Question about vertical centripital force

[barryj]

Homework Statement

What is the tension of the string at these places.
See figure. Vertical circular motion. m = 1kg, assume g = 10 m/s^2, velocity = 10 m/sec
a) what is the tension in the string when the mass is 30 deg below the horizon.
b)what is the tension at the top of the loop
c) what is the tension at the bottom of the loop[/B]

Homework Equations

centripetal force F = mv^2/r = 1(10)^2/10 = 10N
weight = mg = (1)(10) = 10 N[/B]

The Attempt at a Solution

I broke up the weight into two parts, one along the direction of the string, the other tangential to the circle. The component of the weight along the direction of the string is mgcos(60 deg) = 1(10)(.866) = 8.66N. So I figure the tension in the string must be the sum of the centripetal force plus the component of the weight in the direction of the string or T = 10 + 8.66 = 18.66.
The tension at the top is 0 N and at the bottom 20 N.[/B]

 

[BvU]

Is the cosine of 30$^\circ$ 0.866 ?

 

[barryj]

Yes, the cosine of 30 deg is 0.866 but I think I should have used sin of 30 deg = 0.5 instead. This would make the tension at 30 deg = 15 N,yes??

 

[BvU]

You're right. Twice
I looked at the angles and got confued (your 30$^\circ$ angle and 60$^\circ$ angle look distractingly alike)
My approach was that the contribution to T would have to go to 0 if the angle went to 0.

Well done.

 

[barryj]

It is good to know I did it correct. Circular motion in the vertical plane is a bit confusing.

 

[haruspex]

Circular motion in the vertical plane is a bit confusing.

Made more confusing by unrealistic questions, perhaps. Maintaining constant tangential speed for an object tied to a string in a vertical circle would require some careful engineering.