angle of a pendulum?

Question

A simple pendulum, 2 m in length, is released by a push when the support string is at an angle of 25 deg. from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, to what maximum angle will it move in the second half of its swing?

The answer is 30 deg., I just need to figure out how to get it.

Answer 1

u have to use sine,cosine in this type of question
first draw the diagram and draw a line so u get a right triangle
then find the potential energy
to find the potential energy u have to know height
to find the height u need to find the length of ajacent, then use 2m minuse it.
length of adjacent=2 x cos25=1.81m
2m-1.81m=0.19m
Ep=mgh

below is V at the very bottom
mgh=1/2 x m (Vf^2 - 1.2^2)
m can be cancel out

9.8x0.187=1/2 x (Vf^2 -1.2^2)
Vf=2.26

then kinetic energy is transformed back into potential energy again

mgh=1/2 x m x 2.26^2
h=0.26
2-0.26=1.73 m

cosx = 1.73/2
x=30 degree

Answer 2

Hokay. First of all, you have to find how high the mass is above the equilibrium position. You do so by finding the y component of the string at 25 degrees. You know the string is 2m, and that is the hypotenuse, so the y component = 2cos25 = 1.81. However, don't be deceived, that is not the height above the equilibrium, you have to subtract that from 2.o to find the height, because at equilibrium the string is straight down, and has a length of 2.o m. The actual height is .187. Now, since KEi +PEi = KEf + PEf, we can plug the values we have to that equation and solve for the height. Mass is constant, so we can cancel it out, so we have the equation .5v^2 + 9.8(.187) = 0 + 9.8h. The Kinetic energy at the top of the swing is zero because it is at rest at the peak. Now solve for h, and we get .26 But don't be deceived! subtract that from 2.o to find the y component of the string at the peak. (1.74) Now, the length of the string is always 2.o m, so we know that at the peak, the hypotenuse is 2.o, and the y component is 1.74 The cos of the unknown angle = 1.74/2, so find the inverse cosine of that and you get 30 degrees. Viola! I hope you caught all of that, best of luck.

Answer 3

If you calculate this accurately, it isn't 30 deg, it's 29.588 deg. Maye that was why you had a problem getting it.
cos(25d)= .09369
1/m * PE = g*2*(1-cos(25d)) = 1.8376
1/m * KE = .5*1.2^2 = .72
final height = 1/m * (PE+KE)/g = .260804
final angle = arccos((2-final height)/2) = 29.588 deg.
I used g=9.80665; if you use 9.8 the angle only changes a little, to 29.591 deg.

Answer 4

this may well be a concern of geometry on my own. The arc length lined bob interior the 1st case R1* ?1 the place ?1 is in radian. interior the 2d case that's R2 * ?2 the two are equivalent ?2 = (R1/R2) ?1 R1= 11.4 m R2 = 11.4 – 4 = 7.4m R1/R2 = one million.541 ?1 = 40 3 degree = (40 3/a hundred and eighty) ? radian = 0.751 radian. ?2 = (R1/R2) ?1 = one million.157 radian. ?2 = one million.157 * ( a hundred and eighty/ ?) degree = 66.3 degree.

Answer 5

try conservation of energy