suppose a simple pendulum was released from an angle of 58 degrees, with a length of 0.75m and mass of .15kg. what would be the speed of the bob at the bottom of the swing? To what height would the bob get to on the other side. What angle of release would give half the speed of that for 58 degree release angle at the bottom of the swing?
The bob rises to the same height on the opposite sde.
MgL(1 - costheta) + (1/2) M V^2
= M g L[1 -cos(thetamax)]
M cancels out.
At the bottom of the swing,
theta = 0 and
V^2 = 2 g L [1-cos(thetamax)] = Vmax^2
For half the velocity at bottom,
1 - cos(theta)max must be reduced by a factor of 4. Theta becomes 28 degrees.
To find the speed of the bob at the bottom of the swing, we can use the conservation of mechanical energy equation. The mechanical energy of a pendulum is the sum of its potential energy and kinetic energy.
1. Find the potential energy at the highest point of the swing:
The potential energy at the highest point is given by:
PE = mgh
where m is the mass (0.15 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
Since the pendulum is released from an angle, the height is given by:
h = L * (1 - cosθ)
where L is the length (0.75 m) and θ is the release angle (58 degrees converted to radians).
2. Find the speed at the bottom of the swing:
The mechanical energy at the highest point is equal to the mechanical energy at the bottom of the swing since there's no energy loss due to friction.
So, the kinetic energy at the highest point is equal to the kinetic energy at the bottom of the swing:
KE = 1/2 * mv^2
where v is the speed at the bottom.
Equating the mechanical energy equations, we have:
PE = KE
mgh = 1/2 * mv^2
Simplifying and solving for v, we get:
v = √(2gh)
3. Calculate the height the bob will reach on the other side:
The maximum height the bob reaches on the other side is equal to the initial potential energy:
h = PE / (mg)
where PE is the potential energy calculated earlier.
To find the angle of release that gives half the speed at the bottom of the swing, we need to find the release angle (θ') that gives half the velocity compared to the 58-degree release angle.
4. Calculate the new release angle:
Using the conservation of energy, we can set the kinetic energy at the bottom of the swing for the new release angle (θ') to be half of the velocity calculated.
1/2 * m * v'(θ')^2 = 1/2 * m * v^2
Dividing both sides by 1/2 * m, we get:
v'(θ')^2 = v^2
Taking the square root of both sides, we have:
v'(θ') = v
Substituting the formula for the speed of the bob at the bottom, we can solve for θ':
√(2g * L * (1 - cosθ')) = √(2g * L * (1 - cos58°))
Rearranging and solving for θ', we find the angle that gives half the speed.
By following these steps, you can find the speed of the bob at the bottom of the swing, the height it reaches on the other side, and the angle of release that would give half the speed of the 58-degree release angle.