The figure shows a pendulum of length L = 2.6 m. Its bob (which effectively has

Question

The figure shows a pendulum of length L = 2.6 m. Its bob (which effectively has all the mass) has speed v0 when the cord makes an angle 0 = 40° with the vertical. (a) What is the speed of the bob when it is in its lowest position if v0 = 9.1 m/s? What is the least value that v0 can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if 0 is increased by a few degrees?

Explanation / Answer

Here,

L = 2.6 m

theta = 40 degree and speed is v0

a) Using conseravtion of energy

0.5 * m * v^2 = 0.5 *m * v0^2 + m * g * L * (1 - cos(theta))

0.5 * v^2 = 0.5 *v0^2 + g * L * (1 - cos(theta))

0.5 * 9.1^2 = 0.5 * v0^2 + 9.8 * 2.6 * (1 - cos(40))

solving for v0

v0 = 8.41 m/s

the speed v0 must be 8.41 m/s

b)

for reaching to a horizontal position

Using conservation of energy

m * g * L = 0.5 *m * v0^2 + m * g * L * (1 - cos(theta))

g*L = 0.5 *v0^2 + g * L * (1 - cos(theta))

9.8 * 2.6 = = 0.5 * v0^2 + 9.8 * 2.6 * (1 - cos(40))

solving

v0 = 6.24 m/s

the speed v0 must be 6.24 m/s

c)

for string to be straigth

m * v^2/L = m * g

v^2 = L * g

Using conservation of energy

m * g * 2L + 0.5 * m * v^2 = 0.5 *m * v0^2 + m * g * L * (1 - cos(theta))

2 *g*L + 0.5 * v^2 = 0.5 *v0^2 + g * L * (1 - cos(theta))

2 *g*L + 0.5 * L * g = 0.5 *v0^2 + g * L * (1 - cos(theta))

2.5 * 9.8 * 2.6 = 0.5 *v0^2 + g * L * (1 - cos(theta))

solving

v0 = 10.74 m/s

the speed v0 must be 10.74 m/s

d)

as the initial potential energy will increase

the needed speed v0 will decrease

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