A bicycle is turned upside down while its owner repairs a flat tire. A friend sp

Question

A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tangentially. She measures the heights reached by drops moving vertically (see figure below). A drop that breaks loose from the tire on one turn rises vertically 57.0 cm above the tangent point. A drop that breaks loose on the next turn rises 41.0 cm above the tangent point. The radius of the wheel is 0.350 m.

(a) Why does the first drop rise higher than the second drop?

This answer has not been graded yet.

(b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant). (Indicate the direction with the sign of your answer. Take the clockwise direction to be positive.)

Please show each step, and how you got it. Also, where do you get theta and what is it?

Explanation / Answer

(a)

the wheel is rotating with retardation so the speed of the wheel is decreasing.

as the speed decreases the height also decreases

(b)

for first drop

v1 = sqrt(2*g*h1) = sqrt(2*9.8*0.57) = 3.34 m/s

for second drop = v2 = sqrt(2*g*h2) = sqrt(2*9.8*0.41) = 2.83 m/s

distance moved by the point on the rim = s = 2*pi*r

from equation of moetion

v2^2 - v1^2 = 2*a*s

2.83^2-3.24^2 = 2*a*2*pi*0.35

tangential acceleration = -0.565 m/s^2

angular acceleration = alfa = a/R = 0.565/0.35 = 1.614 rad/s^2

direction = out of page

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