The Isolated System—Conservation of Mechanical Energy Problems and Solutions 3

 Problem#1

A circus trapeze consists of a bar suspended by two parallel ropes, each of length l, allowing  performers to swing in a vertical circular arc (Figure 1). Suppose a performer with mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle θ_i with
respect to the vertical. Suppose the size of the performer’s body is small compared to the length l, that she does not pump the trapeze to swing higher, and that air resistance is negligible. (a) Show that when the ropes make an angle θ with the vertical, the performer must exert a force

mg(3 cosθ – 2 cosθ_i)

in order to hang on. (b) Determine the angle θi for which the force needed to hang on at the bottom of the swing is twice the performer’s weight.

Answer:
(a) The force needed to hang on is equal to the force F the trapeze bar exerts on the performer.

From the free-body diagram for the performer’s body, as shown,

F – mg cosθ = mv²/l, or

F = mg cosθ + mv²/l

Apply conservation of mechanical energy of the performer-Earth system as the performer moves between the starting point and any later point:

mgl(1 – cosθ_i) = mgl(1 – cosθ) + ½ mv²

½ mv² = mgl(1 – cosθ_i) – mgl(1 – cosθ)

mv²/l = 2mg(1 – cosθ_i) – 2mg(1 – cosθ)

mv²/l = 2mg cosθ – 2mg cosθ_i

Solve for mv²/l and substitute into the force equation to obtain

F – mg cosθ = 2mg cosθ – 2mg cosθ_i

F = mg(3 cosθ – 2cosθ_i)

(b) At the bottom of the swing, θ = 0⁰, so

F = mg(3 cosθ – 2cosθ_i)

2mg = mg(3 cos0⁰– 2cosθ_i)

cosθ_i = ½

θ_i = 60.0⁰

Problem#2
Two objects are connected by a light string passing over a light frictionless pulley as shown in Figure 3. The object of mass 5.00 kg is released from rest. Using the principle of conservation of  energy, (a) determine the speed of the 3.00-kg object just as the 5.00-kg object hits the ground. (b) Find the maximum height to which the 3.00-kg object rises.

Fig.3

Answer:
Using conservation of energy for the system of the Earth and the two objects

(a) the speed of the 3.00-kg object just as the 5.00-kg object hits the ground is

U_i + K_i = U_f + K_f

m₁gh + 0 = m₂gh + ½ (m₁ + m₂)v²

(5.00 kg)(9.80 m/s²)(4.00 m) = (3.00 kg)(9.80 m/s²)(4.00 m) + ½ (3.00 kg + 5.00 kg)v²

156.8 m²/s² = 8.00kgv²

v = 4.43 m/s

(b) the maximum height to which the 3.00-kg object rises

½ m₂v₂² = m₂g∆y

½ (4.43 m/s)² = (9.80 m/s²)∆y

∆y = 1.00 m

So, ∆y_max = 4.00 m + 1.00 m = 5.00 m

Problem#3
Two objects are connected by a light string passing over a light frictionless pulley as in Figure 3. The object of mass m₁ is released from rest at height h. Using the principle of conservation of energy, (a) determine the speed of m₂ just as m1 hits the ground. (b) Find the maximum height to which m₂ rises.

Answer:
Using conservation of energy for the system of the Earth and the two objects

(a) the speed of the m₂ object just as the m₁ object hits the ground is

U_i + K_i = U_f + K_f

m₁gh + 0 = m₂gh + ½ (m₁ + m₂)v²

(m₁ – m₂)gh = ½ (m₁ + m₂)v²

2(m₁ – m₂)gh = (m₁ + m₂)v²

v = [2(m₁ – m₂)gh/(m₁ + m₂)]^1/2

 (b) Since m₂ has kinetic energy ½ m₂v², it will rise an additional height ∆y determined from

½ m₂v₂² = m₂g∆y

∆y = v²/2g

∆y = (m₁ – m₂)h/(m₁ + m₂)

So, The total height m₂ reaches is

y_max = h + ∆y = h + [(m₁ – m₂)h/(m₁ + m₂)]

y_max = 2m₁h/(m₁ + m₂)

Problem#4
A light rigid rod is 77.0 cm long. Its top end is pivoted on a low-friction horizontal axle. The rod hangs straight down at rest with a small massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?

Answer:

The force of tension and subsequent force of compression in the rod do no work on the ball, since they are perpendicular to each step of displacement. Consider energy conservation of the ballEarth system between the instant just after you strike the ball and the instant when it reaches the top. The speed at the top is zero if you hit it just hard enough to get it there.

K_i + U_gi = K_f + U_gf

½ mv_i² + 0 = 0 + mg(2L)

v_i = √[4gL]

v_i = √[4(9.80 m/s²)(0.770 m)] = 5.49 m/s 

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