Question

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?

Answer 1

Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

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

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

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

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$