A 3.3 kg pendulum bob hangs from a 2 meter string. It is pulled to the side, position A, so that the bob is 103 cm (h on the sketch) above the lowest position, position C. The bob is released and begins to swing.
a) What is its potential energy after it has swung to position B, 27.0 cm below the starting point?
b)How fast is it moving at that point?
Pe at top = m g h = 3.3 * 9.81 * 1.03
a) Pe at 1.03-.27 = 3.3 * 9.81 * 0.76
find difference
that difference is Ke in Joules
so
difference = (1/2) m v^2
a) To find the potential energy of the pendulum bob at position B, we need to calculate the change in height from position A to position B.
Given:
Mass of the pendulum bob (m) = 3.3 kg
Length of the string (L) = 2 m
Height above the lowest position at point A (h_A) = 103 cm = 1.03 m
Height below the starting point at point B (h_B) = -27 cm = -0.27 m
The potential energy (PE) of the pendulum bob at any point is given by the equation:
PE = m * g * h
where m is the mass, g is the acceleration due to gravity, and h is the height.
At position A, the potential energy is:
PE_A = m * g * h_A
At position B, the potential energy is:
PE_B = m * g * h_B
To find the change in potential energy from position A to position B, we subtract the potential energy at position A from that at position B:
ΔPE = PE_B - PE_A
= m * g * h_B - m * g * h_A
Substituting the given values:
ΔPE = 3.3 kg * 9.8 m/s^2 * (-0.27 m) - 3.3 kg * 9.8 m/s^2 * 1.03 m
Simplifying:
ΔPE = - 8.6414 J - 33.9578 J
ΔPE ≈ -42.5992 J
Therefore, the potential energy of the pendulum bob at position B is approximately -42.5992 Joules.
b) The speed of the pendulum bob at a particular point can be calculated using the conservation of mechanical energy. At position B, all of the potential energy has been converted into kinetic energy.
The formula for kinetic energy (KE) is:
KE = 1/2 * m * v^2
where m is the mass of the bob and v is the velocity.
Since we know the change in potential energy (ΔPE) and the change in kinetic energy (ΔKE) are equal, we can set them equal to each other:
ΔPE = ΔKE
Substituting the values:
-42.5992 J = 1/2 * 3.3 kg * v^2
Simplifying:
-42.5992 J = 1.65 kg * v^2
Rearranging the equation:
v^2 = (-42.5992 J) / (1.65 kg)
v^2 ≈ -25.8783 m^2/s^2
Since velocity cannot be negative, we take the positive square root:
v ≈ 5.0885 m/s
Therefore, the pendulum bob is moving at approximately 5.0885 m/s at point B.
To find the potential energy of the pendulum bob at position B, we can use the equation:
Potential energy = mass × acceleration due to gravity × height
The mass of the bob is given as 3.3 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. The height of position B is the difference between the starting height (position A) and the height at position B.
Given that the string length is 2 meters and position A is 103 cm above the lowest position (position C), we need to calculate the height at position B:
Height at position A = 2 m - 1.03 m = 0.97 m
Height at position B = Height at position A - 0.27 m = 0.97 m - 0.27 m = 0.70 m
Now we can plug these values into the potential energy equation:
Potential energy at position B = 3.3 kg × 9.8 m/s^2 × 0.70 m = 22.528 J
Therefore, the potential energy of the pendulum bob at position B is 22.528 J.
To find the speed of the pendulum bob at position B, we can use the principle of conservation of mechanical energy:
Total mechanical energy = Potential energy + Kinetic energy
At position A, the potential energy is maximum, and the kinetic energy is zero. At position B, some of the potential energy is converted into kinetic energy as the bob gains speed.
Therefore, the potential energy at position A is equal to the kinetic energy at position B.
From part (a), we found that the potential energy at position A is 22.528 J.
Now to find the speed at position B, we can set the potential energy at position A equal to the kinetic energy at position B:
Potential energy at position A = Kinetic energy at position B
22.528 J = (1/2) × mass × velocity^2
Rearranging the equation:
Velocity^2 = (2 × 22.528 J) / mass
Velocity^2 = (2 × 22.528 J) / 3.3 kg
Velocity^2 ≈ 30.70 m^2/s^2
Taking the square root of both sides:
Velocity ≈ √30.70 m^2/s^2
Velocity ≈ 5.53 m/s
Therefore, the speed of the pendulum bob at position B is approximately 5.53 m/s.