# Rock on string

1. Jul 15, 2005

### 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?

#### Attached Files:

• ###### showme.gif
File size:
1.7 KB
Views:
93
2. Jul 15, 2005

### Pyrrhus

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.

Last edited: Jul 15, 2005
3. Jul 15, 2005

### huskydc

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. Jul 15, 2005

### Pyrrhus

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.

5. Jul 15, 2005

### huskydc

didn't work cyclovenom,

but how did you come up with that equation?

6. Jul 15, 2005

### Pyrrhus

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!.

Last edited: Jul 15, 2005
7. Jul 15, 2005

### Päällikkö

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 $$sin^{-1} = arcsin$$

Last edited: Jul 15, 2005