A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible.
What is the speed of the rock when the string passes through the vertical position?
Express your answer using two significant figures.
What is the tension in the string when it makes an angle of 45∘ with the vertical?
Express your answer using two significant figures.
What is the tension in the string as it passes through the vertical?
Express your answer using two significant figures.
Physics Help. Answer please?
2015-12-10 4:59 pm
回答 (3)
2015-12-10 11:28 pm
✔ 最佳答案
Δh = L(1 - cosΘ) = 0.80m * (1 - cos45º) = 0.234 mso at the bottom of the arc,
v = √(2*g*Δh) = √(2 * 9.8m/s² * 0.234m) = 2.1 m/s ◄ speed at vertical
tension T = m(g + v²/r) = 0.12kg * (9.8m/s² + (2.1m/s)² / 0.80m)
T = 1.9 N ◄ tension at vertical
At 45º,
T = mgcosΘ = 0.12kg * 9.8m/s² * cos45º = 0.83 N ◄ tension at 45º
Hope this helps!
2015-12-10 6:46 pm
http://www.cpp.edu/~skboddeker/121/exam/04-3sum121e2.htm
If you go to the website above, you will see a drawing and the following equation. This equation is used to determine the vertical distance the rock moves as the pendulum swings from 45˚ to vertical. As this happens, the rock’s potential energy decreases. Since air resistance is negligible, the increase of the rock’s kinetic energy is equal to the decrease of its energy.
∆ h = L – L * cos θ = 0.8 – 0.8 * cos 45
PE = 0.12 * 9.8 * (0.8 – 0.8 * cos 45) = 0.9408 – 0.9408 * cos 45
KE = ½ * 0.12 * v^2 = 0.6 * v^2
0.6 * v^2 = 0.9408 – 0.9408 * cos 45
v = √(1.568 – 1.568 * cos 45)
This is approximately 0.68 m/s. When the string is at this position, the tension is equal to the sum of the rock’s weight and the centripetal force.
Weight = 0.12 * 9.8 = 1.176 N
Fc = m * v^2/r
Fc = 0.12 * (1.568 – 1.568 * cos 45) ÷ 0.8 = 0.2352 – 0.2352 * cos 45
Tension – 1.176 + (0.2352 – 0.2352 * cos 45)
When the string makes an angle of 45∘ with the vertical, the rock’s velocity is 0 m/s. At this position, the tension is equal to the component of the rock’s weight that is parallel to the string.
F = 0.12 * 9.8 * sin 45 = 1.176 * sin 45
If you go to the website above, you will see a drawing and the following equation. This equation is used to determine the vertical distance the rock moves as the pendulum swings from 45˚ to vertical. As this happens, the rock’s potential energy decreases. Since air resistance is negligible, the increase of the rock’s kinetic energy is equal to the decrease of its energy.
∆ h = L – L * cos θ = 0.8 – 0.8 * cos 45
PE = 0.12 * 9.8 * (0.8 – 0.8 * cos 45) = 0.9408 – 0.9408 * cos 45
KE = ½ * 0.12 * v^2 = 0.6 * v^2
0.6 * v^2 = 0.9408 – 0.9408 * cos 45
v = √(1.568 – 1.568 * cos 45)
This is approximately 0.68 m/s. When the string is at this position, the tension is equal to the sum of the rock’s weight and the centripetal force.
Weight = 0.12 * 9.8 = 1.176 N
Fc = m * v^2/r
Fc = 0.12 * (1.568 – 1.568 * cos 45) ÷ 0.8 = 0.2352 – 0.2352 * cos 45
Tension – 1.176 + (0.2352 – 0.2352 * cos 45)
When the string makes an angle of 45∘ with the vertical, the rock’s velocity is 0 m/s. At this position, the tension is equal to the component of the rock’s weight that is parallel to the string.
F = 0.12 * 9.8 * sin 45 = 1.176 * sin 45
2015-12-10 5:19 pm
A massless string cannot have length.
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