Find the tensions in the strings

  • Thread starter Thread starter MyNewPony
  • Start date Start date
  • Tags Tags
    Strings
AI Thread Summary
The discussion focuses on calculating the tensions in two strings attached to objects of equal mass rotating around a shaft. For the first object at distance d, the tension is derived as T = m(ω^2*d), while for the second object at distance 2d, the tension is T = m(ω^2*4d). The key point is that tension opposes centripetal acceleration, and the correct formula for tension is T = m*ω^2*r without squaring the radius. The relationship between tangential and angular acceleration is noted, emphasizing that they are orthogonal. The final conclusion is that the tension in the strings is determined by the mass, angular velocity, and the respective distances from the axis of rotation.
MyNewPony
Messages
30
Reaction score
0

Homework Statement



Two objects of equal mass m are whirling around a shaft with a constant angular velocity ω. The first object is a distance d from the central axis, and the second object is a distance 2d from the axis. You may assume the strings are massless and inextensible. You may ignore the effect of gravity. Find the tensions in the two strings.

Homework Equations



F(net) = ma
α = ω^2*r
a = αr

The Attempt at a Solution



First string:

If gravity is ignored, the only force acting on the object is tension. Thus,

T = ma

and

T = m(ω^2*d^2)

Second string:

T = ma

and

T = m(ω^2*(2d)^2)
T = m(ω^2*4d^2)

Are there any other forces acting on the objects?
 
Physics news on Phys.org
Well you know that the tension will oppose the centripetal acceleration and that the centripetal force will equal the tension. Therefore you are close T=m(w^2*r) where w means omega, or angular velocity, and r is the radius. Just don't square the radius it gets canceled out in the math.
 
Your tangential acceleration |a| is related to |α|*r.

But a is not directed along the string. They are orthogonal. It's α that is directed along the string.

The tension in the string then is given simply by

T = m*ω2*r
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top