Uniform Circular Motion: A plum-bob makes a 70 degree angle...

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SUMMARY

The discussion centers on calculating the period of a merry-go-round using a plumb bob that makes a 70-degree angle with the vertical. The user initially calculated the period (T) as 2.174 seconds but received feedback indicating that the radius of motion for the bob is greater than the stated 1.25 m, which is a critical error. The correct approach involves using the centripetal acceleration formula, \( a_c = g \tan(\theta) \), and recognizing the importance of maintaining proper vector and scalar distinctions in calculations. The mass of the bob is deemed irrelevant for this specific calculation.

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Homework Statement
You are spin a merry-go-round and are trying to use a hanging plum bob attached to the outermost limb of the merry-go-round to measure the period of the merry-go-round. If the plum bob is 20.0 cm long, has a mass of 50.0 g on it, the merry-go-round has a radius of 1.25 m and you observe the plum bob make a 70.0 degree angle with the vertical on the outermost edge of the merry-go-round, what is the period of the merry-go-round?

I attached an image of what I've tried. I don't think this is right though. Shouldn't the centripetal force be equal to the acceleration of the bob * the mass of it?
Relevant Equations
Fc = ma
Fc = (4pi^2rm)/T
unnamed (1).jpg

I also tried to solve for T when the centripetal force is equal to the tension in the x-dir, but I got 2.174s and that's not the right answer either. I'm not sure what I'm doing wrong exactly.
 
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What is your sum of the forces in the ##x## direction( radial)? We strongly prefer that you use latex to present your mathematics of hand written notes. Please see the latex guide.
 
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daisy7777 said:
Homework Statement: You are spin a merry-go-round and are trying to use a hanging plum bob attached to the outermost limb of the merry-go-round to measure the period of the merry-go-round. If the plum bob is 20.0 cm long, has a mass of 50.0 g on it, the merry-go-round has a radius of 1.25 m and you observe the plum bob make a 70.0 degree angle with the vertical on the outermost edge of the merry-go-round, what is the period of the merry-go-round?

I attached an image of what I've tried. I don't think this is right though. Shouldn't the centripetal force be equal to the acceleration of the bob * the mass of it?
Relevant Equations: Fc = ma
Fc = (4pi^2rm)/T

View attachment 341841
I also tried to solve for T when the centripetal force is equal to the tension in the x-dir, but I got 2.174s and that's not the right answer either. I'm not sure what I'm doing wrong exactly.
I would resist the temptation to put in the numbers at the earliest opportunity. Note that the mass of the bob is irrelevant. I would pick things up from:
$$a_c = g\tan x$$Can you find a relationship between ##a_c## and ##T##?

PS personally, I would use ##\theta## as the angle. Especially in a mechanics problem.
 
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daisy7777 said:
If the plum bob is 20.0 cm long,.., the merry-go-round has a radius of 1.25 m and you observe the plum bob make a 70.0 degree angle with the vertical on the outermost edge of the merry-go-round
From the wording in the question, I'd say that the radius of motion (of the bob) is bigger than 1.25m. The upper end of the pendulum is 1.25m from the centre, but the bob itself is further out. You've ignored that - and that's probably the mistake.

Your calculation seems correct apart from the point noted above. But there are some points to note:
- your top drawing shows tension acting vertically upwards; it doesn't!
- you have created unecessary work (and risk of errors) by using values rather than symbols in the inermediate steps (as already noted by @PeroK);
- you have equations which illegally mix scalars and vectors (e.g. you shouldn't equate a scalar to a vector).

Edit: Minor rewording only.
 
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daisy7777 said:
I also tried to solve for T when the centripetal force is equal to the tension in the x-dir, but I got 2.174s
I don't see how that is different from the attempt you posted (assuming you again did not allow for the horizontal component of the string length, as @Steve4Physics points out), so you should have got the same answer.

Btw, it's "plumb bob", from the Latin plumbum=lead (the metal) as in plumber.
 
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