Circular motion question - finding rpm

[jgray]

Homework Statement

A 0.60 kg sphere rotates around a vertical shaft supported by two strings, as shown. If the tension in the upper string is 18 N. Calculate
the tension in the lower string?
the rotation rate (in rev/min) of the system?

Homework Equations

v=2(pi)r/T
Fr= mv^2/r
ω = v/R

The Attempt at a Solution

I believe I have the first part correct, my question is really about the second part. For the second part of the question: I calculated the x components of each tension force to find the centripetal force = 20.96N. I then found the velocity by rearranging the equation: Fr = mv^2/r to get 3.74m/s. I used the equation ω = v/R to get 9.35 rads/sec and converted it to 89.3 rpm by multiplying 9.35 by 60s and dividing it by 2(pi). Am I on the right track?? Just seems like a small answer…

 

[azizlwl]

The upper string has to supply force for the mass to rotate and to stay afloat.

 

[BvU]

[strike]Don't follow the 3.74 m/s. Did you divide by 6 or (0.6 x 9.81) instead of by 0.6 kg ?[/strike]

(Edit: Jgray is fully correct)

 

[azizlwl]

I don't see any workings of the first part.

 

[BvU]

@aziz: I think the poster is happy with the result of the first part (I support him/her in that) and now asks for help in the second part.

[strike]If I fill in v = 3.74m/s in mv^2/r I do not get the 21 N...[/strike]

(Edit: Jgray is fully correct)

 

[jgray]

Hmm, maybe I do have an error in my first part then. I found the centripetal force:
T1x+T2x =
sin θ(T1) + sin θ(T2)=
sin53(18N) + sin53(8.2N) --8.2 N is the answer i found for the second tension force for part 1--
14.37+6.55
20.92N

For the velocity:
v^2= Fr(r)/m
(20.92N)(0.4m)/0.6kg=
square root of 13.95=
3.74m/s

 

[BvU]

Fully correct, it was me that used r=4 (from the 3,4,5 triangle) instead of the 0.4 in the exercise. Sorry.

So the 3.74 is fine too. And the 9.35 and the 89.2.

 

[jgray]

Thanks for the help! :)

 

[BvU]