x: ma=Tsinα
y: mg=Tcosα
ma/mg =Tsinα/ Tcosα
a/g =tanα
v²/R•g =tanα
α=tan⁻¹(v²/R•g) = …
A 20.0 gram mass is attached to a 120 com long string. It moves in a horizontal circle with a constant speed of 1.50 m/s. What is the degree is angle between the pole and the string?
2 answers
Let L=length of string=120cm
Tsin(a)=mv^2/R, R=Lsin(a), T=mg/cos(a)
mgtan(a)=mv^2/Lsin(a), mgsin^2(a)=mv^2cos(a)/L,
g(1-cos^2(a))=v^2/L (cos(a))
gLcos^2(a)+v^2cos(a)-gL=0
let x=cos(a), use quadratic formula to solve for x
x= (-v^2+sqrt(v^2+4L^2*g^2))/2Lg
a=acos(x)=25 degrees
Tsin(a)=mv^2/R, R=Lsin(a), T=mg/cos(a)
mgtan(a)=mv^2/Lsin(a), mgsin^2(a)=mv^2cos(a)/L,
g(1-cos^2(a))=v^2/L (cos(a))
gLcos^2(a)+v^2cos(a)-gL=0
let x=cos(a), use quadratic formula to solve for x
x= (-v^2+sqrt(v^2+4L^2*g^2))/2Lg
a=acos(x)=25 degrees
2 answers