Calculating Angular Velocity of Rock Swinging in a Circle

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A boy swings a 500kg rock tied to a 1.0m string in a horizontal circle, and the goal is to find the angular velocity at a 10-degree angle. The tension in the string is calculated to be 28.22N using the formula T = mg/sin(theta). The velocity of the rock is determined to be 7.4 m/s, leading to an initial angular velocity of 70.7 rpm. However, a correction is made to the angular velocity calculation, resulting in a final value of 7.51 rad/s after accounting for the effective radius. The discussion emphasizes the importance of correctly applying trigonometric functions in circular motion problems.
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A boy ties a 500kg rock to a 1.0m long string and swings it around his head in a horizontal circle. At what angular velocity, in rpm, does the string tilt down at a 10 degree angle?

sintheta = mg / T

sin10 = 0.5kg(9.8m/s^2) / T

T = 4.9N / sin10

T = 28.22N

THEN...

v = SqRt [(rTcostheta) / m] ---> However the radius of the circle is not the length (L) so, r = Lcostheta

v = SqRt [(LTcos^2 theta) / m]

v = SqRt [(1.0m)(28.22N)(cos10^2) / 0.5kg]

v = 7.4m/s

FINALLY...

OMEGA = v / r
= v / Lcostheta

7.4rad / 1sec x 1rev / (2pi)rad x 60sec / 1 min --> rad and sec cancel out, leaving...

= 70.7 rpm


This was my initial attempt to the problem but I am not sure if everything is correct?
 
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Looks fine through v = 7.4 m/s, but then look at \omega = v/r = v/(L cos (10°).

cos 10° = 0.985, so \omega = (7.4 m/s)/(0.985 m) = 7.51 rad/s

The rest is correct in form.
 
Thanks for the correction!
 
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