1.At takeoff a commercial jet has a 65.0 m/s speed. Its tires have a diameter of

Question

1.At takeoff a commercial jet has a 65.0 m/s speed. Its tires have a diameter of 0.700 m.

(a) At how many rpm are the tires rotating?
___ rpm
(b) What is the centripetal acceleration at the edge of the tire?
____ m/s2
(c) With what force must a determined 10-15 kg bacterium cling to the rim?
____N
(d) Take the ratio of this force to the bacterium's weight.
____(force from part (c) / bacterium's weight)

2.(a) A 19.0 kg child is riding a playground merry-go-round that is rotating at 35.0 rpm. What centripetal force must she exert to stay on if she is 1.00 m from its center?
____ N
(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rpm if she is 6.00 m from its center?
____ N
(c) Compare each force with her weight.
____ (force from part (a) / weight)
____ (force from part (b) / weight

Explanation / Answer

1)

a) = v / r = 65m/s / 0.700m = 92.85 rad/s * 1rev/2 rads * 60s/min = 887 rpm

b) a = ²r = (92.85 /s)² * 0.700m = 6034 m/s²

c) F = ma = 1e-15kg * 6034 m/s² = 6.00e-12 N

d) 6034/9.8 = 616

2

a)

  Fc = m * v^2 ÷ r

We need to determine the child’s velocity in m/s from the fact that the merry go round is rotating at 35.0 rpm. As the merry go round rotates one time, the girl moves distance which is equal to the circumference of the circle.

Circumference = 2 * * 1.0 = 3 *
Let’s determine the distance the child moves during 35 rotations.
d = 35 * 3 * = 105 *
To determine the velocity in m/s, divide this distance by 60.
v = 1.75 *
Fc = 19 * (1.75 * )^2 ÷ 1.0 = 573.705475 N

b)

Circumference = 2 * * 6 = 12 *
Let’s determine the distance the child moves during 3 rotations.
d = 3 * 12 * = 36 *
To determine the velocity in m/s, divide this distance by 60.
v = 0.6 *
Fc = 19 * (0.6 * )^2 ÷ 6 = 1.11 * ^2 =10.94N

c) 573.705475/19=30.19

10.94/19=0.57

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