1.At takeoff a commercial jet has a 65.0 m/s speed. Its tires have a diameter of
ID: 1476630 • Letter: 1
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)
w = v/r = 65/0.35 = 185.7 rad/s = 185.7*(60/(2pi)) rpm = 1773.3 rpm
(b)
ac = v^2/r = 65^2/0.35 = 12071.42 m/s^2
(c)
F = m*ac = 10^-15*12071= 1.207*10^-11 N
d)ratio = 1.207*10^-11/(10^-15*9.8) = 1231.63
2)
Fc = m*r*w^2
w = 35 rpm = 35*2pi/60 = 3.67 rad/s
Fc = 19*1*3.67^2 = 255.901 N
b)
w = 3rpm = 3*2*pi/60 = 0.314 rad/s
Fc = 19*6*0.314^2 = 11.24 N
c)
ratio = 255.901/(19*9.8) = 1.37
ratio = 11.24/(19*9.8) = 0.0604
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