Question

A group of kids are racing hula hoops down a hill that is 0.8 m high. One hoop is .7kg with a radius of 67 cm and the other hoop is .5kg with a radius of 43cm. If both hoops are initially traveling at 1.2 m/s right after they are pushed off the top of the hill, what hoop will reach the bottom first use the conservation of energy to support your answer. What is the change in angular momentum for each hoop? (ignore friction) show your work

Answer 1

I1 = mr^2 = 0.7 x 0.67^2 = 0.314 kgm^2

I2 = 0.5 x 0.43^2 = 0.0925 kgm^2

KE = 1/2 mv^2 + 1/2 mv^2 = mv^2

By law of conservation of energy

K1 + P1 = K2 + P2

1/2 x 0.7 x 1.2^2 + 0.7 x 10 x 0.8 = 0.7v^2 + 0

v = 3.53 m/s

p2 - p1 = mr (v2-v1) = 0.7 x 0.67 (3.53-1.2) = 1.093 kgm^2/s

for second ring

1/2 x 0.5 x 1.2^2 + 0.5 x 10 x 0.8 = 0.5v^2 + 0

v = 4.18 m/s

p2 - p1 = 0.5 x 0.43 (4.18 - 1.2) = 0.64 kgm^2/s

Clearly second hoop will reach first.

answered by: anonymous