Question #118121

A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?

1 : 2.0 kg • m2/s towards the west

2 : 5.0 kg • m2/s vertically upwards

3 : 2.0 kg • m2/s towards the east

4 : 5.0 kg • m2/s towards the east

5 : 5.0 kg • m2/s towards the west

Expert's answer

A bicycle is travelling north at 5 m/s. The mass of the wheel, 2 kg, is uniformly distributed along the rim, which has a radius of 20 cm.

moment of inertia of the wheel =mr2=2×(20100)2=225kg m2mr^2=2\times(\frac{20}{100})^2=\frac{2}{25}kg\ m^2



v=ωr5=ω×0.2ω=25rad/sv=\omega r\\5=\omega \times0.2\\\omega =25 rad/s


The angular momentum is, L=Iω = 225×25=2kg m2/sec\frac2{25}\times 25=2kg\ m^2/sec




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