A simple pendulum consists of an object suspended by a string. The object is ass

Question

A simple pendulum consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.90 m long and makes an initial angle of 26.5° with the vertical, calculate the speed of the particle at the following positions.

(a) at the lowest point in its trajectory

(b) when the angle is 15.0°

Explanation / Answer

The total mechanical energy of the pendulum is conserved since we assume that there are no frictional forces. the total mechanical energy of a system is equal to K + U, where K is the kinetic energy of the system and U is its gravitational potential energy.

When the string is pulled back to an angle of 26.5º with the vertical, it has no kinetic energy and only has potential energy. Thus, we can find the total mechanical energy (E) of the system since it equals the potential energy at this "turning point"

a) We will consider the height of the bob to be 0 at the lowest point in its trajectory.

Thus, at this point, the pendulum bob has only kinetic energy. Kinetic energy is given by the formula K = 1/2 • m • v^2, where m is the mass of the object and v is its velocity. Gravitational potential energy is given by U = m • g • h, where m is the object's mass, g is the acceleration due to gravity, and h is the height/altitude of the object in meters. Since energy is conserved, 1/2 • m • v^2 (K at lowest point) = m • g • h (U at highest point).

1/2 • v^2 = g • h (the "m"s cancel)
v^2 = 2 • g • h
v = SQRT(2 • g • h)

We know g equals 9.8 m/s^2. Thus, all we need to find is the height h of the pendulum at its turning point (i.e. its highest point). We know that the height difference between the fixed end and the bob at its highest point equals L • cos(Theta)

h = L - L • cos(Theta) = 2.90 m - 2.90 m • cos(26.5º) = 0.3046 m

v = SQRT(2 • 9.8 m/s^2 • 0.3046 m)
v = 2.44 m/s

b) At 15.0º, the total mechanical energy is a combination of kinetic and potential energy. Thus, E = K + U. Since we want v, we consider K: K = E - U.

We know that E also equals the kinetic energy of the bob at the lowest point in its trajectory (since there is no potential energy at that point). Thus, E = 1/2 • m • 2.44^2 = 2.9768 • m.

All we need now is U. To find U, we need the height of the bob at a 15º angle, which is equal to L(1 - cos(Theta)) as we saw in the last part (nice shortcut!). Thus, substituting L = 2.90 m and Theta = 15º, h = 0.09881 m.

We now find that U = m • 9.8 • 0.09881 = 0.9683 • m.

K = E - U = 2.9768 • m - 0.9683 • m = 2.0085 • m
1/2 • m • v^2 = 2.0085 • m
1/2 • v^2 = 2.0085
v^2 = 4.017
v = 2.0042 m/s

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