Answer:
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Explanation:
Answer:
1) For the swung puck tied to the string that has a mass m₁ and having another mass m₂ attached to the string that passes through a hole in the table, such that m₂ > m₁
The swung puck rotating on the table is kept in the circular motion on the table by the centripetal force acting towards the center of its rotation along the string, which is given by the amount the weight of m₂ exceeds m₁, in the same way a planet moving around the Sun is kept in its orbit by the gravitational attraction between the planet and the Sun.
2) i) The forces acting on the puck are;
a) The centripetal force in the string, acting inward, towards the center of rotation provided by the weight of m₂
b) The centrifugal force of the puck, acting outward, tending to continue its motion in a straight line
ii) The forces acting on the planet are;
a) The force of gravitational attraction between the planet and the Sun, acting towards the Sun, such that, the planet revolves around the Sun
b) The centrifugal force of the planet, acting outward, tending to move in a linear or straight path
Explanation:
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
I think is D cuz all Yoou really need to do is look at the graph
