Calculating Radius & Max Speed of Rock in Figure D

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}}.
  • #1
huskydc
78
0
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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  • #2
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 :smile:. If you have any more questions i will come back later today, after a good night sleep.
 
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  • #3
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
 
  • #4
Yes now you need to apply Newton's 2nd Law

Now with [itex] \theta [/itex]

[tex] T \cos \theta + T \cos \theta = m \frac{v^2}{R} [/tex]

solve for v.
 
  • #5
didn't work cyclovenom,

but how did you come up with that equation?
 
  • #6
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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  • #7
huskydc said:
didn't work cyclovenom,

but how did you come up with that equation?

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

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


In my markings [tex]sin^{-1} = arcsin[/tex]
 
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1. What is the formula for calculating the radius of the rock in Figure D?

The formula for calculating the radius of the rock is: radius = circumference / 2π.

2. How do you measure the circumference of the rock in Figure D?

The circumference of the rock can be measured by wrapping a flexible measuring tape or string around the widest part of the rock and then measuring the length.

3. Is the radius of the rock in Figure D the same as its diameter?

No, the radius is half of the diameter. The diameter is the distance across the widest part of the rock, while the radius is the distance from the center to the edge.

4. How does the radius of the rock affect its maximum speed?

The radius of the rock directly affects its maximum speed. A larger radius means a larger circumference, which means the rock has to travel a greater distance in one rotation. This results in a higher maximum speed.

5. Can the maximum speed of the rock in Figure D be calculated without knowing its radius?

No, the maximum speed cannot be calculated without knowing the radius. The radius is a crucial factor in determining the maximum speed of the rock.

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