Calculating the Angle of Force for 0.140 kg Ball Attached to a 0.600 m String

[Paulbird20]

Estimate the force a person must exert on a string attached to a 0.140 kg ball to make the ball revolve in a circle when the length of the string is 0.600 m. The ball makes 1.00 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?]

I attached the figure below.

Ok so i did all the calculations to get the magnitude of the force and got 3.518 which is correct.

I assumed to get the angle i just do tan inverse of 3.518 which gives me 74.13. Then i minus that from 90 to get angle of 15.86 degrees. Which is not correct . Any help on how to get the angle.

 

[Doc Al]

I assumed to get the angle i just do tan inverse of 3.518 which gives me 74.13.

Why assume this? (At the very least you need two sides of a triangle to come up with a tangent.) Instead, figure out the angle by considering the components of the tension, which you must have figured out to get the tension.

 

[aq1q]

hey, could u explain how u get 3.518.. but to solve for the angle..
draw a free body diagram. Then ask yourself, is the ball accelerating down? its not. So what does this tell u? why isn't the ball accelerating down. what makes the ball stay in the same position (vertically)?

 

[Paulbird20]

hmm,

to get the tension i did.

2 * 3.14* .600 / .25 =15.07

then 15.07/.600 * .140 to get 3.518 at the force.

Should i use those two end results as my components and square both then take square root of the result?

 

[aq1q]

hmm,

to get the tension i did.

2 * 3.14* .600 / .25 =15.07

then 15.07/.600 * .140 to get 3.518 at the force.

Should i use those two end results as my components and square both then take square root of the result?

are you sure u are right, to those exact decimals? I am getting an answer is 3. something.. but not 3.5. does the book say 3.518?.. because i can't follow your work

where do u get that .25 from.

 

[aq1q]

remember Ar(or Ac)=w^2r

 

[Doc Al]

hmm,

to get the tension i did.

2 * 3.14* .600 / .25 =15.07

then 15.07/.600 * .140 to get 3.518 at the force.

Should i use those two end results as my components and square both then take square root of the result?

I'm not following where these expression come from. Or why you think 3.518 N is the correct answer for the tension.

Redo this step by step, applying Newton's 2nd law to both vertical and horizontal force components. Start by specifying what forces act on the ball.

 

[Paulbird20]

Yes 3.518 is the correct answer rounded.

I applied Newtons second to the radial direction. Horizontal of course.

V= 2* 3.14 * R / T
to get the time i divided the 1 revolution by 4 to get .25.
the radius is given as .600.
using the above formula i got 15.07 rounded. ( i used a calculator so i stored the entire data value un rounded)
then next i used
F(tension) = M * V^2 / r
Plugging in my data i arrived at the correct answer of 3.518 rounded 3 decimal places

 

[Paulbird20]

Still having trouble finding the components to arrive at the degree.

 

[Doc Al]

Yes 3.518 is the correct answer rounded.

I applied Newtons second to the radial direction. Horizontal of course.

V= 2* 3.14 * R / T

This makes sense for the speed.

to get the time i divided the 1 revolution by 4 to get .25.

I don't understand this: The time for 1 revolution was given as 1 second. Why divide by 4?

the radius is given as .600.

No, the length of the string was .6 m, not the radius.

using the above formula i got 15.07 rounded. ( i used a calculator so i stored the entire data value un rounded)
then next i used
F(tension) = M * V^2 / r
Plugging in my data i arrived at the correct answer of 3.518 rounded 3 decimal places

The tension acts at an angle; only the horizontal component equals the centripetal force. (Note that you ignored the weight of the ball, despite instructions not to. That's why the angle doesn't enter into your calculations.)

Draw yourself a force diagram for the ball. There are two forces acting: The tension (at some angle theta) and the weight.

 

[aq1q]

well.. I can tell you how to do get the degrees. But i am not sure how/why u divide the time by 4.
lets see it like this... v=wr.. right? w=2pi/1s r is .6cos(theta)--(not .6) therefore v should be 2pi(.6cos(theta))/1s..

 

[aq1q]

ya, doc al's explanation is more detailed.. but its the same thing

 

[aq1q]

i am trying to help you, but u see.. u need the FT. and If the FT is wrong.. then i don't think the degree is right.