∑τ= Iα = - (L/2) mgsin(Θ)
(The net torque equals the negative horizontal component of gravity at the center of mass)(it's negative because the force is oppose of displacement)
sin(Θ) ≈ Θ , at small angles
α (1/3ML^2) = - (L/2) mg θ
2nd order differential equation
Θ = Asin(ωt)
^take the derivative twice
α = -Aω^2sin(ωt) = -θω^2
Go back to α (1/3ML^2) = - (L/2)mg θ
(-θω^2)(1/3ML^2) = - (L/2)mgθ
Cancel stuff out and solve for ω
Plug in L = 1.0m
T = 1.639sec.
L = 0.67m
a) They are in the same point
b) t = 0 s, t = 2.27 s, t = 5.73 s
c) t = 1 s, t = 4.33 s
d) t = 2.67 s
Given equations are:
a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both and depend on t and don't have constant terms.
So both cars A and B are in the same point.
b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).
c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.
d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.
Force is the push or pull on object that can cause different things, like acceleration. According to Newton's 2nd Law of Motion, it is the product of mass and acceleration.
The mass of the wolf is 50.5 kilograms and the acceleration is 5.0 meters per square second. Therefore:
Substitute the values into the formula.
The force required is <u>252.5 Newtons</u>.