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iogann1982 [59]
3 years ago
7

[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.

Physics
2 answers:
ExtremeBDS [4]3 years ago
7 0

Answer:

t = 5.89 s

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 = \frac{4r^2mw}{Tr_P}

Substituting values:

t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s

Lapatulllka [165]3 years ago
7 0

Complete  Question

The complete question is shown on the first uploaded image

Answer:

The value of t is  t = 15.08 \ s

Explanation:

From the question we are told that the angular speed is

         The initial angular speed w_f  = 150 rev/min = \frac{2 \pi }{60s} * 150 = 15.71rad /s

          The force is  T = 20 N

           The radius is r = 400mm = \frac{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 = \frac{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

             

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