if the object was swinging in the opposite direction, because the magnitudes of the tension, radial acceleration, and tangential acceleration would still be the same. In other words, they are independent of the direction of motion.

Thank you for taking the time to read this.

Welcome to PF, Eohlas!

Your solution is correct.

ehild

 

[Eohlas]

Thanks for the welcome, although I actually registered almost two years ago. Heh. And thanks for taking the time to check my solution. :)

 

[ehild]

Thanks for the welcome, although I actually registered almost two years ago. Heh. :)

I saw that this was your first post. Was it? You came out from hide-out just now. .