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natita [175]
3 years ago
13

When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force

F on the cord is increased, the bob rises and then rotates around the horizontal circular path B.a) Determine the speed of the bob around path B
b) Find the work done by force F.
Engineering
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

The speed is the same at 1.5 m/s while

The work done by the force F is 0.4335 J

Explanation:

Here we have angular acceleration α = v²/r

Force = ma = 2.8 × 1.5²/r₁

and ω₁ = v₁/r₁ = ω₂ = v₁/r₂

The distance moved by the force = 600 - 300 = 300 mm = 0.3 m

If the velocity is constant

The speed is 1.5 m/s while the work done is

2.8 × 1.5²1/(effective radius) ×0.3

r₁ = effective radius

2.8*9.81 = 2.8 × 1.5²/r₁

r₁ = 0.229

The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J

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<em>Python code is as follows: </em>

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Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

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(a) the velocity of the particle when x = -1 m

(b) the position where the velocity is maximum

(c) the maximum velocity.

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- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

                          v = +/- 0.323 m/s

- v = v_max when a = 0:

                           -0.1 = sin(x/0.8)

                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

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