A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-loop which has a radius 7 m. The acceleration of gravity is 9.8 m/s 2 . 21.9 m 7 m A What is its speed at point A ? Answer in units of m/s. 042 (part 2 of 2) 10.0 points How large is the normal force on it at point A if its mass is 4 g?

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aristocles aristocles · Answer 1 · 2019-11-17T10:09:40+01:00

Answer:

Part a)

[tex]v = 12.45 m/s[/tex]

Part B)

[tex]F_n = 0.05 N[/tex]

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

[tex]mgH = \frac{1}{2}mv^2 + mg(2R)[/tex]

so the speed at point A is given as

[tex]mg(H - 2R) = \frac{1}{2}mv^2[/tex]

[tex]v = \sqrt{2g(H - 2R)}[/tex]

[tex]v = \sqrt{2(9.81)(21.9 - 2\times 7)}[/tex]

[tex]v = 12.45 m/s[/tex]

Part B)

Now the force equation at point A is given as

[tex]F_n + mg = \frac{mv^2}{R}[/tex]

[tex]F_n = \frac{mv^2}{R} - mg[/tex][/tex]

[tex]F_n = 0.004(\frac{12.45^2}{7} - 9.81)[/tex]

[tex]F_n = 0.05 N[/tex]