We have the equation of motion , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.
Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81
Substituting
A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.
Answer:
the period of the physical pendulum is 0.498 s
Explanation:
Given the data in the question;
= 0.61 s
we know that, the relationship between T and angular frequency is;
T = 2π/ω ---------- let this be equation 1
Also, the angular frequency of physical pendulum is;
ω = √(mgL / ) ------ let this equation 2
where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and is moment of inertia of rod.
Now, moment of inertia of thin uniform rod D is;
=
mD²
since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.
we substitute equation 2 into equation 1
we have;
T = 2π/ω OR T = 2π/√(mgL/) OR T = 2π√(
/mgL)
so we can use =
mD² for moment of inertia of the rod
Since center of gravity of the uniform rod lies at the center of rod
so that L = D.
now, substituting these equations, the period becomes;
T = 2π/√(/mgL) OR T =
OR T = 2π√(2D/3g ) ----- equation 3
length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),
we have;
ω = 2π/
OR ω
= √(g/D) OR ω
= 2π√( D/g )
so we simple solve for D/g and insert into equation 3
so we have;
T = √(2/3) ×
we substitute in value of
T = √(2/3) × 0.61 s
T = 0.498 s
Therefore, the period of the physical pendulum is 0.498 s