Question

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the staring for t seconds. Determine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.

Answer 1

t = 5.89 s

Step-by-step explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley (
`r_p`) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

`\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw`

`t = (4r^2mw)/(Tr_P)`

Substituting values:

`t = (4r^2mw)/(Tr_P)= (4*(0.5)^2*3*15.708)/(20*0.4)=5.89s`

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of t is
`t = 15.08 \ s`

Step-by-step explanation:

From the question we are told that the angular speed is

The initial angular speed
`w_f = 150 rev/min = (2 \pi )/(60s) * 150 = 15.71rad /s`

The force is
`T = 20 N`

The radius is
`r = 400mm = (400)/(1000) = 0.4m`

The total mass of the four sphere is
`m_a = (4*3) = 12kg`

The initial velocity is
`v_i = 0m/s`

The radius of the pully is
`r_p = 100mm = (100)/(1000) = 0.10m`

The initial time is
`t_1 = 0s`

The final time is
`t = t`

Generally the final velocity of the sphere is mathematically represented as

`v_f = r w_f`

`v_f=15.7 r`

The angular impulse momentum principle can be represented methematically as

`(H_O)_i + \int\limits^(t_1)_(t_2){(T \cdot r_(p))} \, dt = (H_O)_f`

`r(m_a v_i ) + \int\limits^(t_2)_(t_1) {T \cdot r_p} \, dt = r(m_a v_f)`

`r(m_a v_1 ) + T \cdot r_p (t_(2) -t_(1)) = r( m_a * 15.7 r )`

Substituting values

`0.4 * (12* 0) + (20 *0.10 * (t-0)) = 0.40 * (12* 0.40 *15.7)`

=>
`0 + 2t = 30.16`

=>
`t = 15.08 \ s`