Suppose the simple pendulum shown below were released from an angle of θ = 74°, with L = 0.73 m and m = 0.25 kg..What angle of release would give half the speed of that for the 74° release angle at the bottom of the swing?
Suppose the simple pendulum shown below were released from an angle of θ = 74°, with L = 0.73 m and m = 0.25 kg..What angle of release would give half the speed of that for the 74° release angle at the bottom of the swing?
You must have been given a formula to calculate this problem. What formulas do you have? You can't just expect to receive an answer without you asking what you need help on and jotting down the formulas, that's being pure lazy.
For half the speed at the bottom, you would need 1/4 the initial potential energy.
The initial P.E. is proportional to
(1 - cos A)
where A is the initial angle from vertical.
1 - cos74 = 0.7244
For 1/4 of that value,
1 - cosA = 0.1811
cosA = 0.8189
A = 35.0 degrees
To find the angle of release that would give half the speed of the 74° release angle, we first need to calculate the speed of the pendulum at the bottom of the swing for the given conditions.
The speed of the pendulum at the bottom of the swing can be determined using the conservation of mechanical energy. The total mechanical energy of the pendulum is the sum of its gravitational potential energy and kinetic energy.
At the highest point (when θ = 74°), all the potential energy is converted to kinetic energy, and at the bottom of the swing, all the potential energy is converted back to kinetic energy.
The gravitational potential energy (PE) of the pendulum is given by:
PE = m * g * h
where m is the mass, g is the acceleration due to gravity, and h is the height above the lowest point.
At the highest point, the height above the lowest point is h = L - L * cos(θ), where L is the length of the pendulum and θ is the angle of release.
The kinetic energy (KE) of the pendulum at the bottom of the swing is given by:
KE = (1/2) * m * v^2
where m is the mass and v is the velocity.
Since the total mechanical energy is conserved, we have:
PE(at highest point) = KE(at bottom of swing)
Using the above equations, we can find the velocity (v) of the pendulum at the bottom of the swing for the given release angle of 74°.
Now, let's calculate the speed (v') for the desired angle of release.
Since the mechanical energy is conserved, we can equate the kinetic energy at the bottom of the swing for the desired angle (KE') to half the kinetic energy at the bottom of the swing for the 74° release angle (KE/2):
KE' = (1/2) * KE
Given that m and L remain constant, we can write:
(1/2) * m * v'^2 = (1/2) * [m * v^2]
Simplifying the equation:
v'^2 = v^2
Taking the square root of both sides:
v' = v
Therefore, the speed at the bottom of the swing for the desired angle of release should be equal to the speed at the bottom of the swing for the 74° release angle.
In conclusion, the angle of release that would give half the speed of that for the 74° release angle at the bottom of the swing does not exist because the speeds would be the same.