Find the tension in the lower string and the rotation rate

• jfnn
And so do your units.You ended up with v^2/r = mg, so your velocity is positive. Since the tangential speed is positive, the angular velocity is positive (even though the acceleration is inward, and the force is inward, and the radial position is decreasing). So, you are fine.Yes, radians per second is the correct unit. Revolutions per second is a possible unit, but it is not the SI unit, so I would avoid it.RPM is a unit of frequency (revolutions per minute). It is equal to 2π rad/s. If you happen to remember that, then you can convert from rad/s to RPM. Otherwise
jfnn

Homework Statement

A 700g ball rotates around a vertical shaft supported by two strings. If the tension in the upper string is 20.0 N, determine (I have attached the file)

a. The tension in the lower string.

b. The rotation rate in rpm of the system.

[/B]
F=ma

F= m v^2/r

The Attempt at a Solution

Sum of forces in y direction = 0 (again, I am not sure if this is zero, but I assume zero acceleration in y direction)

Tension in the y direction of 1 (T1y) - tension in the y direction of 2 (T2y) - weight (mg) = 0

T1sin theta - T2sin theta - mg = 0

20sin(theta) - T2sin(theta) = mg

sin(theta) * (20 - T2) = mg

T2 = 20 - mg/sin(theta)

theta = tan^-1 (0.35/0.50) --> Theta = 35

T2 = 20N - 11.96 N

T2 = 8.04 N ---> Can anyone verify if this is right?

b) w = v^2/r but I need v

sum of forces in x direction = mv^2/r

-T1cos(theta) - T2cos(theta) --> Theta is 35

T1 = 20 N
T2 = 8.04 N

F = -20cos(35) - 8.04cos(35)

F = -22.969 N

-22.969 N = mv^2/r --> -22.969 = (0.7) v^2/(0.5)

V^2 = -16.406 (not sure why this is negative/where to go from here...) I am stuck.

Any help is greatly appreciated! Thank you!

Attachments

• Screen Shot 2017-08-06 at 7.30.21 PM.png
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Your work looks good. The sign issue is related to choosing what you want to call the positive direction for the forces along the radius of the circular motion. If you want to take the positive r direction to be outward, then the acceleration is negative (centripetal). But then, your net force also acts toward the center, so it is also negative.

If you want to choose the positive direction to be inward, towards the center, then both your force and your acceleration will be positive.

jfnn
TSny said:
Your work looks good. The sign issue is related to choosing what you want to call the positive direction for the forces along the radius of the circular motion. If you want to take the positive r direction to be outward, then the acceleration is negative (centripetal). But then, your net force also acts toward the center, so it is also negative.

If you want to choose the positive direction to be inward, towards the center, then both your force and your acceleration will be positive.
TSny said:
Your work looks good. The sign issue is related to choosing what you want to call the positive direction for the forces along the radius of the circular motion. If you want to take the positive r direction to be outward, then the acceleration is negative (centripetal). But then, your net force also acts toward the center, so it is also negative.

If you want to choose the positive direction to be inward, towards the center, then both your force and your acceleration will be positive.

Thank you so much, I am going to attempt this again in the morning with that though. Much appreciated!

jfnn said:

The Attempt at a Solution

Sum of forces in y direction = 0 (again, I am not sure if this is zero, but I assume zero acceleration in y direction)
Tension in the y direction of 1 (T1y) - tension in the y direction of 2 (T2y) - weight (mg) = 0

T1sin theta - T2sin theta - mg = 0
20sin(theta) - T2sin(theta) = mg
sin(theta) * (20 - T2) = mg
T2 = 20 - mg/sin(theta)
theta = tan^-1 (0.35/0.50) --> Theta = 35
T2 = 20N - 11.96 N
T2 = 8.04 N ---> Can anyone verify if this is right?
method is fine, nope, won't check your math
b) w = v^2/r but I need v
what is this?
sum of forces in x direction = mv^2/r
is this with the correct sign?
EDIT: did not see that T scooped me ... but I agree totally.

jfnn
rude man said:
method is fine, nope, won't check your math what is this?is this with the correct sign?
EDIT: did not see that T scooped me ... but I agree totally.

w = rotation rate, which I am asked to find... that is the equation for w.

Thank you, I appreciate your second opinion.

