Problem using centripital accel.... Can you check my work?

[sunnnystrong]

Homework Statement

A student ties a 500 g rock to a 1.0-m-long string and swings it around her head in a horizontal circle.
At what angular velocity in rpm does the string tilt down at a 16 degree angle?

Homework Equations

F = (m*v^2)/r
v = r*w

The Attempt at a Solution

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First I solved for the tension in the string...

∑Fy = 0

T*sin(16°) - mg = 0
T*sin(16°) = mg
T = 17.776 N

Next I solved for the linear velocity

ΣFx = m*a

Tx = 17.776cos(16°)
r = sin(74°)
17.776cos(16°) = ((.5kg)*v^2)/(sin(74°)
Solve for linear velocity --> 5.73 m/s

v = r*w
Solve for omega and I got 5.96 rad/s

Convert to rpm and I got 56.9 rpm.

 

[mfb]

I get the same values.

 

[sunnnystrong]

I get the same values.

Awesome, thank you!

 

[gneill]

You've found a valid solution, well done. Since you've arrived at the answer I'm free to show you an alternative method.

If you make a drawing of the scenario and consider the accelerations experienced by the rock you'll see that the ratio of the gravitational acceleration to the centripetal acceleration will be equal to the tan of the angle. In this diagram the negative of the centripetal acceleration (red vector) is shown for clarity:

So that $tan(θ) = \frac{g}{ω^2 r}$, where: $r = L cos(θ)$. With a bit of rearranging and simplifying this becomes:

$ω = \sqrt{\frac{g}{L sin(θ)}}$

which should yield the same value for ω that you found.

 

[sunnnystrong]

You've found a valid solution, well done. Since you've arrived at the answer I'm free to show you an alternative method.

If you make a drawing of the scenario and consider the accelerations experienced by the rock you'll see that the ratio of the gravitational acceleration to the centripetal acceleration will be equal to the tan of the angle. In this diagram the negative of the centripetal acceleration (red vector) is shown for clarity:

View attachment 114717

So that $tan(θ) = \frac{g}{ω^2 r}$, where: $r = L cos(θ)$. With a bit of rearranging and simplifying this becomes:

$ω = \sqrt{\frac{g}{L sin(θ)}}$

which should yield the same value for ω that you found.

Thank you! I'll remember that :)