The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Answer:
2.26 s
Explanation:
Let's take down to be positive.
Given (in the y direction):
Δy = 25 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
25 m = (0 m/s) t + ½ (9.8 m/s²) t²
25 = 4.9t²
t = 2.26 s
If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s. So the time to fall is still 2.26 s.
Hello.
The answer would be <span> 0.5 s
Have a nice day</span>
Answer:large
Explanation:
As the temperature increases, materials with large coefficients of linear expansion increases a lot in size
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
solving for t
t = 1.075 sec
for horizontal motion
x = 2cos18*1.075
x = 2.044 m
