A pendulum 1.70m long is released (from rest) at an angle 30 degrees.
1. Determine the speed(m/s) of the 60.0g bob at Theta = 15.
2. Determine the tension in the cord at the lowest point Theta = 0.
3. Determine the tension in the cord at Theta = 15.
A formula would be appreciated and a decent explanation would be excellent.
-Thanks.
To solve these problems, we can use the concept of conservation of mechanical energy and Newton's second law of motion. First, let's establish the necessary formulas:
1. The formula to find the speed of the bob at a given angle is:
v = √(2gh)
2. The formula to calculate the tension in the cord is:
T = mgcos(θ) + m * (v^2 / L)sin(θ)
where:
v = speed of the bob
h = height of the bob above the lowest point
g = acceleration due to gravity (approximately 9.8 m/s^2)
T = tension in the cord
m = mass of the bob
L = length of the pendulum
θ = angle from the vertical axis
Let's now solve each problem step by step.
1. To determine the speed of the 60.0g bob at θ = 15 degrees:
- First, calculate the height, h, of the bob above the lowest point.
Since the total length of the pendulum is 1.70m and the bob is released at an angle of 30 degrees, we can use trigonometry to find h:
h = L - L * cos(θ) = 1.70m - 1.70m * cos(30°) = 1.47m
- Substitute the values into the formula:
v = √(2gh) = √(2 * 9.8m/s^2 * 1.47m) ≈ 5.01 m/s
Therefore, the speed of the bob at θ = 15° is approximately 5.01 m/s.
2. To determine the tension in the cord at the lowest point (θ = 0):
- In this case, the height, h, is zero.
- Substitute the values into the formula:
T = m * (v^2 / L)sin(θ) = (0.060kg * (0 m/s)^2 / 1.70m)sin(0°)
Since sin(0°) is zero, the tension is also zero.
Therefore, the tension in the cord at the lowest point is zero.
3. To determine the tension in the cord at θ = 15 degrees:
- Calculate the height, h, using the same method as in question 1:
h = L - L * cos(θ) = 1.70m - 1.70m * cos(15°) ≈ 0.28m
- Calculate the speed, v, using the formula from question 1:
v = √(2gh) = √(2 * 9.8m/s^2 * 0.28m) ≈ 1.64 m/s
- Substitute the values into the formula:
T = mgcos(θ) + m * (v^2 / L)sin(θ)
T = (0.060kg * 9.8m/s^2)cos(15°) + 0.060kg * (1.64m/s)^2 / 1.70m)sin(15°)
T ≈ 0.573 N + 0.206 N ≈ 0.779 N
Therefore, the tension in the cord at θ = 15 degrees is approximately 0.779 N.