A 0.50 kg ball that is tied to the end of a 1.3 m light cord is revolved in a ho

Question

A 0.50 kg ball that is tied to the end of a 1.3 m light cord is revolved in a horizontal plane with the cord making a 30° angle, with the vertical (See Fig. P7.52.) Figure P7.52 (a) Determine the ball's speed. 1.91 (b) If, instead, the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical? 46.3 Your response differs from the correct answer by more than 10%. Double check your calculations (c) If the cord can withstand a maximum tension of 9.4 N, what is the highest speed at which the ball can move? m/s m/s

Explanation / Answer

a) let tension in the cord = T

now Tcos(30 degrees) = mg = 0.50*9.8

=> T = 5.65 N

and Tsin(30 degrees) = mv^2/r

=> 5.65*0.5 = 0.50*v^2/(1.3*sin(30))

=> v = 1.91 m/s

b) if v = 4.0 m/s  

then Tsin(theta) = mv^2/r

=> Tsin(theta) = 0.50*4^2/(1.3sin(theta))

=> Tsin^2(theta) = 6.15 .....eq1

and Tcos(theta) = mg = 0.50*9.8 = 4.9 ...........eq2

by dividing both equations

=> sin^2(theta) / cos(theta) = 1.255

=> tan(theta)*sin(theta) = 1.255

let tan(theta) = x

=> sin(theta) = (1+x^2)^0.5

so x*(1+x^2)^0.5 = 1.255

=> x^2 + x^4 = 1.575

=> x^2(1+x^2) = 1.575

=> x = 0,0,0.758, 0.758

=> tan(theta) = 0.758

=> theta = 37.16 degrees

c) if T = 9.4 N

=> Tcos(theta) = mg =4.9

=> cos(theta) = 4.9/9.4 = 0.521

=> theta = 58.60 degrees

now Tsin(theta) = mv^2/r

=> 9.4*sin(58.60 degrees) = 0.5*v^2/(1.3sin(58.60))

=> v = 4.21 m/s Heighest speed

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