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denpristay [2]
3 years ago
7

A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it

turns through 53.2 rad.
(a) How long had the wheel been turning before the start of the 7.00 s interval?
(b) What was the angular velocity of the wheel at the start of the 7.00 s interval?
Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

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The workdone by Jack is  W_{jack} = -1050J

The workdone by Jill is  W_{Jill} = 0J

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Explanation:

From the question we are given that

          The mass of the boat is m_b = 3300kg

          The initial position of the boat is   P_i  = (2 \r  i  + 0 \r j + 3\r k)m

           The Final position of the boat is  P_f = (4\r i + 0 \r j + 2\r k )\ m

           The Force exerted by Jack \r F = (-420\r i + 0 \r j + 210\r k) \ N

             The Force exerted by Jill  \r F_{Jill} =(180 \r i + 0\r j + 360\r k)

Now to obtain the displacement made we are to subtract the final position from the initial position

                                 \r P = P_f - P_i

                                    = (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k  )

                                     = (2\r i + 0\r j -\r k )m

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  which is mathematically represented as

                                                   W =\r  F * \r P

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                                               W_{jack} =\r  F * \r P

                                                 = [(-420\r i +0\r j +210\r k)(2\r  i + 0\r j - \r k)]

                                                 = (-420) (2) + (210)(-1)

                                                = -840 - 210

                                               =-1050J

  The amount of workdone by Jill would be

                                                 W_{Jill} =\r  F * \r P

                                                        = [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]

                                                       = (180 )(2) +(360)(-1)

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             -1050  = 0.5*3300 [*v^2- (1.1)^2]

            -1050 = 1650 [v^2 -1.21]

               0.6363 = v^2 -1.21

                   v^2 = 0.6363+1.21

                    v^2 =1.846

                    v = 1.36\ m/s

                   

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