Cheers

jfnn said:
w = rotation rate, which I am asked to find... that is the equation for w.

Thank you, I appreciate your second opinion.

Cheers
It's w = v/r, not w=v^2/r, but all in all you surely did a fine job IMO.

jfnn
rude man said:
It's w = v/r, not w=v^2/r, but all in all you surely did a fine job IMO.

Yes, you are right! Thank you for the correction.

TSny said:
Your work looks good. The sign issue is related to choosing what you want to call the positive direction for the forces along the radius of the circular motion. If you want to take the positive r direction to be outward, then the acceleration is negative (centripetal). But then, your net force also acts toward the center, so it is also negative.

If you want to choose the positive direction to be inward, towards the center, then both your force and your acceleration will be positive.

So I attempted the solution again. I decided to say that the inward direction (direction of acceleration and force) is positive. From this I got the same values, just without a negative. I got a velocity of +/- 4.05 m/s.

I had a quick question about the velocity I was hoping you (or someone else) could answer. Since the velocity is +/-, I would choose the positive value because the velocity is related to the acceleration ( a = v^2/r) so it should have the same sign?

After choosing the positive value I used the formula w = v / r where r = 0.5 m... I got 2 = (4.05 m/s)/ (0.5m) --> w is then 8.1 rad/s

My issue here is: Do I assume I have radians here or would it be 8.1 revolutions per second?

I assumed it was 8.1 radians/second and converted to revolutions per minute. Thus obtaining, 77.4 rpm.

Is that the correct way or would I assume I have rps? Thank you.

jfnn said:
So I attempted the solution again. I decided to say that the inward direction (direction of acceleration and force) is positive. From this I got the same values, just without a negative. I got a velocity of +/- 4.05 m/s.
In the equation a = v2/r, v represents speed, not velocity. Speed is always positive, so you would chose the + sign and get v = 4.05 m/s.

After choosing the positive value I used the formula w = v / r where r = 0.5 m... I got 2 = (4.05 m/s)/ (0.5m) --> w is then 8.1 rad/s

My issue here is: Do I assume I have radians here or would it be 8.1 revolutions per second?
It's radians. ω = v/r is derived from θ = s/r where s is arc length, r is radius, and θ is subtended angle in radians.

If you want access to some greek letters and various math symbols, click on the ##\Sigma## symbol on the shaded tool bar at the top of the posting window. This tool bar also has superscript and subscript tools.

I assumed it was 8.1 radians/second and converted to revolutions per minute. Thus obtaining, 77.4 rpm.
Good.

jfnn
TSny said:
In the equation a = v2/r, v represents speed, not velocity. Speed is always positive, so you would chose the + sign and get v = 4.05 m/s.

It's radians. ω = v/r is derived from θ = s/r where s is arc length, r is radius, and θ is subtended angle in radians.

If you want access to some greek letters and various math symbols, click on the ##\Sigma## symbol on the shaded tool bar at the top of the posting window. This tool bar also has superscript and subscript tools.

Good.

Okay. That makes sense. Thank you so much for all your help.

1. What is the purpose of finding the tension in the lower string and the rotation rate?

The purpose of finding the tension in the lower string and the rotation rate is to determine the forces acting on the object, as well as its angular velocity. This information is important for understanding the motion and stability of the object.

2. How is the tension in the lower string calculated?

The tension in the lower string can be calculated using the equation T = mω²r, where T is the tension, m is the mass of the object, ω is the angular velocity, and r is the radius of rotation. This equation is derived from the Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration.

3. What factors affect the tension in the lower string?

The tension in the lower string is affected by the mass of the object, the angular velocity, and the radius of rotation. Other factors such as friction, air resistance, and external forces may also affect the tension.

4. How does the rotation rate affect the tension in the lower string?

The rotation rate, or angular velocity, directly affects the tension in the lower string. As the rotation rate increases, the tension also increases due to the centrifugal force acting on the object. This can cause the string to stretch and potentially lead to its failure if the tension becomes too great.

5. Why is it important to know the tension in the lower string and the rotation rate?

Knowing the tension in the lower string and the rotation rate is important for ensuring the safety and stability of the object in motion. It also allows for accurate predictions and analysis of the object's behavior. In certain applications, such as in engineering and physics experiments, this information is crucial for making precise calculations and designing effective systems.

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