# Homework Help: Beetle On A Pendulum

1. Oct 19, 2009

### Warmacblu

1. The problem statement, all variables and given/known data

A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 170 cm long.

A) If the beetle rides through a swing of 43 degrees, how far has he traveled along the path of the pendulum?

B) What is the displacement experienced by the beetle while moving through the same angle 43 degrees?

C) If the pendulum at some instant is swinging at 3.3 rad/s, how fast is the beetle traveling?

2. Relevant equations

theta = s/r

3. The attempt at a solution

A) 43/360 * (2pi * 170) = 127.584 cm - correct

B) s = r(theta)

s = 170 x 43(pi/180) = 127.584 cm - incorrect

C) theta = s/r

3.3 = s/170

s = 561 cm/s - correct

I don't know what I did wrong for part b. Perhaps I am supposed to multiply by pi/360?

2. Oct 19, 2009

### Staff: Mentor

For displacement, imagine an arrow drawn from the starting position to the final position. What's the length of that arrow?

3. Oct 19, 2009

### Warmacblu

sin43 = x/170

x = sin43 x 170

x = 115.9397

4. Oct 19, 2009

### semc

I don't think its a right angle triangle. Maybe cosine rule?

5. Oct 19, 2009

### Warmacblu

I think it might be a right angle because we are taking the angle measurement from the rest position and if I draw and arrow to the right and connect the hypotenuse (170 cm) then the angle turns out to be 90 degrees on the bottom with a 43 degree angle on top.

6. Oct 19, 2009

### Staff: Mentor

It's not a right triangle; it's an isoceles triangle with the long sides equal to the length of the string. Hint: Law of sines.

7. Oct 19, 2009

### Warmacblu

I looked up the law of sines and law of cosines and I think the law of cosines would be easier to use:

c2 = a2 + b2 – 2abcosC

c2 = 1702 + 1702 - 2*170*170*cos(43)

c = 124.6104

Does that seem okay.

8. Oct 20, 2009

### Staff: Mentor

Looks good.

Using the law of sines, it's just: 170/sin(68.5) = x/sin(43)

9. Oct 20, 2009

### Warmacblu

Thanks for the help.