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Fiesta28 [93]
3 years ago
13

(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a sp

eed (vB)1 = 4.8 ft/s. The attached cord is then pulled down through the hole with a constant speed vr = 2.2 ft/s. a. Determine the ball's speed at the instant r2 = 2 ft. Neglect friction and the size of the ball. Note that particle path is no longer of constant radius, and the particle has velocity components in both tangential and radial directions. b. How much work is done to pull down the cord from the initial instant to the instant when r2 = 2 ft? Neglect friction and the size o
Physics
1 answer:
lisov135 [29]3 years ago
3 0

Answer:

a

 v_r =8.65 \ ft/s

b

  W_{1-2} =  3.24 \  ft \cdot lb

Explanation:

From the question we are told that

  The mass of the ball is m  =  4 \  lb

  The radius is  r= 3 \  ft

   The speed is v_B_1  =  4.8 \ ft /s

    The speed of the attached cord is  v_c =2.2 \  ft

   The position that is been considered is  r_1 =  2 \  fth

Generally according to the law of angular momentum conservation

   L_a =  L_b

Here L_a is the initial  momentum of the ball which is mathematically represented as

      L_a  =  m*  v_B_1 *  r

while  

L_b is the  momentum of the ball  at  r =  2 ft which is mathematically represented as

       L_a  =  m*  v_B_2 *  r_1

So

      m*  v_B_1 *  r = m*  v_B_2 *  r_1

=>      4.8 *  3 =  v_B_2 *  2

=>     v_B_2 =  7.2 \  ft/s

Generally the resultant velocity of the ball is  

      v_r = \sqrt{v_B_2^2 + v_B_1^2   }

=>   v_r = \sqrt{7.2^2 + 4.8^2   }

=>   v_r =8.65 \ ft/s

Generally according to equation for principle of work and energy we have that

    K_1 + \sum W_{1-2} = K_2

Here K_1 is the initial kinetic energy of the ball which is mathematically represented as

K_1  =  \frac{1}{2}  *  m* v_B_1^2

While  \sum W_{1-2} is the sum of the total  workdone by the ball

and  K_2 is the final kinetic energy of the ball  which is mathematically represented as  K_2  =  \frac{1}{2}  *  m* v_r^2

So

     \sum W_{1-2} =  \frac{1}{2}  *  m  (v_r^2  -  v_B_1^2)

Here  m is the mass which is mathematically represented as

     m = \frac{W}{g} here W is the weight in  lb and  g is the acceleration due to gravity which is g =  32 \ ft/s^2

So

    \sum W_{1-2} =  \frac{1}{2}  *  \frac{4}{32} *   (8.65^2  -  4.8^2)

=>   W_{1-2} =  3.24 \  ft \cdot lb

   

   

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