# Calculating Radius & Max Speed of Rock in Figure D

• huskydc
In summary, the conversation is discussing the calculation of the radius and maximum speed of a rock being whirled in a circle by two strings with a specific tension. The conversation also mentions the use of Newton's 2nd Law and trigonometry to calculate these values. The equation used to solve for the maximum speed is v = \sqrt{\frac{2Tcos(sin^{-1}(\frac{d}{2s}))\sqrt{s^2-(\frac{d}{2})^2}}{m}}.

#### huskydc

as shown: in the picture, look at figure d) (attachment) don't worry about figure a) --- given:

720 gm rock

each string has a tension of 29 N

the rock is held by two of the 45 cm strings with ends 56 cm apart and whirled in a circle between them. Neglect gravity.
What is the radius of the circle of motion?

what is the maximum speed the rock can have before the string breaks?

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Hmm, this is a curvilinear trajectory dealing with only normal acceleration component (radial), apply Newton's 2nd Law. Now that i think about it you can calculate the radius throught trigonometry . If you have any more questions i will come back later today, after a good night sleep.

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well, i found the radius to be 35.22 cm. and i also found the respective angle between direction of acceleration and the string tensions. from each string to the r as shown, angle is 38.48 degrees. but i don't know where to go from here

Yes now you need to apply Newton's 2nd Law

Now with $\theta$

$$T \cos \theta + T \cos \theta = m \frac{v^2}{R}$$

solve for v.

didn't work cyclovenom,

but how did you come up with that equation?

It should have worked, well the Tension have 2 components, one of them is acting radially, and because there are two tensions both of them hace 1 component acting as the centripetal force. Rememeber to convert the mass of the rock to kilograms!.

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huskydc said:
didn't work cyclovenom,

but how did you come up with that equation?

Horizontal components of T:
$$T_x = Tcos\theta$$, where $$\theta = sin^{-1}(\frac{d}{2s})$$, where d = 0,56m and s = 0,45m

So, the equation solves into:
$$v = \sqrt{\frac{2Tcos(sin^{-1}(\frac{d}{2s}))\sqrt{s^2-(\frac{d}{2})^2}}{m}}$$

In my markings $$sin^{-1} = arcsin$$

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