Chuck wants to investigate how gas molecules
1 answer:
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Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
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