Part A A 60 g ball is tied to the end of a 70-cm-long string and swung in a vert

Question

Part A A 60 g ball is tied to the end of a 70-cm-long string and swung in a vertical circle. The center of the circle, as shown in the figure, (Figure 1) is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack. If the string is released at the instant the ball is at the top of the loop, how far to the right does the ball hit the ground? Express your answer to two significant figures and include the appropriate units. c + a ? I HÀ Az = 1.41 + m Figure Submit Previous Answers Request Answer X Incorrect; Try Again; 29 attempts remaining Provide Feedback Next! 150 cm

Explanation / Answer

Mass of the ball, m = 60 g = 60*10-3 kg

Length of the string, L = 70 cm = 0.70 m

radius of the circle = 0.7 m

The center of the circle, as shown in the figure is 150 cm above the floor

The minimum velocity at the top is given by,

v2 = r*g

v = Sqrt [gr] = Sqrt [9.81*0.70]

v = 2.620 m/s

Now the ball goes off horizontally at a speed 2.620 m/s and falls through a height is given by

Y = h + r =150 + 70 = 220 cm

Y = 2.20 m

By using,

s = ut + (1/2)a*t2

time is given by

2.20 = 0.5*9.81*t2

t = 0.67 s

If the string is released at the instant the ball is at the top of the loop, the distance to the right the ball hit the ground is given by

x = v*t = (2.620 m/s) (0.67 s)

the distance to the right the ball hit the ground is, x = 1.754 m

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