Maximizing Tension in a Pendulum: Exploring Angles and Equations

[3ephemeralwnd]

there is a pendulum with mass m, attached to the end of a string, oscillating back and forth between angles of -45 degrees and +45 degrees relative to the vertical axis

at which point would the tension be a maximum?
and at what 2 angles would the tension in the string be half of its maximum value?

Attempt:

i think the tension in the string is at maximum when the pendulum is at the very bottom, or when the angle is 0, because T = Fg at this point. (whereas everywhere else, the tension would only equal to the parallel component of the gravitational force, Fg(y))

so then i derived a formula for tension T = mgcos(theta)

if Tmax = mg , then half of the maximum tension must be 0.5 mg, correct?

So then, 0.5 mg = mg cos(theta)
0.5 = cos(theta)
and theta = 60 degrees

but i don't think that answer is correct because the range of angles according to the question is only between -45 and 45 degrees

any thoughts?

 

[tiny-tim]

hi 3ephemeralwnd!

you have to do F = ma …

what is "a" ?​

 

[3ephemeralwnd]

Fnet = fg(x)
ma = mgsin(theta)
a = gsin(theta) ?

 

[tiny-tim]

hi 3ephemeralwnd!

(just got up :zzz: …)

erm … i was expecting you to say that a is the https://www.physicsforums.com/library.php?do=view_item&itemid=27" v²/L plus the tangential acceleration dv/dt

 

[rwisz]

I would like to first clarify that the statement "T = Fg at the bottom" is not entirely accurate. The tension in the string is not equal to the weight of the pendulum (mg) at any point, including the bottom. The tension is actually equal to the net force acting on the pendulum, which includes the weight (mg) and the centripetal force (mv^2/r) where v is the velocity of the pendulum and r is the length of the string.

To answer the first question, at which point would the tension be a maximum, it can be seen from the formula T = mgcos(theta) that the tension will be maximum when cos(theta) is maximum, which occurs at theta = 0 degrees. This aligns with your initial intuition that the tension would be maximum at the bottom of the pendulum's swing.

For the second question, at what 2 angles would the tension in the string be half of its maximum value, your approach is correct. However, the range of angles given in the question (-45 degrees to +45 degrees) does not allow for a solution. This is because the cosine function only has values between -1 and +1, and therefore, the maximum value of cos(theta) will never be 0.5. In fact, the maximum value of cos(theta) in this range would be 1 (at theta = 0 degrees), and the minimum value would be -0.707 (at theta = -45 degrees or +45 degrees). Therefore, there are no angles within the given range where the tension would be half of its maximum value.

If we were to expand the range of angles to include all possible values, then the tension would be half of its maximum value at theta = 60 degrees and theta = 120 degrees. This can be seen by setting 0.5mg = mgcos(theta) and solving for theta, which gives theta = 60 degrees or 120 degrees. However, these angles are outside of the given range and may not be relevant to the original question.

In summary, the tension in the string will be maximum at the bottom of the pendulum's swing (theta = 0 degrees). However, there are no angles within the given range where the tension would be half of its maximum value. I would recommend clarifying the range of angles in the question to get a more accurate answer.